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Right from the sounds we listen to, to the waves in the ocean to even the light we see, waves surround us. The physics involved in such a pattern and the way the universe is structured has been described in this fascinating chapter of Class 11 Physics.
NCERT solution for Class 11 Waves Chapter is very important for both Class 11 board exams and competitive exams like NEET and JEE Main. The NCERT Solutions for Waves Class 11 is like a bible to help students grasp and understand the key concepts covered in the chapter.
The PDF version of Waves NCERT Solutions by Careers360 helps students understand the real-world applications and important concepts of waves for thorough exam preparation.
Free download NCERT solution for Class 11 Physics Waves NCERT solutions PDF for CBSE exam.
Answer:
Mass per unit length of the string is
$\\\mu =\frac{M}{l}\\ =\frac{2.50}{20}\\ =0.125\ kg\ m^{-1}$
The velocity of the transverse wave in the string will be
$\\v=\sqrt{\frac{T}{\mu }}\\ =\sqrt{\frac{200}{0.125}}\\ =\sqrt{1600} \\=40\ m\ s^{-1}$
The time taken by the disturbance to travel from one end to the other is
$\\t=\frac{l}{v}\\ =\frac{20}{40}\\ =0.5\ s$
Answer:
Time taken by the stone to the pond can be calculated using the second equation of motion
$\begin{aligned}
& \mathrm{s}=300 \mathrm{~m} \\
& \mathrm{u}=0 \\
& \mathrm{a}=9.8 \mathrm{~m} \mathrm{~s}^{-2} \\
& \mathrm{~s}=u t_1+\frac{1}{2} a t_1^2 \\
& 300=4.9 t_1^2 \\
& t_1=7.82 \mathrm{~s}
\end{aligned}$
Time taken by the wave to propagate from the pond to the top of the tower is
$\begin{aligned}
& t_2=\frac{300}{340} \\
& t_2=0.88 \mathrm{~s} \\
& \mathrm{t}_1+\mathrm{t}_2=8.7 \mathrm{~s}
\end{aligned}$
The splash is heard at the top of the tower after a time of 8.7 seconds.
Answer:
Mass per unit length of the wire is
$\begin{aligned}
\mu&=\frac{M}{l} \\
& =\frac{2.10}{12} \\
& =0.175 \mathrm{~kg} \mathrm{~m}^{-1}
\end{aligned}$
The speed of a transverse wave in a wire is given by
$\begin{aligned}
& v=\frac{T}{\mu} \\
& 343=\sqrt{\frac{T}{01.175}} \\
& T=343^2 \times 0.175 \\
& T=2.059 \times 10^4 \mathrm{~N}
\end{aligned}$
The tension in the wire should be $2.059\times 10^{4}\ N$ such that the speed of the transverse wave in it is equal to 343 m s -1.
Answer:
$v=\sqrt{\frac{\gamma P}{\rho }}$
Where $\gamma \ and \ \rho$ are the Bulk's modulus and the density respectively
As we know
$\rho =\frac{M}{V}$
where M is the molecular weight of air and V is the volume of 1 mole of air
$\\v=\sqrt{\frac{\gamma P}{\rho }}$
$ v=\sqrt{\frac{\gamma PV}{M}}$
From the ideal gas equation PV=nRT
Since we are talking about 1 mole, we take n = 1
PV=RT
The expression for the speed of sound becomes
$v=\sqrt{\frac{\gamma RT}{M}}$
$\gamma$ , M and R are constant therefore at constant Temperature the speed of sound in the air do not change and it is clear that speed is independent of velocity.
Answer:
From the equation $\\v=\sqrt{\frac{\gamma RT}{M}}\\$ it is clear that the speed of sound is linearly proportional to the square root of the temperature and therefore it will increase with the increase in temperature.
14.4 (c) Use the formula $v=\sqrt{\frac{\gamma P}{\rho }}$ to explain why the speed of sound in air
Answer:
$v=\sqrt{\frac{\gamma P}{\rho }}$
As the humidity of air increases, the proportion of water molecules(M=18) increases and that of Oxygen(M=32) and Nitrogen(M=28) decreases thus reducing the density of air and as the speed of sound is inversely proportional to the square root of density of air, the speed will increase as the density increases and thus it increases with an increase in humidity.
Answer:
No, the given function cannot represent a wave as x or t approach infinity the function won't be converging to a constant value and therefore the converse is not true.
Answer:
No, the given function cannot represent a travelling wave because as x and t become zero the given function won't be converging to a constant value and therefore the converse is not true.
Answer:
No, the given function cannot represent a travelling wave because as x and t become zero, the given function won't be converging to a constant value and therefore the converse is not true.
Answer:
(a) The wavelength of the reflected sound wave which will be travelling in air is
$
\begin{aligned}
\lambda_a & =\frac{v_a}{\nu} \\
\lambda_a & =\frac{340}{10^6} \\
\lambda_a & =3.4 \times 10^{-4} \mathrm{~m}
\end{aligned}
$
(b) The frequency of the transmitted sound wave would not change.
The wavelength of the transmitted sound wave which will be travelling in water is
$
\begin{aligned}
\lambda_w & =\frac{v_w}{\nu} \\
\lambda_w & =\frac{1486}{10^6} \\
\lambda_w & =1.49 \times 10^{-3} \mathrm{~m}
\end{aligned}
$
Answer:
The wavelength of the sound in the tissue is
$
\begin{aligned}
\lambda & =\frac{V}{\nu} \\
\lambda & =\frac{1.7 \times 10^3}{4.2 \times 10^6} \\
\lambda & =4 \times 10^{-4} \mathrm{~m}
\end{aligned}
$
14.8 (a) A transverse harmonic wave on a string is described by
$y(x,t)=3.0\; \sin (36\; t+0.018\; x+\pi /4)$
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?
Answer:
The wave is travelling.
$y(x,t)=Asin(kx+\omega t+\phi )$
The wave is travelling in the negative x-direction i.e. from right to left.
$\\\omega =36\ rad\ s^{-1}$
$ k=0.018\ cm^{-1}$
The speed of the wave is
$
\begin{aligned}
& v=\frac{\omega}{k} \\
& v=\frac{36 \times 10^{-2}}{0.018} \\
& v=20 \mathrm{~ms}^{-1}
\end{aligned}
$
14.8 (b) A transverse harmonic wave on a string is described by
$y(x,t)=3.0\; \sin (36\; t+0.018\; x+\pi /4)$
where x and y are in cm and t in s. The positive direction of x is from left to right.
(b) What are its amplitude and frequency?
Answer:
Amplitude A is 3.0 cm.
Frequency is
$
\begin{aligned}
\nu & =\frac{\omega}{2 \pi} \\
\nu & =\frac{36}{2 \pi} \\
\nu & =5.73 \mathrm{~Hz}
\end{aligned}
$
14.8 (c) A transverse harmonic wave on a string is described by
$y(x,t)=3.0\; \sin (36\; t+0.018\; x+\pi /4)$
where x and y are in cm and t in s. The positive direction of x is from left to right.
(c) What is the initial phase at the origin?
Answer:
The initial phase of the wave at the origin (at x = 0 and t = 0) is $\frac{\pi }{4}$
14.8 (d) A transverse harmonic wave on a string is described by
$y(x,t)=3.0\; \sin (36\; t+0.018\; x+\pi /4)$
where x and y are in cm and t in s. The positive direction of x is from left to right.
(d) What is the least distance between two successive crests in the wave?
Answer:
The difference between two consecutive crests is equal to the wavelength of the wave.
$
\begin{aligned}
& \lambda=\frac{2 \pi}{k} \\
& \lambda=\frac{2 \pi \times 10^{-2}}{0.018} \\
& \lambda=3.49 \mathrm{~m}
\end{aligned}
$
Answer:
$\\y(x,t)=3.0sin(36t+0.018x+\frac{\pi }{4})$
for x = 0
$\\y(t)=3.0sin(36t+\frac{\pi }{4})$
The time period of oscillation is T
$T=\frac{\pi }{18}\ s$
To make the y versus t graph we tabulate values of y(t) at different values of t as follows
$t$ |
$0$ |
$\frac{T}{8}$ |
$\frac{T}{4}$ |
$\frac{3T}{8}$ |
$\frac{T}{2}$ |
$\frac{5T}{8}$ |
$\frac{3T}{4}$ |
$\frac{7T}{8}$ |
$T$ |
$y(t)$ |
$\frac{3}{\sqrt{2}}$ |
$3$ |
$\frac{3}{\sqrt{2}}$ |
$0$ |
$\frac{-3}{\sqrt{2}}$ |
$-3$ |
$\frac{-3}{\sqrt{2}}$ |
$0$ |
$\frac{3}{\sqrt{2}}$ |
The graph of y versus t is as follows
For other values of x, we will get a similar graph. Its time period and amplitude would remain the same, it just will be shifted by different amounts for different values of x.
14.10 (a) For the travelling harmonic wave
$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$
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 $(a)\; 4\; m$
Answer:
The phase difference between two points separated by a distance of $\\\Delta x$ is given by
$\\\Delta \phi =k\Delta x$
$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$
$k=2\pi \times 0.008\ cm^{-1}$
Phase difference for two points separated by a distance of 4 m would be
$\\\Delta \phi =2\pi \times 0.0080\times 4\times 100$
$ \Delta \phi=6.4\pi \ rad$
14.10 (b) For the travelling harmonic wave
$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$
where x and y are in cm and t in s. Calculate the phase difference between the oscillatory motion of two points separated by a distance of
Answer:
The phase difference between two points separated by a distance of $\\\Delta x$ is given by
$\\\Delta \phi =k\Delta x$
$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$
$k=2\pi \times 0.008\ cm^{-1}$
Phase difference for two points separated by a distance of 0.5 m would be
$\\\Delta \phi =2\pi \times 0.0080\times 0.8\times 100$
$ \Delta \phi=0.8\pi \ rad$
14.10 (c) For the travelling harmonic wave
$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$
where x and y are in cm and t in s. Calculate the phase difference between the oscillatory motion of two points separated by a distance of $(c)\lambda /2$
Answer:
The phase difference between two points separated by a distance of $\\\Delta x$ is given by
$\\\Delta \phi =\frac{2\pi }{\lambda }\Delta x$
Phase difference for two points separated by a distance of $\frac{\lambda }{2}$ would be
$\\\Delta \phi =\frac{2\pi }{\lambda }\times \frac{\lambda }{2}$
$ \Delta \phi =\pi\ rad$
14.10 (d) For the travelling harmonic wave
$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$
where x and y are in cm and t in s. Calculate the phase difference between the oscillatory motion of two points separated by a distance of $(d)3\lambda /4$
Answer:
The phase difference between two points separated by a distance of $\\\Delta x$ is given by
$\\\Delta \phi =\frac{2\pi }{\lambda }\Delta x$
Phase difference for two points separated by a distance of $\frac{3\lambda }{4}$ would be
$\\\Delta \phi =\frac{2\pi }{\lambda }\times \frac{3\lambda }{4}$
$ \Delta \phi =\frac{3\pi }{2}\ rad$
14.11 The transverse displacement of a string (clamped at its both ends) is given by
$y(x,t)=0.06\; \sin (\frac{2\pi }{3}x)\cos (120\; \pi t)$
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is $3.0\times 10^{-2}\; kg$ .
Answer the following :
(a) Does the function represent a travelling wave or a stationary wave?
Answer:
The given function is of the following form
$y(x,t)=2Asin\left (kx \right )cos(\omega t)$
which is the general equation representing a stationary wave and therefore the given function represents a stationary wave.
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is $3.0\times 10^{-2}\; kg$ .
Answer the following :
(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?
Answer:
We know that when two waves of the same amplitude, frequency and wavelength travelling in opposite directions get superimposed, we get a stationary wave.
$\\y_{1}=asin(kx-\omega t)\\ \\y_{2}=asink(\omega t+kx)$
$\\y_{1}+y_{2}=asin(kx-\omega t)+asink(kx+\omega t)$
$ =asin(kx)cos(\omega t)-asin(\omega t)cos(kx)+asin(\omega t)cos(kx)+asin(kx)cos(\omega t)$
$=2asink(kx)cos(\omega t)$
Comparing the given function with the above equations we get
$
\begin{aligned}
& k=\frac{2 \pi}{3} \\
& \lambda=\frac{2 \pi}{k} \\
& \lambda=3 \mathrm{~m} \\
& \omega=120 \pi \\
& \nu=\frac{\omega}{2 \pi} \\
& \nu=60 \mathrm{~Hz} \\
& v=\nu \lambda \\
& v=60 \times 3=180 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
$
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is $3.0\times 10^{-2}\; kg$ .
Answer the following :
(c) Determine the tension in the string
Answer:
The mass per unit length of the string is
$\begin{aligned}
\mu & =\frac{M}{l} \\
\mu & =\frac{3 \times 10^{-2}}{1.5} \\
\mu & =0.02 \mathrm{~kg} \mathrm{~m}^{-1}
\end{aligned}$
Speed of a transverse wave is given by
$\begin{aligned}
v & =\sqrt{\frac{T}{\mu}} \\
T & =\mu v^2 \\
T & =0.02 \times(180)^2 \\
T & =648 \mathrm{~N}
\end{aligned}$
Answer:
(i) (a) All the points vibrate with the same frequency of 60 Hz.
(b) They all have the same phase as it depends upon time.
(c) At different points, the amplitude is different and is equal to A(x) given by
$A(x)=0.06sin(\frac{2\pi }{3}x)$
(ii)
$
\begin{aligned}
& A(0.375)=0.06 \sin \left(\frac{2 \pi}{3} \times 0.375\right) \\
& A(0.375)=0.06 \times \sin \left(\frac{\pi}{4}\right) \\
& A(0.375)=0.06 \times \frac{1}{\sqrt{2}} \\
& A(0.375)=0.042 \mathrm{~m}
\end{aligned}
$
$(a)\; y=2\; \cos (3x)\sin (10t)$
$(b)\; y=2\sqrt{x-vt}$
$(c)y=3\sin (5x-0.5t)+ 4\; \cos (5x-0.5t)$
$(d)\; y=\cos x\; \sin t+\cos 2x\: \sin \: 2t$
Answer:
(a) It represents a stationary wave.
(b) The given function does not represent a wave as we can see at certain values of x and t the function would not be defined.
(c) It represents a travelling wave.
(d) It represents a stationary wave. It is a superposition of two stationary waves which ultimately results in another stationary wave.
Answer:
Length of the string is l given by
$
\begin{aligned}
& l=\frac{M}{\mu} \\
& l=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}} \\
& l=0.875 \mathrm{~m}
\end{aligned}
$
Since the wire is vibrating in the fundamental mode
$
\begin{aligned}
& l=\frac{\lambda}{2} \\
& \lambda=2 l \\
& \lambda=2 \times 0.875 \\
& \lambda=1.75 \mathrm{~m}
\end{aligned}
$
Speed of the string $(\mathrm{v})$ is
$
\begin{aligned}
& v=\nu l \\
& v=45 \times 1.75 \\
& v=78.75 \mathrm{~ms}^{-1}
\end{aligned}
$
Answer:
Tension in the string is given by
$\begin{aligned}
& T=\mu v^2 \\
& T=4 \times 10^{-2} \times(78.75)^2 \\
& T=248.0625 N
\end{aligned}$
Answer:
The pipe behaves as a pipe open at one end and closed at one end. Such a pipe would produce odd harmonics i.e.
$
\nu=(2 n-1) \frac{v}{4 l_n}
$
Two consecutive modes of vibration are given in the question
For $\mathrm{I}_1=25.5 \mathrm{~cm}$
$
\nu=(2 n-1) \frac{v}{4 l_1}
$
For $\mathrm{I}_2=79.3 \mathrm{~cm}$
$
\begin{aligned}
\nu & =(2(n+1)-1) \frac{v}{4 l_2} \\
\nu & =(2 n+1) \frac{v}{4 l_2}
\end{aligned}
$
Since at both these modes the system resonates with the same frequency we have
$
\begin{aligned}
& (2 n-1) \frac{v}{4 l_1}=(2 n+1) \frac{v}{4 l_2} \\
& \frac{2 n-1}{2 n+1}=\frac{25.5}{79.3} \\
& \frac{2 n-1}{2 n+1} \approx \frac{1}{3} \\
& n=1 \\
& \nu=\frac{v}{4 l_1} \\
& v=340 \times 4 \times 0.255 \\
& v=346.8 \mathrm{~ms}^{-1}
\end{aligned}
$
Answer:
When the rod is clamped at the middle at is vibrating in the fundamental mode, a node is formed at the middle of the rod and antinodes at the end. i.e.
$
\begin{aligned}
& \frac{L}{2}=\frac{\lambda}{4} \\
& \lambda=2 L \\
& \lambda=0.2 \mathrm{~m}
\end{aligned}
$
where $L$ is the length of the rod.
Speed of sound in steel is
$
\begin{aligned}
& v=\nu \lambda \\
& v=2.53 \times 10^3 \times 0.2 \\
& v=5060 \mathrm{~ms}^{-1}
\end{aligned}
$
Answer:
Let the $\mathrm{n}^{\text {th }}$ harmonic mode of the pipe get resonantly excited by a 430 Hz source.
$
\begin{aligned}
& \frac{(2 n-1) v}{4 l}=\nu \\
& 2 n-1=\frac{430 \times 4 \times 0.2}{340} \\
& n=1
\end{aligned}
$
The pipe resonates with a 430 Hz source in the fundamental mode when one end is open.
Let the $\mathrm{m}^{\text {th }}$ harmonic mode of the pipe get resonantly excited by a 430 Hz source when both ends are open.
$
\begin{aligned}
\lambda & =\frac{2 l}{m} \\
m & =\frac{2 l \nu}{v} \\
m & =\frac{2 \times 0.2 \times 430}{340} \\
m & =0.5
\end{aligned}
$
Since $m$ is coming out to be less than 1 the same source will not be in resonance with the pipe if both ends are open.
Answer:
$
\begin{aligned}
& \nu_A=324 \mathrm{~Hz} \\
& \text { Beat frequency }(b)=6 \mathrm{~Hz} \\
& \nu_B=\nu_A \pm b \\
& \nu_B=318 \mathrm{~Hz} \\
& \text { or } \\
& \nu_B=330 \mathrm{~Hz}
\end{aligned}
$
Since frequency increases with an increase in Tension, the frequency of string A must have decreased. Therefore $\nu _{B}=318Hz$ .(If it were 330 Hz the beat frequency would have increased with a decrease in Tension in string A)
14.19 Explain why (or how) :
(a) in a sound wave, a displacement node is a pressure antinode and vice versa,
Answer:
In the propagation of a sound wave the pressure increases at points where displacement decreases, Therefore maximum pressure at points of minimum displacement and vice-versa i.e. 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”,
Answer:
Bats emit ultrasonic waves and when these waves strike the obstacles and get reflected to the bats they ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”.
Answer:
We can distinguish between the two notes with the same frequency as the harmonics they emit are different.
Answer:
Transverse waves produce shear, gases don't have shear modulus and cannot sustain shear and therefore can only propagate longitudinal waves. Solids have both shear and bulk modulus of elasticity and can propagate both transverse and longitudinal waves.
(e) Explain why (or how): the shape of a pulse gets distorted during propagation in a dispersive medium.
Answer:
As we know a pulse contains waves of different wavelengths, these waves travel at different speeds in a dispersive medium and thus the shape of the pulse gets distorted.
Answer:
$\nu _{o}=\left ( \frac{v\pm v_{o}}{v\pm v_{s}} \right )\nu$
where $\\\nu _{o}$ is the frequency as observed by the observer, $\nu$ is the frequency of the source, v is the speed of the wave, v o is the speed of the observer and v s is the speed of the source.
(i) (a) When the source is moving towards the observer and the observer is stationary.
$\\\nu _{o}=\left ( \frac{v}{v- v_{s}} \right )\nu $
$\nu _{o}=\frac{340}{340-10}\times 400$
$ \nu _{o}=412Hz$
(b)
$\nu _{o}=\left ( \frac{v}{v+ v_{s}} \right )\nu$
$ \nu _{o}=\frac{340}{340+10}\times 400$
$\nu _{o}=389Hz$
(ii) The speed of the sound does not change as it is independent of the speed of observer and source and remains equal to 340 ms -1 .
Answer:
Speed of the wind v w = 10 m s -1
Speed of sound in still air v a = 340 m s -1
Effective speed with which the wave reaches the observer = v = v w + v a = 10 + 340= 350 m s -1
There is no relative motion between the observer and the source and therefore the frequency heard by the observer would not change.
The wavelength of the sound as heard by the observer is
$\\\lambda =\frac{v}{\nu }$
$ \lambda =\frac{350}{400}$
$ \lambda =0.875m$
The above situation is not identical to the case when the air is still and the observer runs towards the yard as then there will be relative motion between the observer and the source and the frequency observed by the observer would change.
$\nu _{o}=\left ( \frac{v\pm v_{o}}{v\pm v_{s}} \right )\nu$
The frequency as heard by the observer is
$\\\nu _{o}=\frac{340+10}{340}\times 400$
$ \nu _{o}=411.76Hz$
$\\\lambda=\frac{340}{400}$
$ \lambda=0.85m$
3. (a) A travelling harmonic wave on a string is described by
$y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)$
(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?
Answer:
$y(x, t)=7.5 \sin (0.0050 x+12 t+\pi / 4)$
The displacement of oscillation of a point at $\mathrm{x}=1 \mathrm{~cm}$ and $\mathrm{t}=1 \mathrm{~s}$ is
$\begin{aligned}
& y(1,1)=7.5 \sin \left(0.0050 \times 1+12 \times 1+\frac{\pi}{4}\right) \\
& y(1,1)=7.5 \sin (12.79) \\
& y(1,1)=7.5 \sin \left(\frac{12.266 \times 180^{\circ}}{\pi}\right) \\
& y(1,1)=7.5 \sin \left(733.18^{\circ}\right) \\
& y(1,1)=1.71 \mathrm{~cm}
\end{aligned}$
The general expression for the velocity of oscillation is
$\begin{aligned}
& v_y(x, t)=\frac{\mathrm{dy}(\mathrm{x}, \mathrm{t})}{\mathrm{d} t} \\
& =\frac{d}{d t}\left[\sin \left(7.5 \sin \left(0.0050 x+12 t+\frac{\pi}{4}\right)\right]\right. \\
& =90 \cos \left(0.0050 x+12 t+\frac{\pi}{4}\right) \\
& v_y(1,1)=90 \cos \left(0.0050 \times 1+121+\frac{\pi}{4}\right) \\
& =90 \cos (12.79) \\
& =90 \cos \left(733.18^{\circ}\right) \\
& =87.63 \mathrm{~cm} \mathrm{~s}^{-1} \\
& y(x, t)=7.5 \sin (0.0050 x+12 t+\pi / 4) \\
& \mathrm{k}=0.005 \mathrm{~cm}^{-1} \\
& \omega=12 \mathrm{rad} / \mathrm{s}
\end{aligned}$
The velocity of propagation of the wave is
$\begin{aligned}
v & =\nu \lambda \\
v & =\frac{\omega}{2 \pi} \times \frac{2 \pi}{k} \\
v & =\frac{\omega}{k} \\
v & =\frac{12}{0.005 \times 100} \\
v & =24 m s^{-1}
\end{aligned}$
The velocity of oscillation of point at $x=1 \mathrm{~cm}$ and $t=1 \mathrm{~cm}$ is not equal to the propagation of the wave.
(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.
Answer:
The wavelength of the given wave is
$\\\lambda =\frac{2\pi }{k}$
$ \lambda =\frac{2\pi }{0.005 cm^{-1}}$
$ \lambda =1256cm$
$ \lambda =12.56m$
The points with the same displacements and velocity at the same instant of time are separated by distances $n\lambda$ .
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 would be at a distance of
$\pm \lambda ,\pm 2\lambda ,\pm 3\lambda ...$ from x = 1cm.
$\lambda =12.56m$
Therefore all points at distances $\pm 12.56m,\pm 25.12m,\pm 37.68m$ from the point x=1cm would have the same transverse displacements and velocity as the $x=1\: cm$ point at $t=2\: s,$ 5 s and 11 s.
Answer:
(a) The pulse does not have a definite frequency or wavelength however the wave has definite speed given the medium is non-dispersive.
(b) The frequency of the note produced by the whistle is not 0.05 Hz. It only implies the frequency of repetition of the pip of the whistle is 0.05 Hz,
Answer:
$\begin{aligned}
& y(x, t)=A \sin (\omega t \pm k x+\phi) \\
& \mathrm{A}=0.05 \mathrm{~m}
\end{aligned}$
Tension in the string is $\mathrm{T}=\mathrm{mg}$
$\begin{aligned}
& T=90 \times 9.8 \\
& T=882 N
\end{aligned}$
The speed of the wave in the string is v
$\begin{aligned}
& v=\sqrt{\frac{T}{\mu}} \\
& v=\sqrt{\frac{882}{8 \times 10^{-3}}} \\
& v=332 m s^{-1}
\end{aligned}$
Angular frequency of the wave is
$\begin{aligned}
\omega & =2 \pi \nu \\
\omega & =2 \pi \times 256 \\
\omega & =1608.5 \mathrm{rad} / \mathrm{s} \\
k & =\frac{2 \pi}{\lambda} \\
k & =\frac{2 \pi \nu}{v} \\
k & =4.84 \mathrm{~m}^{-1}
\end{aligned}$
Since at $\mathrm{t}=0$, the left end (fork end) of the string $\mathrm{x}=0$ has zero transverse displacement ( $\mathrm{y}=0$ ) and is moving along the positive y direction, the initial phase is zero. ( $\phi=0 \mathrm{rad}$ )
Taking the left to the right direction as positive we have
$y(x, t)=0.05 \sin (1608.5 t-4.84 x)$
Here t is in seconds and x and y are in metres.
Answer:
Frequency of $\operatorname{SONAR}(\nu)=40 \mathrm{kHz}$
Speed of enemy submarine $v_0=360 \mathrm{~km} \mathrm{~h}^{-1}=100 \mathrm{~m} \mathrm{~s}^{-1}$
$\begin{aligned}
\nu_o & =\left(\frac{v+v_o}{v}\right) \nu \\
& =\frac{1450+100}{1450} \times 40 \times 10^3 \\
& =42.76 \mathrm{kHz}
\end{aligned}$
This is the frequency which would be observed and reflected by the enemy submarine but won't appear the same to the SONAR(source) as again there is relative motion between the source(enemy submarine) and the observer(SONAR)
The frequency which would be received by the SONAR is
$\begin{aligned}
\nu_o^{\prime} &=\left(\frac{v}{v-v_s}\right) \nu_o \\
& =\frac{1450}{1450-100} \times 42.76 \times 10^3 \\
& =45.92 \mathrm{kHz}
\end{aligned}$
Answer:
Let us assume the earthquake occurs at a distance s.
$
\begin{aligned}
& \Delta t=4 \times 60=240 s \\
& \Delta t=\frac{s}{v_s}-\frac{s}{v_p} \\
& \Delta t=\frac{s}{4}-\frac{s}{8} \\
& 240=\frac{s}{8} \\
& s=1960 \mathrm{~km}
\end{aligned}
$
The origin of the earthquake is at a distance of 1960 km.
Answer:
Apparent frequency striking the wall and getting reflected is
The frequency emitted by the bats is $\nu=40 \mathrm{kHz}$
Speed of sound is v
Speed of bat is 0.03 V
$
\begin{aligned}
& \nu^{\prime}=\left(\frac{v}{v-v_s}\right) \nu \\
& =\frac{v}{v-0.03 v} \times 40 \mathrm{kHz} \\
& =41.237 \mathrm{kHz}
\end{aligned}
$
Frequency of sound as heard by the bat
$
\begin{aligned}
& \nu^{\prime \prime}=\left(\frac{v+v_o}{v}\right) \nu^{\prime} \\
& =\frac{v+0.03 v}{v} \times \nu^{\prime} \\
& =1.03 \times 41.237 \mathrm{kHz} \\
& =42.474 \mathrm{kHz}
\end{aligned}
$
Q.1 A string fixed at both ends is vibrating in three loops. The length of string is 'l' and amplitude of antinode is 'A'. The amplitude of a particle at a distance $\frac{\ell}{4}$ from one end is $\frac{A}{\sqrt{x}}$. Find x.
Answer:
There will be a node at any end as the string is fixed at both the ends.
Let x = 0 is a node. Hence amplitude at any point at a distance x is given as:
$A_x=|A \sin (K x)| \quad \text { where } K=\frac{2 \pi}{\lambda}$
As string is vibrating in 3 loops and each loop length is equal to $\frac{\lambda}{2}$, hence
$\begin{array}{rlr}
& & \frac{3 \lambda}{2}=\ell \\
\Rightarrow & \lambda & =\frac{2 \ell}{3} \\
\Rightarrow & K & =\frac{3 \pi}{\ell}
\end{array}$
Therefore, amplitude at distance $\frac{\ell}{4}$ from one end will be
$A_x=A \sin \left(\frac{3 \pi}{\ell} \times \frac{\ell}{4}\right)=\frac{A}{\sqrt{2}}$
Hence, the answer is 2.
Q.2 The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 60 cm, find the length of the open organ pipe.
Answer:
Given that, the length of the closed pipe, $I_c=60 \mathrm{~cm}$
The first overtone of open organ pipe is, $f_0=\frac{2 \nu}{2 l_0}$
The fundamental frequency of closed organ pipe is, $f_c=\frac{\nu}{4 l_c}$
Given,
$
\begin{aligned}
& f_0=f_c \\
& \frac{2 \nu}{2 l_0}=\frac{\nu}{4 l_c} \\
& l_0=4 l_c \\
& l_0=4 \times 60 \\
& l_0=240 \mathrm{~cm}
\end{aligned}
$
Q.3 Two speakers are separated by a distance of 5 m. A person stands at a distance of 12 m, directly in front of one of the speakers. If $n$ stands for any integer, find the frequencies for Which the listener will hear a minimum sound intensity. (Speed of sound in air is $340 \mathrm{~m} / \mathrm{s}$)
Answer:
$\begin{aligned} & \left(n+\frac{1}{2}\right) \lambda=\text { Path difference } \\ & \left(n+\frac{1}{2}\right) \lambda=(13-12) \\ & \left(n+\frac{1}{2}\right) \lambda=1 \\ & (2 n+1) \lambda=2 \\ & \lambda=\frac{v}{f}=\frac{340}{f} \\ & (2 n+1) \times \frac{340}{f}=2 \\ & f=170(2 n+1) H z\end{aligned}$
Q.4 A source emitting sound of frequency 90 Hz is Placed in front of a wall at a distance 8 m from it. A detector is also placed in front of the wall at the same distance from it. Find the minimum distance between the source and the detector for which the detector detects a maximum of sound speed in air $340 \mathrm{~m} / \mathrm{s}$.
Answer:
$\begin{aligned} & \Delta x=\text { Path difference } \\ & =n \lambda[n=1] \\ & =\lambda \\ & 2 \sqrt{x^2+64}+\frac{\lambda}{2}-2 x=\lambda \\ & \lambda=\frac{v}{f}=\frac{340}{90} \lambda=\frac{V}{f}=\frac{340}{90} \\ & \lambda=\frac{34}{9} \\ & 2 \sqrt{x^2+64}+\frac{34}{18}-2 x=\frac{34}{9} \\ & 2 \sqrt{x^2+64}-2 x=\frac{34}{9}-\frac{34}{18} \\ & 2 \sqrt{x^2+64}-2 x=\frac{17}{9} \\ & 2 \sqrt{x^2+64}=\frac{17}{9}+2 x \\ & 18 \sqrt{x^2+64}=18 x+17 \\ & 324\left(x^2+64\right)=324 x^2+289+612 x \\ & 324 x^2+20736=324 x^2+289+612 x \\ & 20736-289=612 x \\ & 20447=612 x \\ & x=\frac{20447}{612} \\ & x=33.42 \\ & x=2 x \\ & x=2 \times 33.42 \\ & x=66.84 \\ & x=20\end{aligned}$
Q.5 A vibrating tuning fork of frequency $n$ is placed near the open end of a long cylindrical tube. The tube has a side opening and is alto fitted with a movable reflecting Poston. As the piston is moved through 4.65 cm , the intensity of sound changes from a maximum to minimum. If the speed of sound is $400 \mathrm{~m} / \mathrm{s}$ then $n$.
Answer:
$ \begin{aligned} & \text { Path difference }=4.65 \times 2 \\ & \text { Path difference }=\frac{\lambda}{2} \\ & \therefore \quad 9 \cdot 3=\frac{\lambda}{2} \\ & \lambda=18 \cdot 6 \lambda=18.6 \\ & v=n \lambda \\ & n=\frac{u}{\lambda} \\ & n=\frac{400}{18.6} \\ & n=\frac{4000}{186} \\ & n=21.5054 \end{aligned} $
Wave velocity
Wave velocity(W) = Frequency(f) * wavelength (λ) [SI unit- m/s]
velocity of the transverse wave in a stretched string
$V=\sqrt{\frac{T}{\mu}}$
where:
v = Velocity of the wave (in meters per second, m/s), T = Tension in the string (in Newtons, N) and μ = Linear mass density of the string (in kilograms per meter, kg/m)
Velocity of a longitudinal wave in an elastic medium
$V=\sqrt{\frac{E}{\rho}}$
Where:
E = Young's modulus of the material (in Pascals, Pa), ρ = Density of the material (in kilograms per cubic meter, kg/m³)
Waves are classified into mechanical waves (which require a medium) and electromagnetic waves (which do not require a medium).
The main topics covered in the NCERT Class 11 textbook chapter 14 are
In transverse waves, particles move perpendicular to wave direction (e.g., light waves), while in longitudinal waves, particles move parallel to wave direction (e.g., sound waves).
When two or more waves overlap, the resultant displacement is the sum of individual displacements at each point.
Beats occur when two waves of slightly different frequencies interfere, creating periodic variations in loudness.
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