NCERT Solutions for Class 11 Physics Chapter 15

Q.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.

\begin{array}{l}\text{Here,}\text{mass of string,}\text{M}\text{=}\text{2}\text{.50 kg}\\ \text{Tension}\text{produced in the string,}\text{T}\text{=}\text{200 N}\\ \text{Length of string,}\text{l}\text{=}\text{20}\text{.0 m}\\ \text{Mass per unit length,}\text{m}=\frac{M}{l}=\frac{\text{2}.\text{5}0}{\text{2}0.0}=0.125{\text{kgm}}^{-1}\\ \text{Velocity of the transverse wave in the string}\text{is}\text{given}\text{as:}\\ \text{v}\text{=}\sqrt{\frac{T}{\mu }}=\sqrt{\frac{200}{0.125}}=\sqrt{1600}=40{\text{ms}}^{-1}\\ \text{Time taken by the disturbance to reach the other end}\text{is}\text{given}\text{as:}\\ \text{t}\text{=}\frac{\text{Length}\text{of}\text{string}}{\text{Velocity}\text{of}\text{transverse}\text{wave}\text{in}\text{string}}=\frac{l}{v}=\frac{20}{40}=0.50\text{s}\end{array}

Q.2 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 m s^–2)

Ans.

\begin{array}{l}\text{Here,}\text{height of tower,}\text{s}\text{=}\text{300 m}\\ \text{Initial velocity of stone,}\text{u}\text{=}\text{0}\\ \text{Acceleration}\text{due}\text{to}\text{gravity,}\text{g}\text{=}\text{9}{\text{.8 ms}}^{\text{-2}}\\ \text{Speed of sound in air}\text{=}{\text{340 ms}}^{\text{-1}}\\ \text{Let time taken by the stone to hit the surface}\text{of}\text{water}\text{=}{\text{t}}_{\text{1}}\\ \text{According}\text{to second equation of motion,}\text{we}\text{have:}\\ s=u{\text{t}}_{\text{1}}+\frac{1}{2}g{\text{t}}_{\text{1}}{}^2\\ \therefore 300=0+\frac{1}{2}\times \text{9}.\text{8}\times {\text{t}}_{\text{1}}{}^2\\ \therefore {\text{t}}_{\text{1}}=\sqrt{\frac{300\times 2}{\text{9}.\text{8}}}=7.82s\\ \text{Time taken by sound to reach the top of the tower:}\\ t_2=\frac{h}{v}=\frac{300}{340}=0.88s\\ \therefore \text{Total time after which the splash is heard}\text{is}\text{given}\text{as:}\\ t={\text{t}}_{\text{1}}+t_2=\text{7}.\text{82}+0.\text{88}=\text{8}.\text{7 s}\end{array}

Q.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 ms^–1.

Ans.

\begin{array}{l}\text{Here, length of wire,}\text{l}=\text{12 m}\\ \text{Mass of wire},\text{m}=\text{2}.\text{1}0\text{kg}\\ \text{Velocity of transverse wave,}\text{v}\text{=}{\text{343 ms}}^{\text{-1}}\\ \text{Mass per unit length,}\mu =\frac{m}{l}=\frac{\text{2}.\text{1}0}{12}=0.175{\text{kgm}}^{-1}\\ \text{Let}\text{tension}\text{in}\text{the}\text{wire}\text{=}\text{T}\\ \text{Velocity of the transverse wave is}\text{given}\text{by}\text{the}\text{relation:}\\ \text{using the relation:}\\ v=\sqrt{\frac{T}{\mu }}\\ \therefore \text{T}={\text{v}}^{\text{2}}\mu ={\left(\text{343}\right)}^{\text{2}}\times 0.\text{175}\\ =\text{2}0\text{588}.\text{575}\approx \text{2}.0\text{6}\times \text{1}{0}^{\text{4}}\text{N}\end{array}

Q.4

\begin{array}{l}\text{Use the formula}\text{v}=\sqrt{\frac{\gamma \text{P}}{\rho }}\text{to explain why the speed of}\text{sound in air}\\ \left(\text{a}\right)\text{is independent of pressure,}\\ \left(\text{b}\right)\text{increases with temperature,}\\ \left(\text{c}\right)\text{increases with humidity}\text{.}\end{array}

Ans.

\begin{array}{l}\left(\text{a}\right)\text{It}\text{is}\text{given}\text{that:}\\ v=\sqrt{\frac{\gamma P}{\rho }}\to \text{(i)}\\ \text{Here,}\rho =\text{Density}\\ \rho =\frac{\text{Mass}}{\text{Volume}}=\frac{M}{V}\\ \text{Here,}\text{M}\text{and}\text{V}\text{represent}\text{the}\text{molecular}\text{weight}\text{and}\text{volume}\text{of}\text{gas}\text{respectively}\text{.}\\ \therefore \text{Equation}\text{(i)}\text{becomes:}\\ v=\sqrt{\frac{\gamma PV}{M}}\to \text{(ii)}\\ \text{From the ideal gas equation for}\text{n}=\text{1}:\\ \text{PV}=\text{RT}\\ \text{At constant}\text{temperature}\left(T\right),\text{PV}=\text{Constant}\\ \text{As both}\text{M}\text{and}\gamma \text{are constants,}\\ \therefore \text{v}=\text{Constant}\\ \therefore \text{At a constant temperature, the speed of sound in a}\text{gaseous medium is}\text{does}\text{not}\text{depend}\text{on the change in}\\ \text{the}\text{pressure of the gas}\text{.}\\ \left(\text{b}\right)\text{It}\text{is}\text{given}\text{that:}\\ v=\sqrt{\frac{\gamma P}{\rho }}\to \text{(i)}\\ \text{From}\text{the}\text{ideal}\text{gas}\text{equation,}\text{for}\text{n}=1,\text{we}\text{have:}\\ PV=RT\\ P=\frac{RT}{V}\to \text{(ii)}\\ \text{Putting}\text{equation}\text{(ii)}\text{in}\text{equation}\text{(i),}\text{we}\text{obtain:}\\ v=\sqrt{\frac{\gamma RT}{V\rho }}=\sqrt{\frac{\gamma RT}{M}}\to \text{(iii)}\\ \text{Here,}\text{mass}\text{M}\left(=\rho \text{V}\right),\gamma \text{and}\text{R}\text{are constants}\\ \text{From equation}\left(\text{iii}\right)\text{,}\text{it}\text{can}\text{be}\text{observed}\text{that:}\\ v\propto \sqrt{T}\\ \therefore \text{The speed of sound in a gas is directly proportional}\text{to the square root of the}\text{temperature of}\text{the gaseous}\\ \text{medium}\text{.}\text{It}\text{implies}\text{that the speed of the}\text{sound increases}\text{with the}\text{rise in the temperature of the gaseous medium}\text{.}\\ \left(\text{c}\right)\text{It}\text{is}\text{given}\text{that:}\\ v=\sqrt{\frac{\gamma P}{\rho }}\to (i)\\ \text{Let}\text{speed}\text{of}\text{sound}\text{in}\text{the}\text{moist}\text{air}\text{=}{\text{v}}_{\text{m}}\\ \text{Let}\text{speed}\text{of}\text{sound}\text{in}\text{the}\text{dry}\text{air}\text{=}{\text{v}}_{\text{d}}\\ \text{Let}\text{density}\text{of}\text{moist}\text{air}={\rho }_m\\ \text{Let}\text{density}\text{of}\text{dry}\text{air}={\rho }_d\\ \text{Using}\text{equation}\text{(i),}\text{the}\text{speed}\text{of}\text{sound}\text{in}\text{the}\text{moist}\text{air}\text{is}\text{given}\text{as:}\\ v_m=\sqrt{\frac{\gamma P}{{\rho }_m}}\to \text{(ii)}\\ \text{Using}\text{equation}\text{(i),}\text{the}\text{speed}\text{of}\text{sound}\text{in}\text{the}\text{moist}\text{air}\text{is}\text{given}\text{as:}\\ v_d=\sqrt{\frac{\gamma P}{{\rho }_d}}\to \text{(iii)}\\ \text{Dividing}\text{equation}\text{(ii)}\text{and}\text{(iii),}\text{we}\text{obtain:}\\ \frac{v_m}{v_d}=\sqrt{\frac{\gamma P}{{\rho }_m}\times \frac{{\rho }_d}{\gamma P}}=\sqrt{\frac{{\rho }_d}{{\rho }_m}}\\ \text{However,}\text{the water vapour decreases the}\text{density}\text{of}\text{air}\text{.}\\ {\rho }_d<{\rho }_m\\ \therefore v_d>v_m\\ \therefore \text{Speed of sound in moist air is more than it is}\text{in dry air}\text{.}\\ \therefore \text{In the}\text{air,}\text{the}\text{speed of sound increases with humidity}\text{.}\end{array}

Q.5

\begin{array}{l}\text{You have learnt that a travelling wave in one dimension is represented by a function}\text{y}\text{=}\text{f}\left(\text{x, t}\right)\\ \text{where}\text{x}\text{and}\text{t}\text{must appear in the}\text{combination}\text{x}\text{–}\text{vt}\text{or}\text{x}\text{+}\text{vt, i}\text{.e}\text{. y}\text{=}\text{f}\left(\text{x}\text{±}\text{v t}\right)\text{.}\\ \text{Is the converse true? Examine if}\text{the following}\text{functions for}\text{y}\text{can possibly}\\ \text{represent a travelling wave:}\\ \left(\text{a}\right){\left(\text{x}\text{–}\text{vt}\right)}^{\text{2}}\\ \left(\text{b}\right)\text{log}\left[\frac{\text{x}\text{+}\text{vt}}{{\text{x}}_{\text{0}}}\right]\\ \left(\text{c}\right)\frac{\text{1}}{\left(\text{x}\text{+}\text{vt}\right)}\end{array}

Ans.

No, the converse of the given statement is not true. The essential requirement for a function to represent a travelling wave is that it should remain finite for all values of x and t. Out of the given functions, only function (c) satisfies this condition, the remaining functions cannot represent a travelling wave.

Q.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.

\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{frequency of ultrasonic sound,}\nu =\text{1}000\text{kHz}\\ =\text{1}{0}^{\text{6}}\text{Hz}\\ \text{Speed of sound in air,}{\text{v}}_{\text{a}}=\text{34}0{\text{ms}}^{\text{-1}}\\ \text{Wavelength of the reflected sound is given as:}\\ {\lambda }_{\text{r}}=\frac{{\text{v}}_{\text{a}}}{\nu }=\frac{\text{34}0}{\text{1}{0}^{\text{6}}}=3.4\times {10}^{-4}m\\ \left(\text{b}\right)\text{Here,}\text{frequency of ultrasonic sound,}\nu =\text{1}000\text{kHz}\\ =\text{1}{0}^{\text{6}}\text{Hz}\\ \text{Speed of sound in water,}{\text{v}}_{\text{w}}\text{=}{\text{1486 ms}}^{\text{-1}}\\ \text{Wavelength of the transmitted sound is given as:}\\ {\lambda }_{\text{r}}=\frac{{\text{v}}_w}{\nu }=\frac{\text{1486}}{\text{1}{0}^{\text{6}}}\\ =\text{1}.\text{49}\times \text{1}{0}^{–\text{3}}\text{m}\end{array}

Q.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.

\begin{array}{l}\text{Here,}\text{speed of sound in the tissue,}\text{v}=\text{1}.{\text{7 kms}}^{\text{-1}}\\ =\text{1}.\text{7}\times \text{1}{0}^{\text{3}}{\text{ms}}^{\text{-1}}\\ \text{Operating frequency of scanner,}\nu =\text{4}.\text{2 MHz}\\ =\text{4}.\text{2}\times \text{1}{0}^{\text{6}}\text{Hz}\\ \text{Wavelength of sound in the tissue is given by}\text{the}\text{relation:}\\ \lambda =\frac{v}{\nu }=\frac{\text{1}.\text{7}\times \text{1}{0}^{\text{3}}}{\text{4}.\text{2}\times \text{1}{0}^{\text{6}}}=4.1\times {10}^{-4}\text{m}\end{array}

Q.8

\begin{array}{l}\text{A transverse harmonic wave on a string is described by}\text{y}\left(\text{x,t}\right)=\text{3}\text{.0}\text{sin}\left(\text{36t}\text{+}\text{0}\text{.018}\text{x}+\frac{\pi }{4}\right)\\ \text{Where}\text{x}\text{and}\text{y}\text{are in cm andtin s}\text{. The positive}\text{direction ofxis from left to right}\text{.}\\ \left(\text{a}\right)\text{Is this a travelling wave or a stationary wave?}\\ \text{If it is travelling, what are the speed and direction of its propagation?}\\ \left(\text{b}\right)\text{What are its amplitude and frequency?}\\ \left(\text{c}\right)\text{What is the initial phase at the origin?}\\ \left(\text{d}\right)\text{What is the least distance between two successive crests in the wave?}\end{array}

Ans.

\begin{array}{l}\left(\text{a}\right)\text{The equation of a progressive wave travelling}\text{from right to left is given as:}\\ \text{y}\left(\text{x},\text{t}\right)=\text{asin}\left(\omega \text{t}+\text{kx}+\varphi \right)\to \left(\text{i}\right)\\ \text{The given equation is:}\\ \text{y}\left(\text{x},\text{t}\right)=3.0\text{sin}\left(\text{36t}+0.018\text{x}+\frac{\pi }{4}\right)\to \left(\text{ii}\right)\\ \text{On comparing equation}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we observe that equation}\left(\text{ii}\right)\text{is a travelling}\text{wave, propagating from}\\ \text{right to left}\text{.}\\ \text{From equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we have:}\\ \omega ={\text{36 rads}}^{\text{-1}}\\ \text{and}\text{k}=0.0{\text{18 m}}^{–\text{1}}\\ \text{Since}\text{frequency,}\nu =\frac{\omega }{2\pi }\text{and}\text{wavelength,}\lambda =\frac{2\pi }{k}\\ \text{Since},\text{v}=\nu \lambda \\ \therefore v=\left(\frac{\omega }{2\pi }\right)\times \left(\frac{2\pi }{k}\right)=\frac{\omega }{k}=\frac{36}{0.0\text{18}}=2000cm{s}^{-1}=20m{s}^{-1}\\ \therefore \text{Speed of the given travelling wave}\text{=}\text{2}0{\text{ms}}^{\text{-1}}\\ \left(\text{b}\right)\text{Here,}\text{amplitude of the given wave,}\text{a}=\text{3 cm}\\ \text{Frequency of the given wave}\text{is}\text{given}\text{as:}\\ \nu =\frac{\omega }{2\pi }=\frac{36}{2\times 3.14}=5.73Hz\\ \left(\text{c}\right)\text{From equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we observe that the}\text{initial phase angle},\varphi =\frac{\pi }{4}\\ \left(\text{d}\right)\text{Least}\text{distance between two successive crests}\text{or troughs}\text{of}\text{a}\text{wave =}\text{Wavelength of the wave}\\ \text{Wavelength is given as:}\\ \lambda =\frac{2\pi }{k}\\ \therefore \lambda =\frac{2\times 3.14}{0.018}=348.89\text{cm}=3.49\text{m}\end{array}

Q.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.

${\displaystyle \begin{array}{l}\mathrm{H}\mathrm{e}\mathrm{r}\mathrm{e},\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{e}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{c}\mathrm{w}\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{i}\mathrm{s}:\\ \mathrm{y}\left(\mathrm{x},\mathrm{t}\right)=3.0\mathrm{s}\mathrm{i}\mathrm{n}\left(36\mathrm{t}+0.018\mathrm{x}+\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}\right)\to <!--\; \to \; -->(\mathrm{i})\\ \mathrm{F}\mathrm{o}\mathrm{r}\mathrm{x}=0,\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{b}\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{o}:\\ \mathrm{y}\left(0,\mathrm{t}\right)=3.0\mathrm{s}\mathrm{i}\mathrm{n}\left(36\mathrm{t}+\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}\right)\\ \mathrm{A}\mathrm{s},\mathrm{\omega <!--\; \omega \; -->}=\frac{2\mathrm{\pi <!--\; \pi \; -->}}{\mathrm{T}}=36\mathrm{r}\mathrm{a}\mathrm{d}{\mathrm{s}}^{-<!--\; -\; -->1}\\ \therefore <!--\; \therefore \; -->\mathrm{T}=\frac{\mathrm{\pi <!--\; \pi \; -->}}{8}\mathrm{s}\\ \mathrm{P}\mathrm{l}\mathrm{o}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{y}\mathrm{v}\mathrm{s}.\mathrm{t}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}\mathrm{s}\mathrm{u}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{t},\\ \mathrm{a}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{b}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{a}\mathrm{b}\mathrm{l}\mathrm{e}.\end{array}}$

${\displaystyle \begin{array}{l}\mathrm{F}\mathrm{o}\mathrm{r}\mathrm{x}=0,\mathrm{x}=2,\mathrm{a}\mathrm{n}\mathrm{d}\text{x = 4, the phases of the three}\\ \text{waves will}\mathrm{g}\mathrm{e}\mathrm{t}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{d},\text{because amplitude and}\\ \text{frequency are invariant for any}\text{change in}\mathrm{x}.\\ \text{The}\mathrm{y}-<!--\; -\; -->\mathrm{t}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{p}\mathrm{h}\mathrm{s}\mathrm{o}\mathrm{f}\text{the three waves are shown in the}\\ \text{given figure}.\end{array}}$

Q.10 For the travelling harmonic wave
y (x, t) = 2.0 cos 2p (10t – 0.0080x + 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,

\begin{array}{}\end{array} \begin{array}{l}\end{array} \begin{array}{l}\text{(c)}\frac{\lambda }{2},\\ \text{(d)}\frac{3\lambda }{4}.\end{array}

Ans.

\begin{array}{l}\text{The}\text{equation for a travelling harmonic wave is given as:}\\ \text{y}\left(\text{x},\text{t}\right)=\text{2}.0\text{cos 2}\pi \left(\text{1}0\text{t}–0.00\text{8}0\text{x}+0.\text{35}\right)\\ =\text{2}.0\text{cos}\left(\text{2}0\pi \text{t}–0.0\text{16}\pi \text{x}+0.\text{7}0\pi \right)\\ \text{Here,}\text{propagation constant,}\text{k}=0.0\text{16}0\pi \\ \text{Amplitude}\text{of}\text{wave,}\text{a}\text{=}\text{2 cm}\\ \text{Angular frequency}\text{of}\text{wave,}\omega =\text{2}0\pi {\text{rads}}^{\text{-1}}\\ \text{The}\text{phase difference is given as:}\\ \varphi =kx=\frac{2\pi }{\lambda }\\ \left(\text{a}\right)\text{For}\text{distance,}\text{x}\text{=}\text{4 m}\text{=}\text{400 cm}\\ \text{Phase}\text{difference,}\varphi =0.0\text{16}\pi \times \text{4}00=\text{6}.\text{4}\pi \text{rad}\\ \left(\text{b}\right)\text{For distance,}\text{x}\text{=}0.\text{5 m}=\text{5}0\text{cm}\\ \text{Phase}\text{difference},\varphi =0.0\text{16}\pi \times \text{5}0=0.\text{8}\pi \text{rad}\\ \left(\text{c}\right)\text{For}\text{distance,}\text{x}\text{=}\frac{\lambda }{2}\\ \text{Phase}\text{difference},\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{2}=\pi \text{rad}\\ \left(\text{d}\right)\text{For}\text{x}\text{=}\frac{3\lambda }{4}\\ \text{Phase}\text{difference},\varphi =\frac{2\pi }{\lambda }\times \frac{3\lambda }{4}=1.5\pi \text{rad}\end{array}

Q.11 The transverse displacement of a string (clamped at its both ends) is given by

y(x,t)=0.06sin\frac{2\pi }{3}xcos(120\pi 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 are the wavelength, frequency, and speed of each wave?
(c) Determine the tension in the string.

Ans.

\begin{array}{l}\left(\text{a}\right)\text{The general equation representing a stationary wave is given as:}\\ \text{y}\left(\text{x},\text{t}\right)=\text{2a}\text{sin}\text{kx}\text{cos}\omega \text{t}\to \text{(i)}\\ \text{It}\text{is}\text{given}\text{that:}\\ \text{y}\left(\text{x},\text{t}\right)=\text{0}\text{.06}\text{sin}\left(\frac{2}{3}x\right)\text{cos}\left(120\pi t\right)\to \text{(ii)}\\ \text{Comparing}\text{equation}\text{(i)}\text{and}\text{(ii)}\text{,}\text{we}\text{observe}\text{that, the}\text{given function represents a stationary wave}\text{.}\\ \left(\text{b}\right)\text{A wave propagating along the positive}\text{direction}\text{of}\text{x}-\text{axis}\text{is}\text{given as:}\\ y_1=a\sin \left(\omega t-kx\right)\\ \text{The wave propagating along the negative}\text{direction of}\text{x}-\text{axis}\text{is}\text{given as:}\\ y_2=a\sin \left(\omega t+kx\right)\\ \text{Superposition of these two waves gives:}\\ y=y_1+y_2=a\sin \left(\omega t-kx\right)-a\sin \left(\omega t+kx\right)\\ =a\sin \left(\omega t\right)\cos \left(kx\right)-a\sin \left(kx\right)\cos \left(\omega t\right)-a\sin \left(\omega t\right)\cos \left(kx\right)\\ -a\sin \left(kx\right)\cos \left(\omega t\right)\\ =-2a\sin \left(kx\right)\cos \left(\omega t\right)\\ =-2a\sin \left(\frac{2\pi }{\lambda }x\right)\cos \left(2\pi \nu t\right)\to \text{(iii)}\\ \text{Transverse displacement of the string is given by}\text{the}\text{relation:}\\ \text{y}\left(\text{x},\text{t}\right)=\text{0}\text{.06}\text{sin}\left(\frac{2}{3}x\right)\text{cos}\left(120\pi t\right)\to \text{(ii)}\\ \text{Comparing equations}\left(\text{iii}\right)\text{and}\text{(ii),}\text{we}\text{obtain:}\\ \frac{2\pi }{\lambda }=\frac{2\pi }{3}\\ \therefore \text{Wavelength,}\lambda =3m\\ \text{Given}\text{that,}\\ 120\pi =2\pi \nu \\ \therefore \text{Frequency,}\nu =60Hz\\ \text{Wave}\text{speed}\text{is}\text{given}\text{as}:\\ v=\nu \lambda \\ \therefore v=60\times 3=180{\text{ms}}^{-1}\\ \left(\text{c}\right)\text{Velocity of a transverse wave travelling in a string is}\text{given as:}\\ v=\sqrt{\frac{T}{\mu }}\to \text{(i)}\\ \text{Here,}\\ \text{Velocity of transverse wave,}\text{v}=\text{18}0{\text{ms}}^{\text{-1}}\\ \text{Mass of string,}\text{m}=\text{3}.0\times \text{1}{0}^{–\text{2}}\text{kg}\\ \text{Length of string},\text{l}=\text{1}.\text{5 m}\\ \text{Mass per unit length of the string}\text{is}\text{given}\text{as:}\mu =\frac{m}{l}\\ =\frac{\text{3}.0\times \text{1}{0}^{–\text{2}}}{\text{1}.\text{5}}=2.0\times \text{1}{0}^{–\text{2}}{\text{kgm}}^{-1}\\ \text{Let}\text{tension in the string}\text{=}\text{T}\\ \text{Using}\text{equation}\left(\text{i}\right)\text{, tension}\text{can}\text{be}\text{obtained as:}\\ \text{T}={\text{v}}^{\text{2}}\mu ={\left(\text{18}0\right)}^{\text{2}}\times \text{2}\times \text{1}{0}^{–\text{2}}=\text{648 N}\end{array}

Q.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.

\begin{array}{l}\left(\text{i}\right)\left(\text{a}\right)\text{All the points on the string vibrate with the same}\text{frequency, except at the nodes where frequency is zero}\text{.}\\ \left(\text{b}\right)\text{All the points on the oscillating string have the same phase, except at the nodes}\text{.}\\ \left(\text{c}\right)\text{All the points on the oscillating string have different amplitudes of vibration}\text{.}\\ \text{(ii)}\text{It}\text{is}\text{given}\text{that}\\ \text{y}\left(\text{x},\text{t}\right)=\text{0}\text{.06}\text{sin}\left(\frac{2\pi }{3}x\right)\text{cos}\left(120\pi t\right)\\ Atx=0.375m\text{and}t=0,\text{the}\text{amplitude}\text{is}\text{given}\text{as:}\\ \text{Amplitude}=\text{0}\text{.06}\text{sin}\left(\frac{2\pi }{3}x\right)\text{cos}\left(120\pi \times 0\right)\\ =\text{0}\text{.06}\text{sin}\left(\frac{2\pi }{3}\times 0.375\right)\text{cos0}\\ \text{=0}\text{.06}\text{sin}\left(\frac{2\pi }{3}\times 0.375\right)\times 1\\ \text{=0}\text{.06}\text{sin}\left(0.25\pi \right)=\text{0}\text{.06}\text{sin}\left(\frac{\pi }{4}\right)\\ =\text{0}\text{.06}\times \frac{1}{\sqrt{2}}=0.042m\end{array}

Q.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 traveling wave, (ii) a stationary wave or (iii) none at all:
(a) y = 2 cos (3x) sin (10t)

\text{(b)}y=2\sqrt{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) The given relation represents a stationary wave. This is due to the fact that the harmonic terms kx and ωt appear separately in the equation.
(b) Since the given equation does not contain any harmonic term, therefore, it cannot represent travelling or stationary wave.
(c) The given equation represents a travelling wave. This is because the harmonic terms kx and ωt are in the form of of kx – ωt in the equation.
(d) The given equation represents the sum of two functions each of which represents a stationary wave. This is because the harmonic terms kx and ωt appear separately in the equation. Therefore, the given equation represents the superposition of two stationary waves.

Q.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.

\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{mass of wire,}\text{m}=\text{3}.\text{5}\times \text{1}{0}^{–\text{2}}\text{kg}\\ \text{Linear mass density}\text{is}\text{given}\text{as:}\\ \mu =\frac{m}{l}=4.0\times \text{1}{0}^{–\text{2}}{\text{kgm}}^{\text{-1}}\\ \text{Frequency,}\nu =\text{45 Hz}\\ \therefore \text{Length of the wire}\text{is}\text{given}\text{as:}\\ l=\frac{m}{\mu }=\frac{\text{3}.\text{5}\times \text{1}{0}^{–\text{2}}}{4.0\times \text{1}{0}^{–\text{2}}}=0.875\text{m}\\ \text{Wavelength of the stationary wave}\left(\text{λ}\right)\text{is related}\text{to the length of the wire as:}\\ \lambda =\frac{2l}{n}\\ \text{Here,}\text{n}\text{=}\text{Number}\text{of}\text{nodes}\text{in}\text{wire}\\ \text{In}\text{case}\text{of the}\text{fundamental node,}\text{n}=\text{1}\\ \therefore \lambda =\text{2l}\\ \lambda =\text{2}\times 0.\text{875}=\text{1}.\text{75 m}\\ \text{Speed of the transverse wave in the string is given by}\text{the}\text{relation:}\\ \text{v}=\nu \lambda =\text{45}\times \text{1}.\text{75}=\text{78}.{\text{75 ms}}^{\text{-1}}\\ \left(\text{b}\right)\text{The}\text{tension developed in the string is given as:}\\ \text{T}={\text{v}}^{\text{2}}\mu ={\left(\text{78}.\text{75}\right)}^{\text{2}}\times \text{4}.0\times \text{1}{0}^{–\text{2}}=\text{248}.0\text{6 N}\end{array}

Q.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.

\begin{array}{l}\text{Here,}\text{frequency of turning fork},\nu =\text{34}0\text{Hz}\\ \text{As the given tube is joined with a piston at one end, it}\text{will behave as a pipe}\text{with one end closed and the other}\\ \text{end open}\text{.}\text{This}\text{type}\text{of}\text{systems}\text{produce}\text{odd}\text{harmonics}\text{only}\text{.}\\ \text{Here,}\text{length}\text{of}\text{pipe,}{l}_1=25.5cm=0.255m\\ \text{The}\text{fundamental note in a closed pipe is}\text{given as:}\\ l_1=\frac{\lambda }{4}\\ \therefore \lambda =4l_1=4\times 0.255=1.02m\\ \text{The speed of sound in}\text{air}\text{is given as:}\\ v=\nu \lambda =\text{34}0\times \text{1}.0\text{2}=\text{346}.{\text{8 ms}}^{\text{-1}}\end{array}

Q.16 A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel?

Ans.

\begin{array}{l}\text{Here,}\text{distance between two successive nodes}\text{=}\frac{\lambda }{2}\\ \therefore l=\frac{\lambda }{2}\\ \therefore \lambda =2l=2\times 1=2m\\ \text{The speed of sound in steel is given as:}\\ \text{v}=\nu \lambda \\ =\text{2}.\text{53}\times \text{1}{0}^{\text{3}}\times \text{2}\\ =\text{5}.0\text{6}\times \text{1}{0}^{\text{3}}{\text{ms}}^{\text{-1}}=\text{5}.0{\text{6 kms}}^{\text{-1}}\end{array}

Q.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.

\begin{array}{l}\text{Here,}\text{length of pipe,}\text{l}\text{=}\text{20 cm = 0}\text{.2 m}\\ \text{Source frequency}\text{=}\text{Frequency}\text{of}{\text{n}}^{\text{th}}\text{normal}\\ \text{mode,}{\nu }_{\text{n}}=\text{43}0\text{Hz}\\ \text{Speed of sound}\text{in}\text{air,}\text{v}\text{=}{\text{340 ms}}^{\text{-1}}\\ \text{Frequency}\text{of}{\text{n}}^{\text{th}}\text{normal mode}\text{in a closed pipe is given by the relation:}\\ {\nu }_{\text{n}}=\left(2n-1\right)\frac{v}{4l}\\ \text{Here,}\text{n}\text{is}\text{an}\text{integer}\text{.}\\ \therefore \text{43}0=\left(2n-1\right)\frac{340}{4\times 0.2}\\ 2n-1=\frac{\text{43}0\times 4\times 0.2}{340}=1.01\\ 2n=2.01\\ \therefore n\sim 1\\ \therefore \text{The frequency}\text{of}\text{first mode of vibration is}\text{resonantly}\text{excited by the given}\text{source}\text{.}\\ \text{In case}\text{of}\text{a pipe open at both ends, the}\text{frequency}\text{of}{\text{n}}^{\text{th}}\text{mode of vibration is given by the}\text{relation:}\\ {\nu }_n=\frac{nv}{2l}\\ \therefore n=\frac{2l{\nu }_n}{v}=\frac{2\times 0.2\times 430}{340}=0.5\\ \text{As the number of the mode of vibration}\left(\text{n}\right)\text{should be}\text{an integer, the open pipe}\text{cannot}\text{be}\text{in}\text{resonance}\\ \text{with}\text{the}\text{source}\text{.}\end{array}

Q.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.

\begin{array}{l}\text{Here,}\text{frequency of string A,}{\text{f}}_{\text{A}}=\text{324 Hz}\\ \text{Let}\text{frequency of string B}\text{=}{\text{f}}_{\text{B}}\\ \text{Beat}\text{frequency,}\text{n}\text{=}\text{6 Hz}\\ \text{Beat}\text{frequency}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ \text{n}\text{=}\left|{\text{f}}_{\text{A}}\pm {\text{f}}_{\text{B}}\right|\\ \therefore 6=\text{324}\pm {\text{f}}_{\text{B}}\\ \therefore {\text{f}}_{\text{B}}=330\text{Hz}\text{or}\text{318}\text{Hz}\\ \text{When tension in the string}\text{decreases}\text{its}\text{frequency}\text{decreases,}\text{because}\text{frequency is directly proportional to}\\ \text{the square root of tension}\left(\nu \propto \sqrt{T}\right).\\ \therefore \text{Beat}\text{frequency}\text{cannot}\text{be}\text{330}\text{Hz}\text{.}\\ \therefore \text{Beat}\text{frequency,}{\text{f}}_{\text{B}}\text{=}\text{318}\text{Hz}\end{array}

Q.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) Node is a point where the amplitude of oscillation is the minimum and pressure is the maximum. On the contrary, antinode is a point where the amplitude of oscillation is the maximum and pressure is the minimum.
Therefore, in a sound wave, a displacement node is a pressure antinode and vice versa.
(b) Bats emit ultrasonic sound waves of very high-frequency. When these waves are reflected by obstacles in their path, the bat receives a reflected wave and estimates the nature, direction, distance and size of an obstacle.
(c) Although a violin note and a sitar note have the same frequency, but the overtones and their strengthsare different. Thus, we can differentiate the notes produced by a sitar and a violin even if they have the same frequency of vibration.
(d) Solids have both elasticity of shape and elasticity of volume (shear modulus) . Longitudinal waves require the elasticity of shape in the medium for their propagation. Transverse waves propagate in the medium with elasticity of volume. Thus, the solids can support both longitudinal and transverse wave whereas, gases have only the volume elasticity, therefore, the transverse waves cannot propagate through gases.
(e) A sound pulse is a combination of waves having different wavelengths. The shape of wave pulse gets distorted due to propogation of waves in a dispersive medium with different velocities.

Q.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 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.

Ans.

\begin{array}{l}\text{Here,}\text{Frequency of the whistle,}\nu =\text{4}00\text{Hz}\\ \text{Speed of sound}\text{in}\text{still}\text{air,}\text{v}=\text{34}0{\text{ms}}^{\text{-1}}\\ \left(\text{i}\right)\left(\text{a}\right)\text{Here,}\text{speed of train},{\text{v}}_{\text{T}}=\text{1}0{\text{ms}}^{\text{-1}}\\ \text{The apparent frequency of the whistle as the train}\text{approaches the platform is}\text{given as:}\\ \nu ‘=\left(\frac{v}{v-{\text{v}}_{\text{T}}}\right)\nu =\left(\frac{340}{340-10}\right)\times 400=412.12\text{Hz}\\ \left(\text{b}\right)\text{Here,}\text{speed of train,}{\text{v}}_{\text{T}}\text{=}{\text{10 ms}}^{\text{-1}}\\ \text{The apparent frequency of the whistle as the train}\text{recedes from the platform}\text{is given as:}\\ \nu ‘=\left(\frac{v}{v+{\text{v}}_{\text{T}}}\right)\nu =\left(\frac{340}{340+10}\right)\times 400=388.6\text{Hz}\\ \left(\text{ii}\right)\text{The apparent change in the frequency of sound}\text{is due}\text{to the relative motions}\text{of the source and}\\ \text{the observer}\text{. These relative motions do}\text{not}\text{affect}\text{the}\text{speed of sound}\text{.}\\ \therefore \text{The speed of sound in each case remains the}\text{same, i}\text{.e}{\text{., 340 ms}}^{\text{-1}}\text{.}\end{array}

Q.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 at 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 ms^–1? The speed of sound in still air can be taken as 340 m s^–1.

Ans.

\begin{array}{l}\text{Here,}\text{Frequency of the sound produced by the whistle},\nu =\text{4}00\text{Hz}\\ \text{Velocity of wind,}{\text{v}}_{\text{w}}=\text{1}0{\text{ms}}^{\text{-1}}\\ \text{Speed of sound in}\text{still}\text{air,}\text{v}\text{=}{\text{340 ms}}^{\text{-1}}\\ \text{For the stationary observer:}\\ \text{As}\text{the wind is blowing toward the observer,}\\ \therefore \text{Effective speed of the sound,}{\text{v}}_{\text{e}}=v+{\text{v}}_{\text{w}}\text{=}\text{34}0+\text{1}0=\text{35}0{\text{ms}}^{\text{-1}}\\ \text{Since there is no relative motion between the source and}\text{the}\text{observer, the frequency of}\text{the sound heard by the}\\ \text{observer will remain}\text{unchanged,}\text{i}\text{.e}\text{., 400 Hz}\text{.}\\ \text{Wavelength}\left(\lambda \right)\text{of the sound heard by the observer is}\text{given as:}\\ \lambda =\frac{v+{\text{v}}_{\text{w}}}{\nu }=\frac{v+{\text{v}}_{\text{w}}}{\nu }=\frac{350}{400}=0.875\text{m}\\ \text{For the running observer:}\\ \text{Here,}\text{velocity of observer,}{\text{v}}_{\text{o}}=\text{1}0{\text{ms}}^{\text{-1}}\\ \text{Here,}\text{the observer is travelling toward the source,}\text{Due}\text{to}\text{the}\text{relative motions of the}\text{source and the}\\ \text{observer,}\text{change in frequency}\text{takes}\text{place}\text{.}\\ \text{Change in frequency}\left(\nu \text{‘}\right)\text{is given by the relation:}\\ \nu ‘=\left(\frac{v+v_o}{v}\right)\nu =\left(\frac{340+10}{340}\right)\times 400=411.76Hz\\ \text{As the air is still,}\\ \therefore \text{Effective speed of sound}=v+v_w\text{=34}0+0=\text{34}0{\text{ms}}^{\text{-1}}\\ \text{As}\text{the source is at rest,}\text{the wavelength of the sound}\left(\lambda \right)\text{remains 0}\text{.875 m}\text{.}\\ \therefore \text{Above two situations are not exactly}\text{same}\text{.}\end{array}

Q.22

\begin{array}{l}\text{A travelling harmonic wave on a string is described}\\ \text{by}\text{y}\left(\text{x,}\text{t}\right)=\text{7}\text{.5}\text{sin}\left(\text{0}\text{.0050x}\text{+}\text{12t}+\frac{\pi }{\text{4}}\right)\\ \left(\text{a}\right)\text{What are the displacement and velocity of}\text{oscillation of a point at}\text{x}\text{=}\text{1 cm, and}\text{t}\text{=}\text{1 s? Is this}\\ \text{velocity equal to the velocity of wave propagation?}\\ \left(\text{b}\right)\text{Locate the points of the string which have the}\text{same transverse displacements and velocity as the}\text{x}\\ \text{=}\text{1 cm point att =2 s, 5 s and 11 s}\text{.}\end{array}

Ans.

\begin{array}{l}\left(\text{a}\right)\text{The given travelling}\text{harmonic wave is:}\\ y\left(x,t\right)=7.5\text{sin}\left(0.0050x+12t+\frac{\pi }{4}\right)\\ At\text{x}=\text{1 cm and}\text{t}=\text{1s},\\ y\left(1,1\right)=7.5\sin \left(0.0050+12+\frac{\pi }{4}\right)\\ =7.5\sin \left(12.0050+\frac{\pi }{4}\right)=7.5\sin \theta \\ \text{Here},\theta =12.0050+\frac{3.14}{4}=12.79\text{rad}\\ =\frac{180}{3.14}\times 12.79={732.81}^o\\ \therefore y\left(1,1\right)=7.5\sin \left({732.81}^o\right)=7.5\sin \left(90\times 8+{12.81}^o\right)\\ =7.5\sin \left({12.81}^o\right)=7.5\times 0.2217=1.6629\approx 1.663\text{cm}\\ \text{Velocity of the vibration}\text{at}\text{a}\text{given}\text{point}\text{and}\text{time}\text{is}\text{given}\text{as:}\\ \text{v=}\frac{d}{dt}y\left(x,t\right)=\frac{d}{dt}\left[7.5\sin \left(0.0050x+12t+\frac{\pi }{4}\right)\right]\\ =7.5\times 12\cos \left(0.0050x+12t+\frac{\pi }{4}\right)\\ =90\cos \left(0.0050x+12t+\frac{\pi }{4}\right)\\ \text{For}x=1\text{cm}andt=1s\\ \text{Velocity},v=y\left(1,1\right)=90\cos \left(0.005+12+\frac{\pi }{4}\right)\\ =90\cos \left(12.005+\frac{\pi }{4}\right)=90\cos \left({732.81}^o\right)\\ =90\cos \left({720}^o+{12.81}^o\right)=90\cos \left({12.81}^o\right)\\ =90\times 0.975=87.75{\text{cms}}^{-1}\\ \text{The standard}\text{equation of a propagating wave is given as:}\\ y\left(x,t\right)=a\sin \left(kx+\omega t+\varphi \right)\\ \text{Here},k=\frac{2\pi }{\lambda }\\ \therefore \lambda =\frac{2\pi }{k}\\ \text{Also},\omega =2\pi \nu \\ \therefore \nu =\frac{\omega }{2\pi }\\ \text{Velocity}\text{of}\text{wave}\text{propagation,}v=\nu \lambda =\frac{\omega }{k}\\ \text{Here},\omega =12{\text{rads}}^{-1}\\ k=0.0050{\text{m}}^{-1}\\ \therefore v=\frac{12}{0.0050}=2400{\text{cms}}^{-1}\\ \therefore \text{We}\text{observe}\text{that,}\text{the}\text{velocity of wave oscillation at}\text{x}\text{=}\text{1 cm and}\text{t}\text{=}\text{1 s is not equal to velocity of the wave}\text{propagation}\text{.}\\ \left(\text{b}\right)\text{The}\text{relation}\text{connecting}\text{propagation constant to}\text{wavelength is}\text{given}\text{as:}\\ k=\frac{2\pi }{\lambda }\\ \therefore \lambda =\frac{2\pi }{k}=\frac{2\times 3.14}{0.0050}=1256\text{cm}=12.56\text{m}\\ \therefore \text{All the points at distances}\text{n}\lambda \left(n=\pm 1,\pm 2,\mathrm{\dots .}\right),\\ \text{i}.\text{e}.\pm 12.56m,\pm 25.12m,\mathrm{\dots .}\text{and}\text{so}\text{on}\text{for}x=1\text{cm,}\text{will}\text{be}\text{having}\text{the}\text{same}\text{displacements}\text{as}\text{the}\text{x}\text{=}\text{1}\text{cm}\text{points}\text{at}\\ t=2s,5sand11s.\end{array}

Q.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) A narrow sound pulse has neither a definite wavelength nor a definite frequency. However, its speed remains the same, which is equal to the speed of sound in that medium.
(b) If the pulse rate is 1 after every 20 seconds, it does not mean that the frequency of note produced by the whistle is 0.05 Hz. Rather, it implies that, 0.05 Hz is the frequency of the repetition of the short pip of the whistle.

Q.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.

\begin{array}{l}\text{The equation of a travelling wave propagating along the positive}\text{y-direction is given as:}\\ \text{y}\left(\text{x},\text{t}\right)=\text{asin}\left(\text{wt}–\text{kx}\right)\to \left(\text{i}\right)\\ \text{Here,}\text{linear mass density},\mu =8.0\times {10}^{-3}{\text{kgm}}^{-1}\\ \text{Frequency of tuning fork},\nu =\text{256 Hz}\\ \text{Amplitude of wave},\text{a}=\text{5}.0\text{cm}=0.0\text{5 m}\to \left(\text{ii}\right)\\ \text{Mass of pan},\text{m}=\text{9}0\text{kg}\\ \text{Tension produced}\text{in the string}\text{is}\text{given}\text{as:}\\ \text{T}=\text{mg}=\text{9}0\times \text{9}.\text{8}=\text{882 N}\\ \text{Velocity of the transverse wave}\left(\text{v}\right)\text{, is given as:}\\ v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{882}{8.0\times {10}^{-3}}}=332{\text{ms}}^{-1}\\ \text{Angular}\text{frequency}\text{is}\text{given}\text{as:}\\ \omega =2\pi \nu =2\times 3.14\times 256=1608.5=1.6\times {10}^{3}rad{s}^{-1}\to \text{(iii)}\\ \text{wavelength}\text{is}\text{given}\text{as:}\\ \lambda =\frac{v}{\nu }=\frac{332}{256}m\\ \therefore \text{Propagation}\text{constant}\text{is}\text{given}\text{as:}\\ k=\frac{2\pi }{\lambda }=\frac{2\times 3.14}{\frac{332}{256}}=4.84m^{-1}\to \text{(iv)}\\ \text{Substituting the values from equations}\left(\text{ii}\right)\text{,}\left(\text{iii}\right)\text{, and}\left(\text{iv}\right)\text{in equation}\left(\text{i}\right)\text{, we get:}\\ \text{y}\left(\text{x},\text{t}\right)=0.0\text{5 sin}\left(\text{1}.\text{6}\times \text{1}{0}^{\text{3}}\text{t}–\text{4}.\text{84x}\right)\text{m}\end{array}

Q.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.

\begin{array}{l}\text{Here,}\text{operating frequency of the SONAR},\nu =\text{4}0\text{kHz}\\ \text{Speed of the}\text{enemy submarine,}{\text{v}}_{\text{e}}=\text{36}0{\text{kmh}}^{\text{-1}}\\ =\text{1}00{\text{ms}}^{\text{-1}}\\ \text{Speed of sound in the}\text{water,}\text{v}\text{=}{\text{1450 ms}}^{\text{-1}}\\ \text{As}\text{the source is at rest and the observer}\text{(enemy}\text{submarine) is moving toward it,}\\ \therefore \text{Apparent frequency}\left(\nu \text{‘}\right)\text{received and reflected by}\text{the submarine is given}\text{as:}\\ \nu ‘=\left(\frac{v+{\text{v}}_{\text{e}}}{v}\right)\nu =\left(\frac{1450+100}{1450}\right)\times 40=42.76kHz\\ \text{The frequency}\left(\nu ‘‘\right)\text{received by the enemy submarine}\text{is given as:}\\ \nu ‘‘=\left(\frac{v}{v+{\text{v}}_s}\right)\nu ‘\\ \text{Here,}{\text{v}}_{\text{s}}=100{\text{ms}}^{-1}\\ \therefore \nu ‘‘=\left(\frac{1450}{1450+100}\right)\times 42.76=45.93\text{kHz}\end{array}

Q.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.

\begin{array}{l}\text{Let}{\text{v}}_{\text{S}}\text{and}{\text{v}}_{\text{P}}\text{be the velocities of}\text{S}\text{waves}\text{and}\text{P}\text{waves respectively}\text{.}\\ \text{Let distance between the epicentre and seismograph}\text{=}\text{L}\\ \text{Let}{\text{t}}_{\text{1}}\text{and}{\text{t}}_{\text{2}}\text{be the respective times taken by the}\text{S}\\ \text{waves}\text{and}\text{P}\text{waves to travel}\text{to the}\text{position}\text{of}\text{seismograph}\text{.}\\ \text{Then,}\text{we}\text{have:}\\ \text{L}={\text{v}}_{\text{S}}{\text{t}}_1\to \left(\text{i}\right)\\ \text{L}={\text{v}}_{\text{P}}{\text{t}}_2\to \left(\text{ii}\right)\\ \text{Given,}{\text{v}}_{\text{P}}={\text{8 kms}}^{\text{-1}}\\ {\text{v}}_{\text{S}}={\text{4 kms}}^{\text{-1}}\\ \text{From equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we obtain:}\\ {\text{v}}_{\text{S}}{\text{t}}_1={\text{v}}_{\text{P}}{\text{t}}_2\\ {\text{4t}}_1={\text{8t}}_2\\ \therefore {\text{t}}_1={\text{2t}}_2\to \left(\text{iii}\right)\\ \text{Given}\text{that,}\\ {\text{t}}_1–{\text{t}}_2=\text{4 min}=\text{24}0\text{s}\\ {\text{2t}}_2–{\text{t}}_2=\text{24}0\\ \therefore {\text{t}}_2=\text{24}0\to \text{(iv)}\\ \text{From}\text{equations}\text{(iii)}\text{and}\text{(iv),}\text{we}\text{have:}\\ {\text{t}}_1=\text{2}\times \text{24}0=\text{48}0\text{s}\\ \text{From equation}\left(\text{ii}\right)\text{, we obtain:}\\ \text{L}=\text{8}\times \text{24}0=\text{192}0\text{km}\\ \therefore \text{The earthquake occurs at a distance of 1920 km}\text{from}\text{the seismograph}\text{.}\end{array}

Q.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.

\begin{array}{l}\text{Here, frequency of the sound emitted by the bat},\nu =\text{4}0\text{kHz}\\ {\text{Velocity of bat,v}}_{\text{b}}=0.0\text{3v}\\ \text{Here,}\text{v}\text{=}\text{Velocity of sound in air}\\ \text{The apparent frequency of the sound striking the wall is}\text{given by}\text{the}\text{relation:}\\ \nu ‘=\left(\frac{v}{v-v_b}\right)\nu =\left(\frac{v}{v-0.03v}\right)\times 40=\frac{40}{0.97}\text{kHz}\\ \text{This frequency is reflected by the stationary}\text{wall and}\text{is}\text{received}\text{by}\text{the}\text{bat}\text{moving}\text{towards}\text{the}\text{wall}\text{.}\\ \therefore {\text{v}}_{\text{s}}=0\\ \text{The frequency}\left(\nu ‘‘\right)\text{of the sound received}\text{is given as:}\\ \nu ‘‘=\left(\frac{v+v_b}{v}\right)\nu ‘=\left(\frac{v+0.03v}{v}\right)\times \frac{40}{0.97}=\frac{1.03\times 40}{0.97}=42.47kHz\end{array}