Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 1st Lesson Waves Textbook Questions and Answers.

## AP Inter 2nd Year Physics Study Material 1st Lesson Waves

Very Short Answer Questions

Question 1.

What does a wave represent?

Answer:

A wave represents the transport of energy through a medium from one point to another without translation of the medium.

Question 2.

Distinguish between transverse and longitudinal waves.

Answer:

Transverse waves

- The particles of the medium vibrate perpendicular to the direction of wave propagation.
- Crests and troughs are formed alternatively.

Longitudinal waves

- The particles of the medium vibrate parallel to the direction of wave propagation.
- Compressions and rare fractions are formed alternatively.

Question 3.

What are the parameters used to describe a progressive harmonic wave ?

Answer:

Progressive wave equation is given y = a sin (ωt – kx)

Where ω = 2πv = \(\frac{2 \pi}{T}\)

Parameters :

- a = Amplitude
- λ = Wavelength
- T = Time period
- v = Frequency
- k = Propagation constant
- ω = Angular frequency.

Question 4.

Obtain an expression for the wave velocity in terms of these parameters.

Answer:

Let ‘v’ be the velocity of a wave, ‘v’ be frequency and ‘λ’ be the wavelength. If T is the time period, then v = \(\frac{1}{\mathrm{~T}}\)

The distance travelled by the wave in the time T = λ.

Distance travelled in one second = \(\frac{\lambda}{T}\)

which is equal to wave velocity v = \(\frac{\lambda}{T}\).

∴ v = vλ

Question 5.

Using dimensional analysis obtain an expression for the speed of transverse waves in a stretched string.

Answer:

Wave velocity v ∝ T^{a} µ^{b} ⇒ V = K T^{a} µ^{b} ——-> (1)

Dimensions of v = M^{0}L^{1} T^{-1}, Tension T = M^{1}L^{1}T^{-2},

Linear mass µ = M^{1}L^{-1}, Constant K = M^{0}L^{0}T^{0}

Now (1) becomes M^{0}L^{1}T^{-1} = [M^{1}L^{1}T^{-2}]^{a} [M^{1}L^{-1}]^{b}

M^{0}L^{1}T^{1} = M^{a + b}L^{a-b}T^{-2a} Comparing the powers of same physical quantity.

-1 = -2a ⇒ a = \(\frac{1}{2}\)

a + b = 0 ⇒ b = –\(\frac{1}{2}\)

⇒ v = (1)\(T^{\frac{1}{2}} \mu^{\frac{1}{2}}\) [∵ K = 1 Practically]

∴ v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\)

Question 6.

Using dimensional analysis obtain an expression for the speed of sound waves in a medium. .

Answer:

Speed of sound v ∝ B^{a} ρ^{b} ⇒ v = K B^{a} ρ^{b} ——–> (1)

Dimensions of v = M^{0}L^{1}T^{-1},

Elasticity of medium,

B = M^{1}L^{-1}T^{-2}, density ρ = M^{1}L^{-3}, constant K = M^{0}L^{0}T^{0}.

Now (1) becomes M^{0}L^{1}T^{-1} = M^{0}L^{0}T^{0} [M^{1}L^{-1}T^{-2}]^{a} [M^{1}L^{-3}]^{b}

0 = a + b

1 = -a – 3b

-1 = -2a ⇒ a = \(\frac{1}{2}\)

b = –\(\frac{1}{2}\)

v = K \(B^{\frac{1}{2}} \rho^{\frac{1}{2}}\)

∴ v = \(\sqrt{\frac{B}{\rho}}\) [∵ K = 1, practically]

Question 7.

What is the principle of superposition of waves ?

Answer:

When two or more waves, are acting simultaneously on the particle of the medium, the resultant displacement is equal to the algebraic sum of individual displacements of all the waves. This is the principle of superposition of waves.

If y_{1}, y_{2}, …… y_{n} be the individual displacements of the particles,then resultant displacement y = y_{1} + y_{2} + ……. + y_{n}

Question 8.

Under what conditions will a wave be reflected ?

Answer:

- When the medium ends abruptly at any point.
- If the density and rigidity modulus of the medium changes at any point.

Question 9.

What is the phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary.

Answer:

π Radian or 180°.

Question 10.

What is a stationary or standing wave ?

Answer:

When two identical progressive (Transverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are superimposed then the resultant wave is called stationary waves or standing wave.

Question 11.

What do you understand by the terms ‘node ‘and’ antinode’?

Answer:

Node : The points at which the amplitude is zero, are called nodes.

Antinodes : The points at which the amplitude is maximum, are called antinodes.

Question 12.

What is the distance between a node and an antinode in a stationary wave ?

Answer:

The distance between node and antinode is \(\frac{\lambda}{4}\)

Question 13.

What do you understand by ‘natural frequency’ or ‘normal mode of vibration’ ?

Answer:

When a body is set into vibration and then left to itself, the vibrations made by it are called natural or free vibrations. Its frequency is called natural frequency or normal mode of vibration.

Question 14.

What are harmonics ?

Answer:

The frequencies in which the standing waves can be formed are called harmonics.

(Or)

The integral multiple of fundamental frequencies are called harmonics.

Question 15.

A string is stretched between two rigid supports. What frequencies of vibration are possible in such a string ?

Answer:

The possible frequencies of vibrations in a stretched string between two rigid supports is given by

v_{n} = (n + \(\frac{1}{2}\))\(\frac{v}{21}\) where n = 0, 1, 2, 3, ……

Question 16.

The air column in a long tube, closed at one end, is set in vibration. What harmonics are possible in the vibrating air column ?

Answer:

The possible harmonics in the vibrating air column of a long closed tube is given by

v_{n} = [2n + 1]\(\frac{v}{4 l}\) where n = 0, 1, 2, 3, ……..

Question 17.

If the air column in a tube, open at both ends, is set in vibration; what harmonics are possible?

Answer:

The possible harmonics in vibrating air column of a long open tube is given by

V_{n} = \(\frac{n v}{21}\)

where n = 1, 2, 3, ……….

Question 18.

What are ‘beats’ ?

Answer:

Beats : When two sound notes of nearly frequency travelling in the same direction and interfere to produce waxing and waning of sound at regular intervals of time is called “Beats”.

Question 19.

Write down an expression beat frequency and explain the terms there in.

Answer:

Expression of beat frequency, Δv = v_{1} ~ v_{2}

where v_{1} and v_{2} are the frequencies of two waves.

Question 20.

What is ‘Doppler effect’? Give an example.

Answer:

Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between source of sound and observer is called “Doppler effect”.

E.g.: When the whistling railway engine approaches the stationary observer on the platform, the frequency of sound appears to increase above the actual frequency. When it moves away from the observer, the apparent frequency decreases.

Question 21.

Write down an expression for the observed frequency when both source and observer are moving relative to each other in the same direction.

Answer:

Apparent frequency of sound heard by an observer,

v’ = \(\left[\frac{v-v_0}{v-v_s}\right] v\)

where v = frequency of sound

v = velocity of sound

v_{0} = velocity of observer

v_{S} = velocity of source

Short Answer Questions

Question 1.

What are transverse waves ? Give illustrative examples of such waves.

Answer:

Transverse waves: In a wave motion, the vibration of the particles and the direction of the propagation of the waves are perpendicular to each other, the waves are said to be transverse waves.

Illustration:

- Waves produced in the stretched strings are transverse.
- When a stretched string is plucked, the waves travel along the string.
- But the particles in the string vibrate in the direction perpendicular to the propagation of the wave.
- They can propagate only in solids and on the surface of the liquids.
- Ex : Light waves, surface water waves.

Question 2.

What are longitudinal waves ? Give illustrative example of such waves.

Answer:

Longitudinal waves : In a wave motion, the direction of the propagation of the wave and vibrations of particles are in the same direction, the waves are said to be longitudinal waves.

Illustration:

- Longitudinal waves may be easily illustrated by releasing a compressed spring.
- A series of compressions and rarefactions (expansions) propagate along the spring.

C = Compression; R = Rarefaction. - They can travel in solids, liquids and gases.
- Ex : Sound waves.

Question 3.

Write an expression for a progressive harmonic wave and explain the various parameters used in the expression.

Answer:

The expression of a progressive harmonic wave is written as y = a sin(ωt – \(\frac{2 \pi}{\lambda} x\))

or y = a sin(ωt – kx) where ω = 2πv, k = \(\frac{2 \pi}{\lambda}\)

Parameters:

- Amplitude (a) : It is the maximum displacement of a vibrating particle from its mean position.
- Frequence (v): It is the number of complete vibrations made by a vibrating body in one second.
- Wave length (λ) : It is defined as the distance covered by a wave while it completes one vibration, (or) It is the distance between two consecutive points in the same phase.
- Phase of vibration (ϕ) : The phase of vibration of a vibrating particle gives its state of displacement at a given instant. It is generally given by the phase angle.

Question 4.

Explain the modes of vibration of a stretched string with examples.

Answer:

Modes of vibrations of a stretched string :

- In sitar or Guitar, a stretched string can vibrate, in different frequencies and form stationary waves. This mode of vibrations are known as harmonics.
- If it vibrates in one segment, which is known as fundamental harmonic. The higher harmonics are called the overtones.

- It vibrates in two segments then the second harmonic is called first overtone. Similarly the pattern of vibrations are shown fig.
- If a stretched string vibrates with P ’Seg’ ments (loop) then frequency of vibration v = \(\frac{\mathrm{P}}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\) where T = tension in the string, µ = linear density = \(\frac{\text { mass }}{\text { length }}\)
- In first mode of vibration, P = 1, then v = \(\frac{1}{2l} \sqrt{\frac{T}{\mu}}\) (1
^{st}hamonic (or) fundamental frequency) - second mode of vibration, P = 2, then v
_{1}= \(\frac{2}{2l} \sqrt{\frac{T}{\mu}}\) = 2v (2^{nd}harmonic (or) 1^{st}overtone) - In third mode of vibration, P = 3, then v
_{2}= \(\frac{3}{2l} \sqrt{\frac{T}{\mu}}\) = 3v (3^{rd}harmonic (or) 2^{nd}overtone)

The ratio of the frequency of Harmonics are, v : v_{1}: v_{2}= v : 2v : 3v = 1 : 2 : 3

Question 5.

Explain the modes of vibration of an air column in an open pipe.

Answer:

Modes of vibration of an air column in an open pipe :

1) For a open pipe both the ends are open. So antinodes will be formed at both the ends. But two antinodes cannot exist without a node between them.

2) The possible harmonics in vibrating air column of a open pipe is given by .

Where n = 1, 2, 3

(1^{st} harmonic or fundamental frequence)

3) In first normal Mode of vibrating air column in a open pipe v_{1} = \(\frac{v}{2l}\) = v

(2nd harmonic 1^{st} overtone)

4) In second normal Mode of vibrating air column in a open pipe, v_{2} = \(\frac{2 v}{2l}\) = 2v

5) In third, normal Mode of vibrating air column in a open pipe, v_{3} = \(\frac{3 v}{21}\) = 3u

(3^{rd} harmonic 2^{nd} overtone)

6) In open pipe the ratio of frequencies of harmonics is

v_{1} : v_{2} : v_{3} = v : 2v : 3v = 1 : 2 : 3

Question 6.

What do you understand by ‘resonance’ ? How would you use resonance to determine the velocity of sound in air ?

Answer:

Resonance: If the natural frequency of a vibrating body is equal to the frequency of external periodic force then the two bodies are said to be in resonance. At resonance the bodies will vibrate with increasing amplitude.

Determination of velocity of sound in air using resonance :

1) In resonance tube, an air column is made to vibrate by means of vibrating fork. At certain length of air column, the air column would have the same frequency as that of the fork. Then the air column vibrates with the maximum amplitude and the intense sound is produced.

2) The vibrating fork of known frequency (v) is placed above the open end of the tube.

3) The length of air column is gradually increased until the booming sound can be heard at two different lengths of air column.

4) In first resonance, l_{1} be the length of air column, then \(\frac{\lambda}{4}\) = l_{1} + C …….. (1)

Where λ is the wavelength of sound emitted by the fork and C is the end correction of the tube.

5) In second resonance, l_{2} be the length of air column, then \(\frac{3 \lambda}{4}\) = l_{2} + C …… (2)

(2) – (1) ⇒ \(\frac{\lambda}{2}\) = l_{2} – l_{1}

λ = 2 (l_{2} – l_{1})

Speed of sound is given by

υ = vλ = v[2(l_{2} – l_{1})]

∴ υ = 2v (l_{2} – l_{1})

6) Hence by knowing v, l_{1}, l_{2} the speed of sound is calculated.

Question 7.

What are standing waves ? Explain how standing waves may be formed in a stretched string.

Answer:

Standing wave or stationary: When two identical progressive (Transverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are super-imposed then the resultant wave is called standing wave.

Formation of standing wave in a stretched string : –

- If a string of length ‘l’ is stretched between two fixed points and set into vibration, a transverse progressive wave begins to travel along the string.
- The wave is get reflected at the other fixed end.
- The incident and reflected waves interfere and produce a stationary wave.
- The stationary wave with nodes and antinodes is shown below.

Question 8.

Describe a procedure for measuring the velocity of sound in a stretched string.

Answer:

The velocity of a transverse wave travelling along a stretched string in fundamental mode is given by v = 2vl, where v = frequency, l = resonating length.

Measurement of velocity of sound in a stretched string using sonometer :

- The wire is subjected to a fixed tension with suitable load.
- A tuning fork of known frequency (v), is excited and the stem is held against the sono – meter box.

- The distance between the two bridges is adjusted such that a small paper rider at the middle of B
_{1}B_{2}vibrates vigorously and flies off due to resonance. - The resonating length ‘l’ can be measured between two bridges with scale.
- By knowing v and l; we can find the velocity of a wave using υ = 2vl.

Question 9.

Explain, using suitable diagrams, the formation of standing waves in a closed pipe. How may this be used to determine the frequency of a source of sound ?

Answer:

Formation of standing waves in a closed pipe :

- In closed pipe one end is closed and the other end is open. So antinode is formed at open end and antinode is formed at closed end.
- The possible harmonics in vibrating air column in a closed pipe v
_{n}= \(\frac{(2 n+1) v}{4 l}\) where v = 0, 1, 2, 3, ……. - In first normal mode of vibrating air column in a closed pipe, v
_{1}= \(\frac{v}{41}\)

[first harmonic (or) fundamental frequency]

- In second normal mode of vibrating air column in a closed pipe,

v_{3}= \(\frac{3 \mathrm{v}}{4l}\) [Third harmonic (or) first overtone] - In third normal mode of vibrating air column in a closed pipe,

v_{5}= \(\frac{5 \mathrm{v}}{4 \mathrm{l}}\) [Fifth harmonic (or) second overtone]

Determination of frequency of a source of sound :

- The vibrating fork of unknown frequency (v) is placed above the open end of the tube.

- Reservoir is slowly lowered, until a large booming sound is heard. Measure 1st resonating, air column length l
_{1}. - Further lower the reservoir, until second time a large booming sound is heard. Measure 2
^{nd}resonating air column length l_{2}. - Velocity of a wave at 0°C is v = 331 m/s.
- By knowing v, l
_{1}and l_{2}we can find unknown frequency of a tuning fork using

v = \(\frac{v}{2\left(l_2-l_1\right)}\)

Question 10.

What are ‘beats’ ? When do they occur ? Explain their use, if any.

Answer:

Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing (maximum) and waning (minimum) in the intensity of the resultant sound waves at regular intervals of time is called beats.

It two vibrating bodies have slightly difference in frequencies, beats can occur.

No. of beats can be heard Δυ = υ_{1} – υ_{2}

Importance :

- It can be used to tune musical Instruments.
- Beats are used to detect dangerous gases

Explanation-for tuning musical instruments with beats :

Musicians use the beat phenomenon in tuning their musical instruments. If an instrument is sounded against a standard frequency and tuned until the beats disappear. Then the instrument is in tune with the standard frequency.

Question 11.

What is ‘Doppler effect’? Give illustrative examples.

Answer:

Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between the observer and the source of sound is called doppler effect.

Examples:

- The frequency of whistling engine heard by a person standing on the platform appears to increase, when the engine is approaching the platform and it appears to decrease when the engine is moving away from the platform.
- Due to Doppler effect the frequency of sound emitted by the siren of an approaching ambulance appears to increase. Similarly the frequency of sound appears to drop when it is moving away.

Long Answer Questions

Question 1.

Explain the formation of stationary waves in stretched strings and hence deduce the laws of transverse wave in stretched strings. (A.P. Mar. ’19)

Answer:

A string is a metal wire whose length is large when compared to its thickness. A stretched string is fixed at both ends, when it is plucked at mid point, two reflected waves of same amplitude and frequency at the ends are travelling in opposite direction and overlap along the length. Then the resultant waves are known as the standing waves (or) stationary waves.

Let two transverse progressive waves of same amplitude a, wave length λ and frequency ‘v’, travelling in opposite direction be given by

y_{1} = a sin (kx – ωt) and y_{2} = a sin (kx + ωt)

where ω’ = 2πv and k = \(\frac{2 \pi}{\lambda}\)

The resultant wave is given by y = y_{1} + y_{2}

y = a sin (kx – ωt) + a sin (kx + ωt)

y = (2a sin kx) cos ωt

2a sin kx = Amplitude of resultant wave.

It depends on ‘kx’. If x = 0, \(\frac{\lambda}{2}\), \(\frac{2 \lambda}{2}\), \(\frac{3 \lambda}{2}\),……. etc, the amplitude = zero

These positions are known as “Nodes”.

If x = \(\frac{\lambda}{4}\), \(\frac{2 \lambda}{2}\), \(\frac{3 \lambda}{2}\) ……… etc, the amplitude = zero

The positions are known as “Nodes”

If x = \(\frac{\lambda}{4}\), \(\frac{3 \lambda}{4}\), \(\frac{5 \lambda}{4}\) ……. etc, the amplitude = maximum (2a).

These positions are called “Antinodes”.

If the string vibrates in ‘P’ segments and ‘l’ is its length then length of each segment = \(\frac{l}{p}\)

Which is equal to \(\frac{\lambda}{2}\)

∴ \(\frac{l}{\mathrm{p}}\) = \(\frac{\lambda}{2}\) ⇒ λ = \(\frac{2 l}{\mathrm{P}}\)

Harmonic frequency v = \(\frac{v}{\lambda}=\frac{v \mathrm{P}}{2 l}\)

v = \(\frac{v P}{2 l}\) ——- (1)

If ‘ T’ is tension (stretching force) in the string and ‘μ’ is linear density then velocity of transverse wave (v) in the string is v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\) —– (2)

From the Eqs (1) and (2)

Harmonic frequency v = \(\frac{p}{2 l} \sqrt{\frac{T}{\mu}}\)

If P = 1 then it is called fundamental frequency (or) first harmonic frequency

∴ Fundamental Frequency v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\) —— (3)

Laws of Transverse Waves Along Stretched String:

Fundamental frequency of the vibrating string v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\)

First Law : When the tension (T) and linear density (μ) are constant, the fundamental frequency (v) of a vibrating string is inversely proportional to its length.

∴ v ∝ \(\frac{1}{l}\) ⇒ vl = constant, when T and ‘μ’ are constant.

Second Law: When the length (I) and its, linear density (m) are constant the fundamental frequency of a vibrating string is directly proportional to the square root of the stretching force (T).

∴ v ∝ \(\sqrt{\mathrm{T}}\) ⇒ \(\frac{v}{\sqrt{T}}\) = constant, when ‘l’ and ‘m’ are constant.

Third Law: WHien the length (J) and the tension (T) are constant, the fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (m).

∴ v ∝ \(\frac{1}{\sqrt{\mu}}\) ⇒ \(v \sqrt{\mu}\) = constant, when ‘l’ and ‘T’ are constant.

Question 2.

Explain the formation of stationary waves in an air column enclosed in open pipe. Derive the equations for the frequencies of the harmonics produced. (T.S. Mar. ’16, A.P. Mar. ’15)

Answer:

A pipe, which is opened at both ends is called open pipe. When a sound wave is sent through a open pipe, which gets reflected by the earth. Then incident and reflected waves are in same frequency, travelling in the opposite directions are super – imposed stationary waves are formed.

Harmonics in open pipe : To form the stationary wave in open pipe, which has two antinodes at two ends of the pipe with a node between them.

∴ The vibrating length (l) = half of the wavelength \(\left(\frac{\lambda_1}{2}\right)\)

l = \(\frac{\lambda_1}{2}\) ⇒ λ_{1} = 2l

fundamental frequency v_{1} = \(\frac{\mathrm{v}}{\lambda_1}\) where v is velocity of sound in air v_{1} = \(\frac{v}{2 l}\) = v —— (1)

For second harmonic (first overtone) will have one more node and antinode than the fundamental.

If λ_{2} is wavelength of second harmonic l = \(\frac{2 \lambda_2}{2}\) ⇒ λ_{2} = \(\frac{2l}{2}\)

If ‘v_{2}‘ is frequency of second harmonic then v_{2} = \(\frac{v}{\lambda_2}\) = \(\frac{v \times 2}{2 l}\) = 2v

v_{2} = 2v —– (2)

Similarly for third harmonic (second overtone) will have three nodes and four antinodes as shown in above figure.

If λ_{3} is wave length of third harmonic l = \(\frac{3 \lambda_3}{2}\)

λ_{3} = \(\frac{2l}{3}\)

If ‘v_{2}’ is frequency of third harmonic then

v_{3} = 3v —– (3)

Similarly we can find the remaining or higher harmonic frequencies i.e v_{3}, v_{4} etc, can be determined in the same way.

Therefore the ratio of the harmonic frequencies in open pipe can be written as given below.

v : v_{1} : v_{2} = 1 : 2 : 3 ………

Question 3.

How are stationary waves formed in closed pipes ? Explain the various modes of vibrations and obtain relations for their frequencies. (A.P. & T.S. Mar. ’15)

Answer:

A pipe, which is closed at one end and the other is opened is called closed pipe. When a sound wave is sent through a closed pipe, which gets reflected at the closed end of the pipe. Then incident and reflected waves are in same frequency, travelling in the opposite directions are superimposed stationary waves are formed.

To form the stationary wave in closed pipe, which has atleast a node at closed end and antinode at open end of the pipe, it is known as first harmonic in closed pipe. Then length of the pipe (l) is equal to one fourth of the wave length.

∴ l = \(\frac{\lambda_1}{4}\) ⇒ λ_{1} = 4l

If ‘v_{1}‘ is fundamental frequency then

v_{1} = \(\frac{v}{\lambda_1}\) where ‘υ’ is velocity of sound in air

v_{1} = \(\frac{v}{4l}\) = v ——- (1)

To form the next harmonic in closed pipe, two nodes and two antinodes should be formed. So that there is possible to form third harmonic in closed pipe. Since one more node and antinode should be included.

Then length of the pipe is equal to \(\frac{3}{4}\) of the wavelength.

∴ l = \(\frac{3 \lambda_3}{4}\) where ‘λ_{3}‘ is wave length of third harmonic.

λ_{3} = \(\frac{4l}{3}\)

If ‘v_{3}‘ is third harmonic frequency (first overtone)

∴ v_{3} = \(\frac{v}{\lambda_3}\) = \(\frac{3 v}{41}\)

v_{3} = 3v ——- (2)

Similarly the next overtone in the close pipe is only fifth harmonic it will have three nodes and 3 antinodes between the closed end and open end.

Then length of the pipe is equal to \(\frac{5}{4}\) of wave length (λ_{5})

∴ l = \(\frac{5 \lambda_5}{4}\) where ’λ_{5}‘ is wave length of fifth harmonic. .

λ_{5} = \(\frac{4l}{5}\)

If ‘V_{5}‘ is frequency of fifth harmonic (second overtone)

V_{5} = \(\frac{v}{\lambda_5}=\frac{5 v}{4 I}\)

v_{5} = 5v —– (3)

∴ The frequencies of higher harmonics can be determined by using the same procedure. Therefore from the Eq (1), (2) and (3) only odd harmonics are formed.

Therefore the ratio of the frequencies of harmonics in closed pipe can be written as

v_{1} : v_{3} : v_{5} = v : 3v : 5v

v_{1} : v_{3} : v_{5} = 1 : 3 : 5

Question 4.

What are beats ? Obtain an expression for the beat frequency ? Where and how are beats made use of ?

Answer:

Beats : Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing and waning in the intensity of the resultant sound waves at regular intervals of time are called Beats.

If v_{1} and v_{2} are the frequencies of two sound notes superimposed along the same direction, no of beats heard per second = Δv = v_{1} – v_{2}.

Maximum no. of beats heard per sec is 10 due to persistence of hearing.

Expression for the beat frequency :

- Consider the two wave trains of equal amplitude but of nearly equal frequencies.
- Let the frequencies of the waves be v
_{1}and v_{2}. Say v_{1}is slightly greater than v_{2}. - Let the beat period be T seconds.
- No.of vibrations, made by the first wave train in T seconds – v
_{1}T

[∵ no.of oscillations in 1 sec = v]

[∵ no.of oscillations in T sec = vt] - No.of vibrations, made by the second wave train in T seconds = v
_{2}T - During the time interval T, the first wave train would have completed one vibration more than the second wave train.
- Hence, v
_{1}T – v_{2}T = 1 or v_{1}– v_{2}= \(\frac{1}{\mathrm{~T}}\) - Since, T is the beat period, no.of beats per seconds = \(\frac{1}{\mathrm{~T}}\)
- Hence the beat frequency = \(\frac{1}{\mathrm{~T}}\) = v
_{1}– v_{2}= Δv - That is the beat frequency is the difference between the frequencies of the two wave trains.

Practical applications of beats:

- Determination of an unknown frequency: Out of two tuning forks, one is loaded with wax and the other is filed. The excited tuning forks are close together and no.of beats can be heard. Then after unknown frequencies of them will be found practically.
- For tuning musical instruments : Musicians use the beat phenomenon in tuning their musical instruments.
- For producing colourful effects in music: Sometimes, a rapid succession of beats is knowingly introduced in music. This produces an effect similar to that of human voice and is appreciated by the audience.
- For detection of Marsh gas (dangerous gases) in mines.

Question 5.

What is Doppler effect? Obtain an expression for the apparent frequencý of sound heard when the source is in motion with respect to an observer at rest. (A.P. Mar. ’16, Mar. ’14)

Answer:

Doppler effect : The apparent change in the frequency heard by the observer due to the relative motion between the observer and the source of sound is called doppler effect.

When a whistling railway engine approaches an observer standing on the platform, the frequency of sound appears to increase. When it moves away the frequency appear to decrease.

Expression for apparent frequency when source is in motion and listener at rest:

Let S = Source of sound

O = listener

Let ‘S be the source, moving with a velocity ‘υ_{s}‘ towards the stationary listener.

The distance travelled by the source in time period T’ = υ_{s}. T

Therefore the successive compressions and rarefactions are drawn closer to listener.

∴ Apparent wavelength = λ’ = λ – υ_{s}T.

If ‘v’ “is apparent frequency heard by the listener

then v’ = \(\frac{v}{\lambda^{\prime}}\) where ‘υ’ is velocity of sound in air

v’ = \(\frac{v . V}{v-v_s}\)

Therefore the apparent frequency is greater than the actual frequency.

Similarly, if the source is away from the stationary listener then apparent frequency

v’ = \(\frac{v . V}{v+v_s}\), which is less than the actual frequency.

Limitation : Doppler effect is applicable when the velocities of the source and listener are much less than that of sound velocity

Question 6.

What is Doppler shift? Obtain an expression for the apparent frequency of sound heard when the observer Is In motion with respect to a source át rest.

Answer:

Doppler Shift: Due to the relative motion, when the source comes closer to listener, the apparent frequency is greater than actual frequency and source away from listener; the apparent frequency is less than actual frequency So the difference in apparent and actual frequencies is known as Doppler shift.

Expression for the apparent frequency heard by a moving observer:

Case (1) : When observer Is moving towards source:

Let ‘υ_{0}’ be velocity of listener ‘O’, moving towards the stationary source ‘s’ as shown in figure. So observer will receive more number of waves in each second.

The distance travelled by observer in one second = υ_{0}

The number of extra waves received by the observer = \(\frac{v_0}{\lambda}\)

We know v = vλ ⇒ λ = \(\frac{v}{v}\)

Where υ = Velocity of sound

v = Frequency of sound

If ‘v’ is apparent frequency heard by him then

Therefore the apparent frequency is greater than actual frequency.

Case (2) : When observer Is moving away from rest source:

If the observer is moving away from the stationary source then he loses the number of waves \(\frac{v_0}{\lambda}\)

∴ Apparent frequency v’ = v – \(\frac{v_0}{\lambda}\) = v – \(\frac{v_0 \cdot v}{v}\)

v’ = \(\left[\frac{v-v_0}{v}\right] \cdot v\)

Hence the apparent frequency is less than actual frequency.

Problems

Question 1.

A stretched wire of length 0.6m is observed to vibrate with a frequency of 30Hz in the fundamental mode. If the string has a linear mass of 0.05 kg/m find

(a) the velocity of propagation of transverse waves in the string

(b) the tension in the string.

Solution:

v = 30 Hz; l = 0.6 m ; μ = 0.05 kg m^{-1} υ = ? ; T = ?

a) υ = 2vl = 2 × 30 × 0.6 = 36 m/s

b) T = υ^{2}μ = 36 × 36 × 0.05 = 64.8 N

Question 2.

A steel cable of diameter 3 cm is kept under a tension of 10kN. The density of steel is 7.8 g/cm^{3}. With what speed would transverse waves propagate along the cable ?

Solution:

T = 10 kN = 10^{4} N

D = 3 cm; r = \(\frac{\mathrm{D}}{2}\) = \(\frac{3}{2}\)cm

= \(\frac{3}{2}\) × 10^{-2}m;

A = πr² = \(\frac{22}{7} \times\left[\frac{3}{2} \times 10^{-2}\right]^2\)

Question 3.

Two progressive transverse waves given by y_{1} = 0.07 sinπ(12x-500t) and y_{2} = 0.07 sinπ(12x + 500t) travelling along a stretched string from nodes and antinodes. What is the displacement at the (a) nodes (b) antinodes ? What is the wavelength of the standing wave ?

Solution:

A_{1} = 0.07; A_{2} = 0.07; K = 12π

a) At nodes, displacement

y = A_{1} – A_{2} = 0.07 – 0.07 = 0

b) At antinodes, displacement

y = A_{1} + A_{2} = 0.07 + 0.07 = 0.14 m

c) Wavelength λ = \(\frac{2 \pi}{\mathrm{K}}=\frac{2 \pi}{12 \pi}\) = 0.16 m

Question 4.

A string has a length of 0.4m and a mass of 0.16g. If the tension in the string is 70N, what are the three lowest frequencies it produces when plucked ?

Solution:

I = 0.4 m; M = 0.16g = 0.16 × 10^{-3} kg;

Question 5.

A metal bar when clamped at its centre resonantes in its fundamental frequency with longitudinal waves of frequency 4 kHz. If the clamp is moved to one end. What will be its fundamental resonance frequency ?

Solution:

When a metal bar of length l is clamped in the middle, it has one node in the middle and two antinodes at its free ends. In the fundamental mode. l = \(\frac{\lambda}{2}\) ⇒ λ = 21

In fundamental mode of frequency of bar

= frequency of wave = 4 kHz.

∴ v = \(\frac{\mathrm{v}}{\lambda}\) = \(\frac{\mathrm{v}}{2l}\) = 4kHz —– (1)

When clamp is moved to one end,

l = \(\frac{\lambda^{\prime}}{4}\) ⇒ λ’ = 4l

∴ V^{1} = \(\frac{\mathrm{v}}{\lambda}\) = \(\frac{\mathrm{v}}{4 \mathrm{l}}\) = \(\frac{4 \mathrm{kHz}}{2}\) = 2kHz

[∵ from (1)]

Question 6.

A closed organ pipe 70 cm long is sounded. If the velocity of sound is 331 m/s, what is the fundamental frequency of vibration of the air column ?

Solution:

I = 70 cm = 70 × 10^{-2}m; v = 331 m/s ; v = ?

v = ?

v = \(\frac{v}{4 l}\) = \(\frac{331}{4 \times 70 \times 10^{-2}}\) = 118.2 Hz

Question 7.

A vertical tube is made to stand in water so that the water level can be adjusted. Sound waves of frequency 320 Hz are sent into the top of the tube. If standing waves are produced at two successive water levels of 20 cm and 73 cm, what is the speed of sound waves in the air in the tube ?

Solution:

v = 320 Hz; l_{1} = 20cm = 20 × 10^{-2}

l_{2} = 73 cm = 73 × 10^{-2}m; v = ?

v = 2v(l_{2} – l_{1})

= 2 × 320 (73 × 10^{-2} – 20 × 10^{-2})

∴ v = 339 m/s .

Question 8.

Two organ pipes of lengths 65cm and 70cm respectively, are sounded simultaneously. How many beats per second will be produced between the fundamental frequencies of the two pipes ? (Velocity of sound = 330 m/s).

Solution:

l_{1} = 65 cm = 0.65 m

_{2} = 70 cm = 0.7 m

v = 330 m/s

No. of beats per second ∆υ = υ_{1} – υ_{2}

= \(\frac{v}{2 h}\) – \(\frac{\mathrm{v}}{2 l_2}\) = \(\frac{330}{2 \times 0.65}\) – \(\frac{330}{2 \times 0.7}\)

∴ ∆v = 253.8 – 235.8 = 18Hz

Question 9.

A train sounds its whistle as it approaches and crosses a level crossing. An observer at the crossing measures a frequency of 219 Hz as the train, approaches and a frequency of 184 Hz as it leaves. If the speed of sound is taken to be 340 m/s, find the speed of the train and the frequency of its whistle.

Solution:

When a whistling train approaches to rest observer,

v’ = \(\left[\frac{v}{v-v_s}\right] v\) ——– (1)

When a whistling train away from rest observer

v” = \(\left[\frac{v}{v+v_{\mathrm{S}}}\right] v\) —— (2)

Here v’ = 219 Hz; V” = 184 Hz;

v = 340 m/s

\(\frac{(1)}{(2)}\) ⇒ \(\frac{v^{\prime}}{v^{\prime \prime}}\) = \(\frac{\left(v+v_s\right)}{\left(v-v_s\right)}\)

\(\frac{219}{184}\) = \(\frac{340+v_{\mathrm{s}}}{340-v_{\mathrm{s}}}\)

219(340 – υ_{s}) = 184(340 + υ_{s})

219 × 340 – 219 υ_{s} = 184 × 340 + 184 υ_{s}

403 υ_{s} = 35 × 340

∴ Velocity of train υ_{s} = 29.5 m/s

Frequency of whistle, v = v’ × \(\left[\frac{v-v_{\mathrm{S}}}{v}\right]\)

= 219 × \(\left[\frac{340-29.5}{340}\right]\)

= 199.98

∴ v = 200 Hz.

Question 10.

Two trucks heading in opposite directions with speeds of 60 kmph and 70 kmph respectively, approach each other. The driver of the first truck sounds his horn of frequency 400Hz. What frequency does the driver of the second truck hear ? (Velocity of sound = 330 m/ s). After the two trucks have passed each other, what frequency does the driver of the second truck hear ?

Solution:

v_{s} = 60 kmph = 60 × \(\frac{5}{18}\) m/s = \(\frac{300}{18}\) m/s

v_{0} = 70 kmph = 70 × \(\frac{5}{18}\) m/s = \(\frac{350}{18}\) m/s

v = 400 Hz

When two trucks approach each other

When two trucks crossed each other,

Textual Exercises

Question 1.

A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is caused at one end of the string, how long does the disturbance take to reach the other end ?

Answer:

Here M = 2.50 kg, T = 200 N, l = 20.0M

Mass per unit length; μ = \(\frac{\mathrm{M}}{l}\) = \(\frac{2.5}{20.0}\)

= 0.125 kg/m

Velocity V = \(\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{200}{0.125}}\) = 40 m/s

Time taken by disturbance to reach the other end

t = \(\frac{l}{\mathrm{~V}}\) = \(\frac{20}{40}\) = 0.5s.

Question 2.

A stone dropped from the top of a tower of height 300 m high splashes into the pond of water near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s^{-1}, (g = 9.8m s^{-2})

Answer:

Here, h = 300m, g = 9.8 m/s^{2}, V = 340 m/s. If t_{1} = time taken by stone to strike the surface of water in the pond, then from

S = ut + \(\frac{1}{2}\) at^{2}

300 = 0 + \(\frac{1}{2}\) × 9.8 \(\mathrm{t}_1^2\)

t_{1} = \(\sqrt{\frac{300}{4.9}}\) = 7.82s.

Time taken by sound to reach the top of tower t_{2} = \(\frac{\mathrm{h}}{\mathrm{v}}\)

= \(\frac{300}{400}\) = 0.88s

Total time after which splash of sound is heard = t_{1} + t_{2} = 7.82 + 0.88 = 8.70s

Question 3.

A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20° C = 343 m s^{-1}.

Answer:

Here, l = 12.0M, M = 2.10kg, T = ?

V = 343 m/s

Mass per unit length μ = \(\frac{\mathrm{M}}{l}\) = \(\frac{2.10}{12.0}\)

= 0.175 kg/m

As V = \(\sqrt{\frac{\mathrm{T}}{\mu}}\)

T = V^{2} . μ = (343)^{2} × 0.175 = 2.06 × 10^{4}N.

Question 4.

Use the formula υ = \(\sqrt{\frac{\gamma P}{\rho}}\) to explain why the speed of sound in air

a) is independent of pressure,

b) increases with temperature,

c) increases with humidity.

Answer:

a) Effect of pressure:

The speed of sound in gases, υ = \(\sqrt{\frac{\gamma \mathrm{P}}{\rho}}\)

At constant temperature, PV = constant

P\(\frac{m}{\rho}\) = constant ⇒ \(\frac{\mathrm{P}}{\rho}\) = constant

If P increases, ρ also increases. Hence speed of sound in air is independent of pressure.

b) Effect of temperature:

As PV = nRT, P\(\frac{\mathrm{m}}{\rho}\) = \(\frac{m}{M} R T\)

⇒ \(\frac{P}{\rho}\) = \(\frac{\mathrm{RT}}{\mathrm{M}}\)

∴ υ = \(\sqrt{\frac{R T}{M}}\)

Since R, M are constants υ ∝ \(\sqrt{\mathrm{T}}\)

∴ Velocity of sound in air depends on temperature.

c) Effect of humidity:

As υ = \(\sqrt{\frac{\gamma \mathrm{P}}{\rho}}\) ∴ υ ∝ \(\frac{1}{\sqrt{\rho}}\)

As the density of water vapour is less than density of dry air at STP. So the presence of moisture in air decreases the

density of air. Since the speed of sound is inversely proportional to the square root of density. So sound travels faster in moist air than dry air. Hence velocity of sound

V ∝ humidity

Question 5.

You have learnt that a travelling wave in one dimension is represented by a function y = f(x, t) where x and t must appear in the combination x – υt or x + υt, i.e. y = f(x ± υ t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:

Answer:

No, the converse is not true. The basic requirement for a wave function to represent a travelling wave is that for all values of x & t, wave function must have a finite value.

Out of the given functions y, no one satisfies this condition therefore, none can represent a travelling wave.

Question 6.

A bat emits ultrasonic sound a frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of

(a) the reflected sound,

(b) the transmitted sound?

Speed of sound in air is 340 m s^{-1} and in water 1486 m s^{-1}.

Answer:

Here V = 100 KHz = 10^{5}Hz, V_{a} = 340m/s, V_{w} = 1486 m/s^{-1}

Wavelength of reflected sound, λ_{a} = \(\frac{\mathrm{V}^{\mathrm{a}}}{\mathrm{V}}\)

= \(\frac{340}{10^5}\) = 3.4 × 10^{-3} m

Wavelength of transmitted sound,

λ_{w} = \(\frac{V_w}{V}\) = \(\frac{1486}{10^5}\) = 1.486 × 10^{-2} m

Question 7.

A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s^{-1}? The operating frequency of the scanner is 4.2 MHz.

Answer:

λ = ? υ = 1.7 Km/s = 1700 ms^{-1}

y = 4.2 MHz = 4.2 × 10^{6}Hz

λ = \(\frac{v}{v}\) = \(\frac{1700}{4.2 \times 10^6} \mathrm{~m}\) = 0.405 × 10^{-3} m

= 0.405 mm

Question 8.

A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right.

a) Is this a travelling wave or a stationary wave?

If it is travelling, what are the speed and direction of its propagation?

b) What are Its amplitude and frequency?

C) What is the initial phase at the origin?

d) What is the least distance between two successive crests in the wave?

Answer:

Compare the given equation with that of plane progressive wave of amplitude r, travelling with a velocity V from right to left.

y(x, t) = rsin\(\left[\frac{2 \pi}{\lambda}(v t+x)+\phi_0\right]\) ……… (1)

We find that

a) The given equation represents a transvërse harmonic wave travelling from right to left. It is ñot a stationary wave.

b) The given equation can be written as

Y(x, t) = 3.0sin[0.018(\(\frac{36}{0.018}\) + x) + \(\frac{\pi}{4}\)] ……… (2)

equating coefficient of t in the two

(1) & (2) we get. :

V = \(\frac{.36}{0.018}\) = 2000 cm/sec.

Obviously, r = 3.0 cm

Also, \(\frac{2 \pi}{\lambda}\) = 0.018

λ = \(\frac{2 \pi}{0.018} \mathrm{~cm}\)

Frequency, v = \(\frac{v}{\lambda}\) = \(\frac{2000}{2 \pi}\) × 0.018

= 5.7351.

c) Intial phase, φ_{0} = \(\frac{\pi}{4}\) radian.

d) Least distance between two successive crests of the wave =

Wave length, λ = \(\frac{2 \pi}{0.018 \mathrm{~cm}}\) = 349 cm,

Question 9.

For the wave described in the last problem plot the displacement (y) versus (t) graphs for x = 0.2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

Answer:

The transverse harmonic wave is

y(x,t) = 3.0 sin[36t + 0.018x + \(\frac{\pi}{4}\)]

For x = 0

y(0, t) = 3.0 sin(36t + \(\frac{\pi}{4}\)) —— (i)

Here w = \(\frac{2 \pi}{T}\) = 36, T = \(\frac{2 \pi}{36}\) = \(\frac{\pi}{18}\)-sec.

For different values of t, we calculate y using eq(i). These values are tabulated below.

On plotting y versus t graph, we obtain a sinusoidal curve as shown in fig.

Similar graphs are obtained for x = 2cm & x = 4cm. The oscillary motion in travelling wave differs from one point to another only in terms of phase. Amplitude and frequency of oscillatory motion remain the same in all the three areas.

Question 10.

For the travelling harmonic wave

y(x, t) = 2.0 cos 2 π (10t – 0.0080 x + 0.35)

where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of

a) 4 m,

b) 0.5 m,

c) λ/2,

d) 3λ/4,

Answer:

The given equation be written as

y = 2.0 cos[2π(10t – 0.0080x) + 2π × 0.35]

y = 2.0 cos[2π × (0.0080(\(\frac{10 \mathrm{t}}{0.0080}\) – x) + 0.7π]

Compare it with the standard equation of a travelling harmonic, we have

y = r.cos[\(\frac{2 \pi}{\lambda}(v t-x)+\phi_0\)

We get, \(\frac{2 \pi}{\lambda}\) = 2π × 0.0080

Further we know that phase diff. φ = \(\frac{2 \pi}{\lambda} \mathrm{x}\)

a) When x = 4m = 400 cm

φ = \(\frac{2 \pi}{\lambda}\) . x = 2π × 0.0080 × 400

= 6.4 π rad.

b) When x = 0.5 = 50 cm

φ = \(\frac{2 \pi}{\lambda}\) . x = 2π × 0.0080 × 50

= 0.8π rad.

c) When x = \(\frac{\lambda}{2}\)

φ = \(\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\) = π rad.

d) When x = \(\frac{3 \lambda}{4}\)

φ = \(\frac{2 \pi}{\lambda} \times \frac{3 \lambda}{4}\) = \(\frac{3 \lambda}{2} \mathrm{rad}\)

Question 11.

The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin\(\left(\frac{2 \pi}{3} x\right)\)cos (120 πt)

where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10^{-2} kg

Answer the following:

a) Does the function represent a travelling wave or a stationary wave?

b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

c) Determine the tension in the string.

Answer:

The given equation is

y(x, t) = 0.06 sin \(\frac{2 \pi}{3} \mathrm{x} \cos 120 \pi \mathrm{t}\)

a) As the equation involves harmonic functions of x and t seperately, it represents a stationary wave.

b) We know that when a wave pulse

y_{1} = r sin \(\frac{2 \pi}{\lambda}(v t+x)\) —– (i)

travelling along + direction of x-axis is super imposed by the reflected wave

y = y_{1} + y_{2} = -2rsin\(\frac{2 \pi}{\lambda}\) xcos \(\frac{2 \pi}{\lambda}\) vt is formed. ——- (ii)

Comparing (i) & (ii) we find that

\(\frac{2 \pi}{\lambda}\) = \(\frac{2 \pi}{3}\) ⇒ λ = 3m.

Also \(\frac{2 \pi}{\lambda} v\) = 120π (Or)

V = 60λ = 60 × 3 = 180m/s.

frequency, v = \(\frac{v}{\lambda}\) = \(\frac{180}{3}\) = 60 hertz.

Note that both the waves have same wave length, same frequency and same speed.

c) Velocity of transverse waves is

υ = \(\sqrt{\frac{T}{\mu}}\) (or) υ^{2} = T/μ

T = V^{2} × μ where μ = \(\frac{3 \times 10^{-2}}{1.5}\)

= 2 × 10^{-2} kg/m

T = (180)^{2} × 2 × 10^{-2} = 648 N.

Question 12.

i) For the wave on a string described in previous problem do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers.

(ii) What is the amplitude of a point 0.375 m away from one end?

Answer:

i) All the points on the string

a) have the same frequency except at the nodes (where frequency is cos θ),

b) have the same phase every where in one loop except at the nodes,

c) however, the amplitude of vibration at different points is different.

ii) From y(x, t) = 0.06 sin\(\left(\frac{2 \pi}{3} x\right)\) cos (120 πt)

The amplitude at x = 0.375 m is 0.06

sin \(\frac{2 \pi}{3} x \times 1\) = 0.06 × sin \(\frac{2 \pi}{3} \times 0.375\)

= 0.06sin\(\frac{\pi}{4}\) = \(\frac{0.06}{\sqrt{2}}\) = 0.042 M

Question 13.

Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave State which of these represent

(i) a travelling wave,

(ii) a stationary wave or

(iii) none at all:

a) y = 2 cos (3x) sin (10t)

b) y = \(2 \sqrt{x-v t}\)

c) y = 3 sin(5x – 0.5t) + 4 cos(5x – 0.5t)

d) y = cos x sin t + cos 2x sin 2t

Answer:

a) It represents a stationary wave as harmonic functions of x & t are contained separetely in the equation.

b) It cannot represent any type of wave.

c) It represents a progressive / travelling harmonic wave.

d) This equation is sum of two functions each representing a stationary wave. Therefore it represents superposition of two stationary waves.

Question 14.

A wire stretched between two right supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10^{-2} kg and its linear mass density is 4.0 × 10^{-2} kg m^{-1}. What is (a) the speed of transverse wave on the string, and (b) the tension in the string?

Answer:

Here, v = 45Hz, μ = 3.5 × 10^{-2} kg

Mass/length = μ = 4.0 × 10^{-2} kg/m

l = \(\frac{\mu}{\mu}\) = \(\frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}}\) = \(\frac{7}{8}\)

As \(\frac{\lambda}{2}\) = l = \(\frac{7}{8}\) ∴ λ = \(\frac{7}{4}\)m = 1.75m

a) The speed of transverse wave

υ = vλ = 45 × 1.75 = 78.75 m/s.

b) As υ = \(\sqrt{\frac{\mathrm{T}}{\mu}}\)

∴ T = υ^{2} × μ = (78.75)^{2} × 4.0 × 10^{-2}

= 248.06 N.

Question 15.

A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length Is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.

Answer:

As there is a piston at one end of the tube, it behaves as a closed organ pipe, which produces odd harmonics only. Therefore the pipe is in resonance with the fundamental note at the third harmonic (79.3 cm is about 3 times 25.5 cm)

In the fundamental note = \(\frac{\lambda}{4}\) = l_{1} = 25.5

λ = 4 × 25.5 = 102 cm = 1.02m

Speed of sound in air.

υ = vλ = 340 × 1.02

= 346.0 m/s

Question 16.

A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?

Answer:

Here, l = 100 cm = 1 m, y = 2.53 KHz

= 2.53 × 10^{3} Hz

When the rod is clamped at the middle, then in the fundamental mode of vibration of the rod, anode is formed at the middle and antinode is formed at each end.

Therefore, it is clear from fig

l = \(\frac{\lambda}{4}\) + \(\frac{\lambda}{4}\) + \(\frac{\lambda}{2}\)

λ = 2l = 2m

As v = λl

v = 2.53 × 10^{3} × 2

= 5.06 × 10^{3} ms^{-1}

Question 17.

A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s^{-1}).

Answer:

Here l = 20 cm = 0.2m, v_{n} = 430 Hz,

υ = 340 m/s

The frequency of n^{th} normal mode of vibration of closed pipe is

v_{n} = (2n – 1)\(\frac{v}{4l}\)

∴ 430 = (2n – 1)\(\frac{340}{4 \times 0.2}\)

2n – 1 = \(\frac{430 \times 4 \times 0.2}{340}\) = 1.02

2n = 2.02, n = 1.01

Hence it will be the 1^{st} normal mode of vibration. In a pipe, open at both ends we have

v_{n} = n × \(\frac{\mathrm{v}}{2l}\) = \(\frac{\mathrm{n} \times 340}{2 \times 0.2}\) = 430.

∴ n = \(\frac{430 \times 2 \times 0.2}{340}\) = 0.5

As n has to be an integer, therefore open organ pipe cannot be in resonance with the source.

Question 18.

Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324Hz. What is the frequency of B?

Answer:

Let original frequency of sitar string A be n_{a} & original frequeny of sitar string B be n_{b}.

As number of beats / sec = 6

∴ n_{b} = n_{a} ± 6 = 330 (or) 318Hz.

When tension in A is reduced, its frequency reduces (∴ n ∝ \(\sqrt{T}\))

As number of beats/sec decreases to 3 therefore, frequency of B = 324 – 6

= 318Hz.

Question 19.

Explain why (or how):

a) in a sound wave, a displacement node is a pressure antinode and vice versa,

b) bats can ascertain distances, directions, nature and sizes of the obstacles without any ‘eyes”.

c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,

d) Soils can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases and

e) The shape of a pulse gets distorted during propagation In a dispersive medium.

Answer:

a) Node (N) is a point where the amplitude of oscillation is 0. (and pressure is maximum)

Antinode (A) is a point where the amplitude of oscillation is maximum (and pressure is min).

These nodes & antinodes do not coincide with pressure nodes & antinodes.

Infact, N coincides with pressure antinode and A coincides with pressure node, as is clear from the definitions.

b) Bats emit ultrasonic wave of large frequencies, when these waveš are reflected from the obstacles in their path,

they give them the idea about the distance, direction, size & nature of the obstacle.

c) Though the violin note and sitar note have the same frequency, yet the over tones produced and their reactive strengths are different in the two flotes that is why we can distinguish between the two notes.

d) This is because solids have both, the elasticity of volume and elasticity of shape where as gases have only the volume elasticity.

e) A sound pulse is a combination of waves of different wavelengths. As waves of different wavelengths travel in a disperse medium with different velocities, therefore the shape of the pulse gets distorted.

Question 20.

A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.

i) What is the frequency of the whistle for a platform observer when the train

(a) approaches the platform with a speed of 10ms^{-1},

(b) recedes from the platform with a speed of 10 m s^{-1} ?

ii) What is the speed of sound in each case ? The speed of sound in still air can be taken as 340 m s^{-1}.

Answer:

i) Here, y = 400 Hz, υ = 340 m/s

a) Train approaches the platform

υ_{s} = 10m/s

v’ = \(\frac{v}{v-v_s}\) = \(\frac{340 \times 400}{340-10}\) = 412.12 Hz.

b) Train recedes from the platform

υ_{s} = 10m/s

v’ = \(\frac{v \times v}{v \times v_s}\) = \(\frac{340 \times 400}{340+10}\)

= 388.6Hz

ii) The speed of sound in each case is the same = 340 m/s

Question 21.

A train, stañdingin a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10m s^{-1}. What are the frequency, wavelength and speed of sound for an observer standing on the station’s platform ? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s^{-1} ? The speed of sound in still air can be taken as 340 m^{-1}

Answer:

Here y = 400 Hz, υ_{m} = 10ms^{-1}, υ = 340m/s

As the wind is blowing in the direction of sound, therefore effective speed of sound

= υ + υ_{m} = 340 + 10 = 350m/s

As the source & Iistner both are at rest, therefore, frequency remains unchanged

i.e. v = 400 Hz.

Wavelength, λ = \(\frac{v+v_m}{v}\) = \(\frac{350}{400}\)

= 0.875 M.

When air is still, υ_{m} = 0

Speed of observer υ_{1} = 10m/s υ_{s} = 0

As observer moves toward the source

υ’ = \(\frac{\left(v+v_l\right)}{v} \times v\) = \(\frac{(340+10)}{340} \times 400\)

= 411.76 Hz.

As source is at rest, wavelength does not change

i.e, λ’ = λ = 0.875M.

Also, speed of sound = υ + υ_{m} = 340 + 0

= 340 m/s

The situations in the two cases are entirely different.

Additional Exercises

Question 1.

A travelling harmonic wave on a string is described by

y(x, t) = 7.5 sin (0.0050x + 12t + π/4)

a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?

b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1cm

point at t = 2 s, 5 s and 11 s.

Answer:

a) The travelling harmonic wave is y(x, t)

From (1), y(1, 1) = 7.5 sin(732.55°)

= 7.5 sin (720 + 12.55°)

= 7.5 sin 12.55° = 7.5 × 0.2173 = 1.63 cm

velocity of oscillation, v = \(\frac{d y}{d t} y(1,1)\)

= \(\frac{\mathrm{d}}{\mathrm{dt}}\)[7.5 sin (0.005x + 12t + \(\frac{\pi}{4}\))

= 7.5 × 12 cos (0.005x + 12t + \(\frac{\pi}{4}\))

At x = 1 cm, t = 1 sec

= 7.5 × 12 cos (o.oo5 + 12 + \(\frac{\pi}{4}\))

= 90 cos (732.55°).

= 90 cos(720 + 12.55°)

= 90 cos (12.55°)

= 90 × 0.9765

= 87.89 cm/s.

Comparing the given equation with the standard form

y(x, t) = r sin\(\left[\frac{2 \pi}{\lambda}(v t+x)+\phi_0\right]\)

We get r = 7.5 cm, \(\frac{2 \pi v}{\lambda}\) = 12 (or) 2πV = 12

V = \(\frac{6}{\pi}\)

2\(\frac{\pi}{\lambda}\) = 0.005

∴ λ = \(\frac{2 \pi}{0.005}\) = \(\frac{2 \times 3.14}{0.005}\) = 1256 cm

= 12.56 m.

Velocity of wave propagation, υ = Vλ

= \(\frac{6}{\pi}\) × 12.56 .

= 24 m/s.

We find that velocity at x = 1 cm t = 1 sec is not equal to velocity of wave propagation.

b) Now, all points which are at a distance of ±λ, ± 2λ, ± 3λ from x = 1 cm will have same transverse displacement and velocity. As λ = 12.56 m, therefore, all points at distances ± 12.6m, ± 25.2 m displacement and velocity As λ = 12.56m, therefore all points at distances ± 12.6m, ± 25.2m, ± 37.8m from x = 11m will have same displacement & velocity at x = 1 cm point at t = 25.55 & 115s.

Question 2.

A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (I) frequency, (ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s (that is the whistle blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?

Answer:

a) A short pip by a whistle has neither a definite wavelength nor a definite frequency. However its speed of propagation is fixed, being equal to speed of sound in air.

b) No, frequency of the note produced by whistle is not 1/20 = 0.05 Hz. Rather 0.05 Hz is the frequency of repetition of the short pipe of the whistle.

Question 3.

One end of a long string of linear mass density 8.0 × 10^{-3} kg m^{-1} is connected to an electrically driven tuniúg fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the Incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.

Answer:

Here, μ = 8.0 × 10^{-3} kg/m, y = 256 Hz, T = 90kg = 90 × 9.8 = 882N.

Amplitude of wave, r = 5.0 1m = 0.05m.

As the wave propagation along the string is a transverse travelling wave, the velocity of the wave is given by

As the wave is propagating along x direction, the equation of the wave is

y(x, t) = r sin (ωt – kx)

= 0.05 sin (1.61 × 10^{3}t – 4.84x)

Here x, y are in mt & t in sec

Question 4.

A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^{-1}. What is the frequency of sound reflected by the submarine ? Take the speed of sound in water to be 1450 m s^{-1}.

Answer:

Here, frequency of SONAR,

v = 40 KHz = 40 × 10^{3} Hz.

Speed of observer / enemy’s submarine

υ_{1} = 360km/h .

= 360 × \(\frac{5}{18}\)m/s = 100m/s.

Speed of sound wave in water; υ = 1450 m/s.

As the source is at rest & observer is moving towards the source, therefore, apparent frequency received by enemy submarine

v’ = \(\frac{\left(v+v_1\right) v}{v}\)

= \(\frac{(1450+100) 40 \times 10^3 \mathrm{~Hz}}{1450}\)

= 4.27 × 10^{4} Hz.

This frequency is reflected by enemy submarine (source) and is observed by SONAR. Therefore in this case,

υ_{s} = 360 km/s = 100 m/s, υ_{1} = 0

∴ Apparent frequency, v^{11} = \(\frac{v \times v}{v_i-v_s}\)

= \(\frac{1450 \times 4.276 \times 10^4}{1450-10}\)

= 4.59 × 10^{4} Hz = 45.9 Hz.

Question 5.

Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s^{-1} and that of P wave is 8.0 km s^{-1}. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?

Answer:

Let υ_{1}, υ_{2} be the velocity of S waves & P waves & t_{1}, t_{2} be the time taken by these waves to travel to the position of seismograph. If l is the distance of occurrence of earth quake from the seismograph, then

l = υ_{1}t_{1} = υ_{2}t_{2} ——- (i)

now υ_{1} = 4 km/s & υ_{2} = 8 km/s .

∴ 4t_{1} = 8t_{2} (or) t_{1} = 2t_{2} ——- (ii)

Also t_{1} – t_{2} = 4min = 240s.

using (iii), 2t_{2} – t_{2} = 240s, t_{2} = 240s

t_{1} = 2t_{2} = 2 × 240 = 480s.

Now from (i) l = υ_{1}t_{1} = 4 × 480 = 1920 km.

Hence earthquake occurs 1920 km away from the seismograph.

Question 6.

A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

Answer:

Here, the frequency of sound emitted by the bat, v = 40 kHz.

velocity of bat, υ_{s} = 0.03υ, where υ is velocity of sound.

Apparent frequency of sound striking the wall.

v’ = \(\frac{v \times v}{v-v_s}\) = \(\frac{v}{v-0.03 v}\) × 40 kHxz

= \(\frac{40}{0.97}\) kHZ.

This frequency is reflected by the wall & is received by the bat moving towards the wall, So υ_{s} = 0.

υ_{1} = 0.03 υ

v’ = \(\frac{\left(v+v_1\right) v^{\prime}}{v}\) = \(\frac{(v+0.03 v)}{v}\left(\frac{40}{0.97}\right)\)

= \(\frac{1.03}{0.97} \times 40 \mathrm{kHz}\)

= 42.47 kHz