Question:

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( v = 343 m)s^−1:

Answer:

1. Determine the linear mass density of the wire: Linear mass density = Mass of wire/Length of wire = 2.10 kg/12.0 m = 0.175 kg/m

2. Calculate the wave speed on the wire: Wave speed = Square root of (Tension/Linear mass density)

3. Calculate the tension required in the wire: Tension = (Wave speed)^2 x Linear mass density = (343 m/s)^2 x 0.175 kg/m = 8.6 x 10^4 N

Question:

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 struck at one end of the string, how long does he disturbance take to reach the other end ?

Answer:

1. Convert the mass of the string into its linear density, using the formula linear density = mass/length: Linear density = 2.50 kg / 20.0 m = 0.125 kg/m

2. Calculate the wave speed of the disturbance using the formula wave speed = √(tension/linear density): Wave speed = √(200 N / 0.125 kg/m) = 20 m/s

3. Calculate the time it takes for the disturbance to reach the other end using the formula time = distance/speed: Time = 20.0 m / 20 m/s = 1.0 s

Question:

For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x=0,2 and 4cm. 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:

1. Exercise 15.8: A travelling wave is described by the equation y = 0.2sin(2π(x/2 + t/2))

2. Plot the displacement (y) versus (t) graphs for x=0,2 and 4cm:

y(t) for x=0 cm: y = 0.2sin(2πt/2) y(t) for x=2 cm: y = 0.2sin(2π(2/2 + t/2)) y(t) for x=4 cm: y = 0.2sin(2π(4/2 + t/2))

3. The shapes of these graphs are sinusoidal.

4. In travelling wave, the amplitude remains the same, but the frequency and phase differ from one point to another.

Question:

The transverse displacement of a string (clamped at its both ends) is given by y(x,t)=0.06sin(2πx​/3)cos(120πt) 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 following : (a) Does 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:

(a) The function represents a travelling wave.

(b) The wavelength of the wave is 3 m, the frequency is 40 Hz, and the speed is 40 m/s. The wave can be interpreted as a superposition of two waves travelling in opposite directions, each with a wavelength of 3 m, a frequency of 40 Hz, and a speed of 40 m/s.

(c) The tension in the string can be determined using the formula T = μLω^2, where μ is the mass per unit length of the string (3.0 × 10^−2 kg/1.5 m = 0.02 kg/m), L is the length of the string (1.5 m), and ω is the angular frequency of the wave (120π rad/s). Therefore, the tension in the string is T = 0.02 × 1.5 × (120π)^2 = 5184.48 N.

Question:

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:

1. Calculate the wavelength of the sound wave in the tube by using the equation λ = v/f, where λ is the wavelength, v is the velocity of the sound wave, and f is the frequency of the tuning fork.

λ = v/340 Hz = 340 m/s/340 Hz = 1 m

2. Determine the length of the tube by subtracting the length of the piston from the total length of the tube.

Tube Length = 1 m - 0.255 m = 0.745 m

3. Calculate the speed of sound in air by using the equation v = λ/T, where v is the velocity of the sound wave, λ is the wavelength, and T is the time it takes for the sound wave to travel the length of the tube.

v = 0.745 m/T = 340 m/s

Question:

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 - v t or x + v t i.e. y=f(x±vt). Is the converse true ? Examine if the following functions for y can possibly represent a travelling wave : (a) (x+vt)^2 (b) log[(x+vt)/xo​] (c) 1/(x+vt)

Answer:

a) No, (x+vt)^2 cannot represent a travelling wave.

b) Yes, log[(x+vt)/xo] can represent a travelling wave.

c) No, 1/(x+vt) cannot represent a travelling wave.

Question:

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=2cos(3x)sin(10t) (b) y=2√x−vt (c) y=3sin(5x−0.5t)+4cos(5x−0.5t) (d) y=cosx sint+cos2x sin2t

Answer:

(a) Travelling wave (b) None at all (c) Travelling wave (d) Stationary wave

Question:

A wire stretched between two rigid 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 a transverse wave on the string, and (b) the tension in the string ?

Answer:

a) The speed of a transverse wave on the string can be calculated using the formula v=√(T/μ), where T is the tension in the string and μ is the linear mass density.

b) The tension in the string can be calculated using the formula T=μv^2, where v is the speed of a transverse wave on the string and μ is the linear mass density.

Therefore,

a) v=√(T/μ)=√(45 Hz/4.0×10^−2 kg m^−1)=7.07 m/s

b) T=μv^2=4.0×10^−2 kg m^−1*7.07 m/s^2=0.29 N

Question:

A steel rod 100 cm long is clamped at its mid-point. the fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel? A 5.06 km/s B 6.06 km/s C 7.06 km/s D 8.06 km/s

Answer:

Step 1: Determine the formula needed to solve the problem. The formula needed to solve the problem is the formula for the speed of sound in a solid, which is given by v = (F x L)/2π, where v is the speed of sound, F is the fundamental frequency, and L is the length of the rod.

Step 2: Substitute the given values into the formula. v = (2.53 kHz x 100 cm)/2π

Step 3: Calculate the answer. v = 8.06 km/s

Step 4: Compare the answer to the given options. The answer is D, 8.06 km/s.

Question:

A train, standing in 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 10ms^−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 10ms^−1? The speed of sound in still air can be taken as 340 m s^−1

Answer:

Frequency: 400 Hz

Wavelength: 0.85 m

Speed of sound: 330 m s^−1

No, the situation is not exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10ms^−1. The frequency and wavelength remain the same, but the speed of sound is decreased by 10 m s^−1 due to the wind blowing in the direction from the yard to the station.

Question:

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:

1. Determine the speed of the P wave: 8.0 km s^−1

2. Determine the speed of the S wave: 4.0 km s^−1

3. Determine the time difference between the arrival of the P and S waves: 4 min

4. Calculate the distance traveled by each wave by multiplying the speed of each wave by the time difference: P wave: 8.0 km s^−1 x 4 min = 32 km S wave: 4.0 km s^−1 x 4 min = 16 km

5. Calculate the distance of the earthquake by adding the distances traveled by the P and S waves: 32 km + 16 km = 48 km

6. Answer: The earthquake occurs at a distance of 48 km.

Question:

Use the formula v=√γP​/ρ ​ to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature,
(c) Increases with humidity .

Answer:

(a) The speed of sound in air is independent of pressure because the pressure (P) is in the denominator of the equation and therefore the pressure cancels out when the equation is solved.

(b) The speed of sound in air increases with temperature because the ratio of the specific heat capacity of air (γ) to the density of air (ρ) increases as temperature increases. Therefore, as the temperature increases, the speed of sound also increases.

(c) The speed of sound in air increases with humidity because the density of air (ρ) decreases as the humidity increases. Therefore, as the humidity increases, the speed of sound also increases.

Question:

A bat emits ultrasonic sound of 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 ms^−1 and in water 1486 ms^−1.

Answer:

a) Wavelength of reflected sound = (Speed of sound in air / Frequency of sound in air) = (340 ms^−1 / 1000 kHz) = 0.34 m

b) Wavelength of transmitted sound = (Speed of sound in water / Frequency of sound in air) = (1486 ms^−1 / 1000 kHz) = 1.486 m

Question:

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 324 Hz what is the frequency of B ?

Answer:

1. The original frequency of string A is 324 Hz.

2. The tension in string A is reduced, and the beat frequency is found to reduce to 3 Hz.

3. The frequency of string B can be calculated using the formula: fB = fA ± (beat frequency/2).

4. Therefore, the frequency of string B is: fB = 324 Hz ± (3 Hz/2) = 321 Hz.

Question:

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 10 ms^−1, (b) recedes from the platform with a speed of 10 ms^−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) (a) The frequency of the whistle for the platform observer when the train is approaching the platform with a speed of 10 ms^−1 is: f = 400 Hz * (340 m/s + 10 m/s) / 340 m/s f = 402 Hz

(b) The frequency of the whistle for the platform observer when the train is receding from the platform with a speed of 10 ms^−1 is: f = 400 Hz * (340 m/s - 10 m/s) / 340 m/s f = 398 Hz

(ii) The speed of sound in each case is 340 m/s.

Question:

A narrow sound pulse (for example, a short pip by a whistle) is sent across a

Answer:

1. A narrow sound pulse is created (for example, a short pip by a whistle).

2. The sound pulse is sent across a distance.

3. The sound pulse reaches its destination.

4. The sound pulse is detected and measured at the destination.

Question:

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 is 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) (i) Yes, the pulse has a definite frequency. (ii) Yes, the pulse has a definite wavelength. (iii) Yes, the pulse has a definite speed of propagation.

(b) No, the frequency of the note produced by the whistle is not equal to 1/20 or 0.05 Hz. The frequency of the note produced is equal to 1 Hz, since it is blown for a split second after every 20 seconds.

Question:

One end of a long string of linear mass density 8.0 ×10^−3 kg m^−1 is connected to an electrically driven tuning 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:

y(x,t) = 5 cm * sin (2π * (x/λ) - 2π * (256 * t))

where λ is the wavelength of the wave.

Question:

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:

1. Calculate the speed of sound in air: Speed of sound in air = 340 m/s

2. Calculate the speed of the bat: Speed of bat = 0.03 x 340 m/s = 10.2 m/s

3. Calculate the frequency of the sound reflected off the wall: Frequency of sound reflected off the wall = (2 x 10.2 m/s) / 0.03 m = 680 kHz

Question:

(i) For the wave on a string described in Exercise 15.11, do all the points c n the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What Is the amplitude of a point 0.375m away from one end?

Answer:

(i) (a) No, different points on the string oscillate with different frequencies. (b) No, different points on the string oscillate with different phases. (c) No, different points on the string oscillate with different amplitudes.

(ii) The amplitude of a point 0.375m away from one end is equal to the amplitude of the wave at that point.

Question:

A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond 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 = 9.8 ms^−2)

Answer:

1. Calculate the time taken for the stone to reach the water near the base of the tower: t = √(2h/g) = √(2 x 300/9.8) = 5.6 s

2. Calculate the distance traveled by the sound of the splash: Distance = Speed x Time = 340 x 5.6 = 1908 m

3. Calculate the time taken for the sound of the splash to reach the top of the tower: Time = Distance/Speed = 1908/340 = 5.6 s

4. The splash will be heard at the top of the tower after 5.6 s.

Question:

In a hospital an ultrasonic scanner is used to locate tumours in a tissue. The operating frequency of the scanner is 4.2 MHz. The speed of sound in a tissue is 1.7kms^−1. The wavelength of sound in the tissue is approximately A 4×10^−4 m B 2×10^−4 m C 8×10^−6 m D 2×10−6 m

Answer:

Answer: A 4×10^−4 m

Explanation: The wavelength of sound in a tissue can be calculated using the formula:

Wavelength = Speed of Sound / Frequency

Therefore, Wavelength = 1.7kms^−1 / 4.2 MHz

Wavelength = 4×10^−4 m

Question:

A transverse harmonic wave on a string is described by

Answer:

1. y(x,t)=A sin(kx-ωt)

This equation describes a transverse harmonic wave on a string, where y(x,t) is the displacement of the string at a given position x and time t, A is the amplitude of the wave, k is the wave number, and ω is the angular frequency of the wave.

Question:

y(x,t)=3.0sin(36t+0.018x+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:

(a) This is a travelling wave. The speed of propagation is 0.018 cm/s and the direction of propagation is from left to right.

(b) The amplitude is 3.0 cm and the frequency is 36 s-1.

(c) The initial phase at the origin is π/4 radians.

(d) The least distance between two successive crests in the wave is 0.018 cm.

Question:

For the travelling harmonic wave y(x,t)=2.0cos 2π (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 x A x=4m, Δϕ=6.4π rad B 0.5m, Δϕ=0.6πrad C λ/2, Δϕ=.6π rad D 3λ/4, Δϕ=2.5π rad.

Answer:

A. Δϕ=2π (4/0.008)=6.4π rad

B. Δϕ=2π (0.5/0.008)=0.6π rad

C. Δϕ=2π (λ/2/0.008)=0.6π rad

D. Δϕ=2π (3λ/4/0.008)=2.5π rad

Question:

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:

1. Calculate the wavelength of the source (λ): λ = 340 m s^−1 / 430 Hz = 0.79 m

2. Determine the fundamental frequency (f1) of the pipe: f1 = v/2L = 340 m s^−1 / (2 x 0.2 m) = 1700 Hz

3. Calculate the wavelength of the fundamental frequency (λ1): λ1 = v/f1 = 340 m s^−1 / 1700 Hz = 0.2 m

4. Determine the harmonic mode of the pipe that is resonantly excited by the source: The source (λ = 0.79 m) is twice the wavelength of the fundamental frequency (λ1 = 0.2 m) of the pipe, so the second harmonic mode (2f1) of the pipe is resonantly excited by the source.

5. Determine if the same source will be in resonance with the pipe if both ends are open: No, because the wavelength of the fundamental frequency of the pipe (λ1) will be doubled if both ends are open, so the source will not be in resonance with the pipe.

Question:

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) solids 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) Pressure antinodes and displacement nodes are points in a sound wave where the pressure and displacement of the wave are at their maximum and minimum, respectively. As sound waves move, the pressure and displacement of the wave fluctuate in opposite directions, so where the pressure is at a maximum, the displacement is at a minimum, and vice versa.

(b) Bats use echolocation to determine distances, directions, nature and sizes of obstacles. Echolocation is the process of emitting sound waves and then listening for the echoes of those sound waves off of objects. By measuring the time it takes for the sound waves to return to the bat, they can determine the distance, direction, nature and size of the object.

(c) Even though a violin note and sitar note may have the same frequency, we can distinguish between the two notes because of the difference in their timbre. Timbre is the quality of a sound that distinguishes it from other sounds of the same pitch and loudness. The timbre of a sound is determined by its harmonic content, which is the number and relative strength of its overtones.

(d) Solids can support both longitudinal and transverse waves because the particles in a solid are strongly connected and can vibrate in multiple directions. On the other hand, gases are composed of particles that are far apart and can only vibrate in one direction, so only longitudinal waves can propagate in gases.

(e) The shape of a pulse gets distorted during propagation in a dispersive medium because different frequencies travel at different speeds in a dispersive medium. This causes the different frequencies to spread out over time, leading to a distorted shape.

Question:

A travelling harmonic wave on a string is described by y(x,t)=7.5sin(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=1 cm point at t=2s , 5 s and 11s

Answer:

a) Displacement of the point at x=1 cm and t=1 s: y(1,1) = 7.5sin(0.0050(1)+12(1)+π/4) = 7.5sin(0.1275π) = 6.5 cm

Velocity of oscillation of the point at x=1 cm and t=1 s: v(1,1) = dy/dt = 7.5(0.0050)(12)cos(0.1275π) = 0.9 cm/s

This velocity is not equal to the velocity of wave propagation, which is given by v = (2π/T)λ, where T is the period of the wave and λ is the wavelength.

b) Points with the same transverse displacement and velocity as the x=1 cm point at t=2s: y(x,2) = 7.5sin(0.0050x+12(2)+π/4) = 6.5 cm v(x,2) = 7.5(0.0050)(12)cos(0.0050x+12(2)+π/4) = 0.9 cm/s

Therefore, the points with the same transverse displacement and velocity as the x=1 cm point at t=2s are x = 0.1275π/0.0050 = 25.5 cm and x = (2π+0.1275π)/0.0050 = 51 cm.

Points with the same transverse displacement and velocity as the x=1 cm point at t=5s: y(x,5) = 7.5sin(0.0050x+12(5)+π/4) = 6.5 cm v(x,5) = 7.5(0.0050)(12)cos(0.0050x+12(5)+π/4) = 0.9 cm/s

Therefore, the points with the same transverse displacement and velocity as the x=1 cm point at t=5s are x = (3π+0.1275π)/0.0050 = 76.5 cm and x = (5π+0.1275π)/0.0050 = 102 cm.

Points with the same transverse displacement and velocity as the x=1 cm point at t=11s: y(x,11) = 7.5sin(0.0050x+12(11)+π/4) = 6.5 cm v(x,11) = 7.5(0.0050)(12)cos(0.0050x+12(11)+π/4) = 0.9 cm/s

Therefore, the points with the same transverse displacement and velocity as the x=1 cm point at t=11s are x = (9π+0.1275π)/0.0050 = 178.5 cm and x = (11π+0.1275π)/0.0050 = 204 cm.

Question:

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:

1. Convert the speed of the enemy submarine from km h^−1 to m s^−1: 360 km h^−1 = 100 m s^−1

2. Calculate the Doppler shift of the sound: Doppler shift = (speed of sound - speed of object) / (speed of sound + speed of object) = (1450 m s^−1 - 100 m s^−1) / (1450 m s^−1 + 100 m s^−1) = 0.93

3. Calculate the frequency of the reflected sound: Reflected frequency = (original frequency) * (1 + Doppler shift) = 40.0 kHz * (1 + 0.93) = 77.2 kHz