# NCERT Solutions for Class 11 Physics Chapter 15 - Waves

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

**Ans.** Given,

Mass of the string, M = 2.50 kg

Tension in the string, T = 200 N

Length of the string, l = 20.0 m

Mass per unit length,

$$\mu=\frac{\text{M}}{l}=\frac{\text{2.5}}{20}=0.125\space\text{kg/m}\\\text{We know}\\\text{Velocity of the transverse wave,}\\v=\sqrt{\frac{\text{T}}{\mu}}=\sqrt{\frac{200\text{N}}{0.125\text{kg m}^{\normalsize-1}}}\\v=\sqrt{1600}\space\text{ms}^{\normalsize-1}=40\text{ms}^{\normalsize-1}\\t=\frac{l}{v}=\frac{20.0}{40}\text{S}=\frac{1}{2}=0.5\space\text{s}$$

**15.2. A stone dropped from the top of a tower of height 300 m 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 m s^{–2})**

**Ans.** Height of the bridge, s = 300 m

Initial velocity of the stone, u = 0

Acceleration, a = g = 9.8 m/s^{2}

Speed of sound in air = 340 m/s

Let t be the time taken by the stone to hit the water’s surface

We know,

$$s=ut+\frac{1}{2}gt^{2}\\300=0+\frac{1}{2}×9.8×t^{2}\\\text{therefore, t = 7.82 s}\\\text{The time taken by the sound to reach the bridge,}\\\text{t'}=\frac{300}{340}=0.88\space\text{s}$$

Therefore, from the moment the stone is released from the bridge, the sound of it splashing the water is heard after = t + t′ = 7.82 s + 0.88 s = 8.7s

**15.3. 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}.**

**Ans.** Given,

Length of the steel wire, l = 12 m

Mass of the steel wire, m = 2.10 kg

Velocity of the transverse wave, = 343 m/s

Mass per unit length,

μ = M/l = 2.10/12 = 0.175 kg / m

We know,

Velocity of the transverse wave ,

$$v=\sqrt{\frac{\text{T}}{m}}$$

T = v^{2}.m = (343)^{2} × 0.175

= 2.06 × 10^{4} N

$$\textbf{15.4. Use the formula v =}\sqrt{\frac{\gamma \textbf{P}}{\rho}}\textbf{to explain why the speed of sound in air}$$

**(a) Is independent of pressure**

**(b) Increases with temperature**

**(c) Increases with humidity.**

**Ans.** We are given that

$$v=\sqrt{\frac{\gamma \text{P}}{\rho}}\\\text{We know PV = nRT (for n moles of ideal gas)}\\\Rarr\space\text{PV}=\frac{m}{\text{M}}\text{RT}\\\text{where m is the total mass of M is the molecular mass of the gas.}\\\therefore\space\text{P}=\frac{m}{\text{M}}.\frac{\text{RT}}{\text{M}}=\frac{\rho\text{RT}}{\text{M}}\\\Rarr\space\frac{\text{P}}{\rho}=\frac{\text{RT}}{\text{M}}$$

(a) For a gas at constant temperature,

$$\frac{\text{P}}{\rho}=\text{constant}\\\therefore\text{As P increase, r also increased and vice versa. This implies that v =}\sqrt{\frac{\gamma\text{P}}{\rho}}=\text{constant,}\\\text{i.e., velocity is independent of pressure of the gas.}\\\text{(b)}\space\text{Since}\frac{\text{P}}{\rho}=\frac{\text{RT}}{\text{M}},\text{therefore,}\\v=\sqrt{\frac{\gamma\text{P}}{\rho}}=\sqrt{\frac{\gamma\text{RT}}{\text{M}}}\\\text{Clearly v ∝}\sqrt{\text{T}}\space\text{i.e., speed of sound in air increases in temperature.}$$

(c) Increase in humidity decrease the effective desnity of air. Therefore the velocity

$$\bigg(v ∝ \frac{1}{\sqrt{\rho}}\bigg)\text{increases.}$$

**15.5. 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 – vt or x + vt, 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)/x0]**

**(c) 1/(x + vt)**

**Ans.** No, the converse is not true, because a wave function representing a travelling wave must have a finite value for all x and t values.

None of the functions above satisfy the criteria, hence none of them describe a travelling wave.

**15.6. 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 m s ^{–1} and in water 1486 m s^{–1}.**

**Ans.** Here, n = 1000 × 10^{3} Hz = 10^{6} Hz, v_{a} = 340 ms^{–1},

v_{w} = 1486 ms^{–1}

Wavelength of reflected sound,

$$\lambda_a=\frac{v_a}{\text{v}}=\frac{340}{10^{6}}\text{m}=3.4×10^{\normalsize-4}\text{m}\\\text{Wavelength of transmitted sound,}\\\lambda_\omega=\frac{v_\omega}{\text{v}}=\frac{1486}{10^{6}}\text{m}=1.486×10^{\normalsize-3}\text{m}$$

**15.7. 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.**

**Ans.** Here, speed of sound = v = 1.7 kms^{–1} = 1700 ms^{–1}

and frequency v = 4.2 MHz = 4.2 × 10^{6} Hz

$$\therefore\text{Wavelength,}\space\lambda=\frac{v}{\text{v}}=\frac{1700}{4.2×10^{6}}=4.1×10^{\normalsize-4}\space\text{m}.$$

**15.8. A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (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 ?**

**Ans.** The given equation is y(x, t)

$$=3.0\space\text{sin}\bigg(36t+0.018x+\frac{\pi}{4}\bigg),$$

where x and y are in cm and t in sec.

(a) The equation is the equation of a travelling wave, travelling from right to fleft (i.e., along –ve direction of x because it is an equation of the type

y(x, t) = A sin (ωt + kx + Φ)

$$\text{Here, A = 3.0 cm,}\space\omega = 36 \text{rad s}^{\normalsize–1}, k = 0.018 \text{cm}^{\normalsize–1} \text{and}\space\phi=\frac{\pi}{4}.\\\therefore\text{Speed of wave propagation,}\\v=\frac{\omega}{k}=\frac{36\space\text{rad}\space s ^{\normalsize-1}}{0.018\space\text{cm}^{\normalsize-1}}\\=\frac{36\space\text{rad}s^{\normalsize-1}}{0.018×10^{2}\text{m}^{\normalsize-1}}=20\space\text{ms}^{\normalsize-1}$$

(b) Amplitude of wave,

A = 3.0 cm = 0.03 m

Frequency of wave

$$v=\frac{\omega}{2\pi}=\frac{36}{2\pi}=5.7\space\text{Hz}\\\text{(c) Initial phase at the origin,}\\\phi=\frac{\pi}{4}\\\text{(d) Least distance between two successive crests in the wave}\\=\lambda=\frac{2\pi}{k}=\frac{2\pi}{0.018}$$

= 349 cm = 3.5 m

**15.9. For the wave described in Exercise 15.8, 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 ?**

**Ans.** The transverse harmonic wave is

$$y(x,t)=3.0\space\text{sin}\bigg(36t+0.018x+\frac{\pi}{4}\bigg)\\\text{For x = 0,}\\\text{y(0, t) = 3\space\text{sin}}\bigg(36t+0+\frac{\pi}{4}\bigg)\\=3\space\text{sin}\bigg(36t+\frac{\pi}{4}\bigg)\space...\text{(1)}\\\text{Here}\space\omega=\frac{2\pi}{\text{T}}=36\\\Rarr\space\text{T}=\frac{2\pi}{36}$$

To plot a (y) versus (t) graph, different values of y corresponding to the values of t may be tabulated as under (by making use of equation (i)).

t | y |

0 | $$\frac{3}{\sqrt{2}}$$ |

$$\frac{\text{T}}{8}$$ | 3 |

$$\frac{\text{2T}}{8}$$ | $$\frac{\text{3}}{\sqrt{2}}$$ |

$$\frac{\text{3T}}{8}$$ | 0 |

$$\frac{\text{4T}}{8}$$ | $$\frac{\text{-3}}{\sqrt{2}}$$ |

$$\frac{\text{5T}}{8}$$ | -3 |

$$\frac{\text{6T}}{8}$$ | $$\frac{-3}{\sqrt{2}}$$ |

$$\frac{\text{7T}}{8}$$ | 0 |

T | $$\frac{3}{\sqrt{2}}$$ |

Using the values of t and y (as in the table), a graph is plotted as under the graph obtained is sinusoidal.

Similar graphs are obtained for x = 2 cm and x = 4 cm. All the waves will have a different initial phase. Amplitude and frequency of oscillatory motion remains the same in all the cases.

**15.10. For the travelling harmonic wave y(x, t) = 2.0 cos 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**

**(a) 4 m,**

**(b) 0.5 m,**

**(c) λ/2,**

**(d) 3 λ/4**

**Ans.** The given equation can be rewritten as under :

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

$$\text{or f(x, t) = 2.0 cos }[2\pi × 0.0080\bigg(\frac{10t}{0.0080}-x\bigg)+0.7\pi]\\\text{Comparing this equation with the standard equation of a travelling harmonic wave.}\\\frac{2\pi}{\lambda}=2\pi×0.0080\\\text{or}\space\lambda=\frac{1}{0.0080}\text{cm}$$

= 125 cm

The phase difference between oscillatory motion of two points separated by a distance Δx is given by

$$\Delta\phi=\frac{2\pi}{\lambda}\Delta x\\\text{(a) When} \Delta \text{x = 4 m = 400 cm, then}\\\Delta\phi=\frac{2\pi}{125}×400=6.4\pi\space\text{rad}\\\text{(b) When Δx = 0.5 m = 50 cm, then}\\\Delta\phi=\frac{2\pi}{125}×50=0.8\pi\space\text{rad}\\\text{(c) When}\space\Delta x=\frac{\lambda}{2}=\frac{125}{2}\text{cm, then}\\\Delta\phi=\frac{2\lambda}{125}×\frac{125}{2}=\pi\text{rad}\\\text{(d) When}\space\Delta x =\frac{3\lambda}{4}=\frac{3×125}{4}\text{cm, then}\\\Delta\phi=\frac{2\pi}{125}×\frac{3×125}{4}=\frac{3\pi}{2}\text{rad}.$$

**15.11. The transverse displacement of a string (clamped at its both ends) is given by**

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

**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.**

**Ans.** The given equation is

$$\text{y(x, t) = 0.06 sin}\bigg(\frac{2\pi}{3}x\bigg)\text{cos 120 πt}\space...\text{(1)}$$

(a) As the equation involves harmonic func-tions of x and t separately, it represents a stationary wave.

(b) We know that when a wave pulse

$$\text{y}_1=\text{r sin}\frac{2\pi}{\lambda}(vt-x)\\\text{travelling along + direction of x-axis is superimposed by the reflected wave}\\\text{y}_2=-r\text{sin}\frac{2\pi}{\lambda}(vt-x)\\\text{travelling in opposite direction, a stationary wave}\\y = y_1 + y_2\\=- 2r\space\text{sin}\frac{2\pi}{\lambda}x\text{cos}\frac{2\pi}{\lambda}\space vt\space\text{is formed}...(2)\\\text{Comparing equn.s (1) and (2), we find that}\\\frac{2\pi}{\lambda}=\frac{2\pi}{3}\Rarr\lambda=3 \text{m}\\\text{Also,}\space\frac{2\pi}{\lambda}v=120\pi$$

or v = 60λ = 60 × 3 = 180 ms^{–1}

$$\text{Frequency, v =}\frac{v}{\lambda}=\frac{180}{3}=60\space\text{Hz}$$

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

(c) Velocity of transverse waves is

$$v=\sqrt{\frac{\text{T}}{m}}\space\text{or}\space v^{2}=\frac{\text{T}}{m}\\=\text{T = mv}^{2}, \text{where m} =\frac{3×10^{\normalsize-2}}{1.5}$$

= 2 × 10^{–2} kg/m

∴ T = (180)^{2} × 2 × 10^{–2}

= 648 N.

**15.12. (i) For the wave on a string described in Exercise 15.11, 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 ?**

**Ans.** (i) For the wave on the string described in questions 15.11 we have seen that l = 1.5 m and λ = 3 m. So, it is clear that

$$\lambda'=\frac{\lambda}{2}$$

and for a string clamped at both ends, it is possible only when both ends behave as nodes and there is only one antinode in between i.e., whole string is vibrating in one segment only.

(a) Yes, all the string particles, except nodes, vibrate with the same frequency v = 60 Hz.

(b) As all string particles lie in one segment, all of them are in same phase.

(c) Amplitude varies from particle to particle. At antinode, amplitude = 2A = 0.06 m. It gradually falls on going towards nodes and at nodes, it is zero.

(ii) Amplitude at a point x = 0.375 m will be obtained by putting cos (120 *π*t) as + 1 in the wave equation.

$$\therefore\space\text{A (x) = 0.06 sin}\bigg(\frac{2\pi}{3}×0.375\bigg)×1\\=0.06\space\text{sin}\frac{\pi}{4}=0.042\space\text{m.}$$

**15.13. 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)**

$$\textbf{(b) y = 2}\sqrt{\textbf{x-vt}}$$

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

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

**Ans.** (a) This equation describes a stationary wave because the harmonic terms ωt and kx appear separately.

(b) This equation does not contain any harmonic term. Thus, it is neither a travelling wave nor a stationary wave.

(c) This equation describes a travelling wave as the harmonic terms ωt and kx are in the combination of kx – ωt.

(d) This equation describes a stationary wave because the harmonic terms ωt and kx appear separately in the equation. In fact, this equation describes the superposition of two stationary waves.

**15.14. 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?**Ans.** Here, v = 45 Hz, M = 3.5 × 10^{–2} kg

Mass per unit length

= m = 4.0 × 10^{–2} kg m^{–1}

$$\therefore\space l=\frac{\text{M}}{m}=\frac{3.5×10^{\normalsize-2}}{4.0×10^{\normalsize-2}}=\frac{7}{8}\\\text{As}\space\frac{l}{2}=\lambda=\frac{7}{8}\\\therefore\lambda=\frac{7}{4}\text{m}=1.75\space\text{m}\\\text{(a) The speed of the transverse wave,}\\\text{v = vλ = 45 × 1.75 = 78.75 m/s}\\\text{(b)}\space\text{As}\space v=\sqrt{\frac{\text{T}}{\text{m}}}$$

∴ T = v^{2} × m

= (78.75)^{2} × 4.0 × 10^{–2}

= 248.06 N

**15.15. 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 effects may be neglected.**

**Ans.** Frequency of n^{th} mode of vibration of the closed organ pipe of length

$$l_1=(2n-1)\frac{v}{4l_1}\\\text{Frequency of (n + 1)}^{\text{th}} \text{mode of vibration of closed pipe of length}\\\\'l_2\space'=[2(n+1)-1]\frac{v}{4l_2}\\=(2n+1)\frac{v}{4l_2}\\\text{Both the modes are given to resonate with a frequency of 340 Hz.}\\\therefore(2n-1)\frac{v}{4l_1}=(2n+1)\frac{v}{4l_2}\\\text{or}\space\frac{2n-1}{2n+2}=\frac{l_1}{l_2}=\frac{25.5}{79.3}=\frac{1}{3}$$

[Approximation has been used because the edge effect is being ignored. Moreover, we know that in the case of a closed organ pipe, the second resonance length is three times the first resonance length.]

On 6n – 3 = 2n + 2

⇒ 4n = 5

$$\Rarr\space n=\frac{5}{4}=1.25\\\text{Now,}\space \frac{(2n-1)v}{4l_v}=340.\\\therefore\frac{(2×1-1)v×100}{4×25.5}=340\\\text{or}\space v=346.8\text{ms}^{\normalsize-1}.$$

**15.16. 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?**

**Ans.** Here, L = 100 cm = 1m, n = 2.53 k Hz

= 2.53 × 10^{3} Hz

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

Therefore, as is clear from Fig.

$$\text{L}=\frac{\lambda}{4}+\frac{\lambda}{4}=\frac{\lambda}{2}$$

λ = 2L = 2 m

As v = nλ

∴ v = 2.53 × 10^{3} × 2

= 5.060 × 10^{3} ms^{–1}

**15.17. 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}).**

**Ans.** Here length of pipe, l = 20 cm = 0.20 m, frequency v = 430 Hz and speed of sound in air υ = 340 ms^{-1}

For closed end pipe,

$$v=\frac{(2n-1)v}{4l},\text{where n = 1, 2, 3.....}\\\therefore\space(2n-1)=\frac{4vl}{v}=\frac{4×430×0.20}{340}=1.02\\\Rarr\space2n = 1.02 + 1 = 2.02\\\Rarr\space n=\frac{0.20}{2}=1.01\\\text{Hence, resonance can occur only for first (or fundamental) mode of vibration.}\\\text{ As for an open pipe}\\\text{v}=\frac{nv}{2l},\text{where n = 1, 2, 3,....}\\\therefore\space n=\frac{2lv}{v}=\frac{2×430×0.20}{340}=0.51.$$

As n < 1, hence, in this case resonance position cannot be obtained.

**15.18. 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 324 Hz, what is the frequency of B?**

Ans. Let v_{1} and v_{2 }be the frequencies of strings A and B respectively.

Then, v_{1} = 324 Hz, v_{2} = ?

Number of beats,

b = 6

v_{2} = v_{1} ± b = 324 ± 6

i.e., v_{2} = 330 Hz or 318 Hz

Since, the frequency is directly proportional to square root of tension, on decreasing the tension in the string A, its frequency v_{1} will be reduced i.e., number of beats will increase if v2 = 330 Hz. This is not so because the number of beats becomes 3.

Therefore, it is concluded that the frequency v_{2} = 318 Hz. Because of reducing the tension in the string A, its frequency may be reduced to 321 Hz, thereby giving 3 beats with v_{2} = 318 Hz.

**15.19. 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) 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.**

**Ans.** (a) A decrease in displacement, or displacement node, causes an increase in pressure, or pressure antinode, to form a sound wave. A drop in pressure also causes an increase in displacement.

(b) Bats produce high-frequency ultrasonic vibrations from their lips. The bats view these waves after they have been reflected back from the barriers in their path. These waves offer them a notion of the barriers' distance, direction, type and size.

(c) The sound quality of a violin note differs from that of a sitar tone. As a result, they generate different harmonics that the human ear can detect and utilise to distinguish between the two notes.

(d) This is because gases only have the bulk modulus of elasticity, but solids have both the shear modulus and the bulk modulus of elasticity.

(e) A sound pulse is made up of many waves with varying wavelengths. In a dispersive medium, these waves travel at different speeds, causing the wave to be distorted.

**15.20. 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 10 m s ^{–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}.**

**Ans.** Frequency of whistle,

v = 400 Hz; speed of sound,

v = 340 ms^{–1} speed of train

v_{s} = 10 ms^{–1}

(i) (a) When the train approaches the platform (i.e., the observer at rest),

$$v'=\frac{v}{v-v_s}×v\\=\frac{340}{340-10}×400\\text{= 412 Hz.}\\\text{(b) When the train recedes the platform}\\\text{v'}=\frac{v}{v+v_s}×v=\frac{340}{340+10}×400$$

= 389 Hz.

(ii) The speed of sound in each case does not change. It is 340 ms^{–1} in each case.

**15.21. 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 10 m 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 s^{–1}.**

**Ans.** Here actual frequency of whistle of train

v = 400 Hz,

Speed of sound in still air

v = 340 ms^{–1}.

As wind is blowing in the direction from the yard to the station with a speed of

v_{m }= 10ms^{–1}

∴ For an observer standing on the platform, the effective speed of sound

v′ = v + v_{m} = 340 + 10 = 350 ms^{–1}

As there is no relative motion between the sound source (rail engine) and the observer, the frequency of sound for the observer,

v = 400 Hz.

∴ Wavelength of sound heard by the observer

$$\lambda'=\frac{v'}{\text{v}}=\frac{350}{400}=0.875\space \text{m}.$$

The situation is not identical to the case when the air is still and observer runs towards the yard at a speed of v_{0} = 10 ms^{–1}. In this situation as medium is at rest.Hence, v′ = v = 340 ms^{–1}.

$$\text{v'}=\frac{v+v_0}{v}=\frac{340+10}{340}×400\\=412\space\text{Hz}\\\text{and}\space\lambda'=\lambda=\frac{v}{\text{v}}=\frac{340}{400}=0.85\text{m}$$

**15.22. 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 = 1 cm point at t = 2 s, 5 s and 11 s.**

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

= 7.5 sin (0.0050x + 12t + *π*/4)

At x = 1 cm and t = 1 sec,

y (1, 1) = 7.5 sin (0.005 × 1 + 12 × 1 + *π*/4)

= 7.5 sin (12.005 + *π*/4) ...(i)

Now, θ = (12.005 + *π*/4) radian

$$=\frac{180}{\pi}(12.005+\pi/4)\text{degree}\\=\frac{12.005×180}{\frac{22}{7}}+45$$

= 732.55°.

∴ From (i), y (1, 1)

= 7.5 sin (732.55°)

= 75 sin(720 + 12.55°)

= 7.5 sin 12.55°

= 7.5 × 0.2173 = 1.63 cm.

Velocity of oscillation,

$$v=\frac{dy}{dt}(1,1)\\=\frac{d}{dt}\bigg[7.5\text{sin}\bigg(0.005x+12t+\frac{\pi}{4}\bigg)\bigg]\\=7.5×12\text{cos}\bigg[0.005x+12t+\frac{\pi}{4}\bigg]$$

At x = 1 cm, t = 1 sec.

v = 7.5 × 12 cos (0.005 + 12 + *π*/4)

= 90 cos (732.35°)

= 90 cos (720 + 12.55)

v = 90 cos (12.55°)

= 90 × 0.9765 = 87.89 cm/s.

Comparing the given eqn. with the standard

$$\text{form y (x,t) = t sin}\bigg[\frac{\pi}{4}(vt+x)+\phi_0\bigg]\\\text{We get r = 7.5 cm,}\\\frac{2\pi v}{\lambda}=12\text{or 2} \pi v = 12\\v=\frac{6}{\pi}\\\frac{2\pi}{\lambda}=0.005.\\\therefore\space\lambda=\frac{2\pi}{0.005}=\frac{2×3.14}{0.005}$$

= 1256 cm = 12.56 m

Velocity of wave propagation,

$$v=\text{v}\lambda=\frac{6}{\pi}×12.56=24\text{m/s}$$

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 will have same transverse displacement and velocity. As

λ = 12.56 m, therefore, all ponits at distances ± 12.6 m, ± 25.2 m, ± 37.8 m.... from x = 1 cm will have same displacement and velocity, as at x = 1 point t = 2s, 5 s and 11 s.

**15.23. 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 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 ?**

**Ans.** (a) The speed of propagation is known, and it is the same as the speed of sound in air. The wavelength and frequency are not set in stone.

(b) The sound produced by a whistle does not

$$\frac{1{20}=0.05\space\text{Hz}$$

as frequency. It means that 0.05 Hz is the frequency of the repetition of the pip of the whistle.

**15.24. 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.**

**Ans. **Frequency, v = 256 Hz

and amplitude, A = 5.0 cm = 0.05 cm

As the wave propagating along the string is a transverse travelling wave, the velocity of the wave,

$$v=\sqrt{\frac{\text{T}}{\mu}}=\sqrt{\frac{882}{8×10^{\normalsize-3}}}\text{ms}^{\normalsize-1}=3.32×10^{2}\text{ms}^{\normalsize-1}\\\text{Now} , ω = 2\pi v = 2 × 3.142 × 256 = 1.61 × 10^{3} \text{rad s}^{\normalsize–1}\\\text{\text{Also,}}\space v=v\lambda\space\text{or}\space\lambda=\frac{v}{\text{v}}=\frac{3.32×10^{2}}{256}\text{m}\\\text{Propagation constant,}\\\text{k =}\frac{2\pi}{\lambda}=\frac{2×3.142×256}{3.32×10^{2}}$$

= 4.84 m^{–1}

∴ The equation of the wave is,

v(x, t) = A sin (ωt – kx)

= 0.05 sin (1.61 × 10^{3} t – 4.84 x)

Here, x, y are in metre and t is in second.

**15.25. 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}.**

**Ans.** Frequency of the SONAR system,

f = 40 kHz = 40 × 10^{3} Hz

Speed of sound in water, v = 1450 m/s

Speed of the enemy submarine,

v_{0} = 360 km/h= 360 × (5/18) = 100 m/s

The SONAR is at rest and the energy submarine moves towards it. Therefore, the apparent frequency is given by the relation

$$f'=\bigg[\frac{v+v_0}{v}\bigg]f\\=\bigg[\frac{1450+100}{1450}\bigg]×40×10^{3}$$

= 42.75 × 10^{3} Hz

This frequency (f ′) is reflected by the enemy submarine and it is observed by the SONAR.

Therefore,

v_{s} = 360 km/s = 100 m/s

$$f'=\bigg(\frac{v}{v-v_s}\bigg)×f'\\=\bigg(\frac{1450}{1450-100}\bigg)×42.75×10^{3}\\=\bigg(\frac{1450}{1350}\bigg)×42.75×10^{3}$$

= 45.91 × 10^{3} Hz

**15.26. 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 ?**

**Ans.** Let the speeds of S and P be v_{1} and v_{2} respectively. The time taken by the S and P waves to reach the position of the seismograph is t_{1} and t_{2} respectively

l = v_{1}t_{1} = v_{2}t_{2}

The speed of S wave, v_{1} = 4.0 km s^{–1}

The speed of P wave, v_{2} = 8.0 km s^{–1}

4t_{1} = 8t_{2}

t_{1} = 2t_{2}

The first P wave arrives 4 min before the S wave.

t_{1} – t_{2} = 4 min = 4 × 60 s = 240 s

2t_{2} – t_{2} = 240 s

t_{2} = 240 s

t_{1} = 2t_{2} = 2 × 240 = 480 s

Distance at which earthquake occur,

l = v_{1}t_{1 }= 4 × 480 = 1920 km

**15.27. 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 ?**

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

Velocity of bat, v_{s} = 0.03 v, where v is velocity of sound. Apparent frequency of sound striking the wall

$$\text{v}'=\frac{v}{v-v_s}×\text{v}\\=\frac{v}{v-0.03v}×40\text{kHz}\\=\frac{40}{0.97}\text{kHz}$$

This frequency is reflected by the wall and is received by the bat moving towards the wall.

So,

v_{s} = 0

v_{L} = 0.03 v

$$\text{v}''=\frac{(v+v_L)}{v}×\text{v}'=\frac{(v+0.03v)}{v}\bigg(\frac{40}{0.97}\bigg)\\=\frac{1.03}{0.97}×40\text{kHz}=42.47\text{kHz.}$$

## NCERT Solutions for Class 11 Physics Chapter 15 Free PDF Download

Please Click on Free PDF Download link to Download the NCERT Solutions for Class 11 Physics Chapter 15 Waves