Sample Exercises

Hint

Tension, \(\quad T=200 \mathrm{~N}\);
Length, \(\quad l=20.0 \mathrm{~m}\); Mass, \(M=2.50 \mathrm{~kg}\)
Mass per unit length, \(\mu=\frac{2.50}{20.0} \mathrm{~kg} \mathrm{~m}^{-1}=0.125 \mathrm{kgm}^{-1}\)
Wave velocity, \(\quad v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{200 \mathrm{~N}}{0.125 \mathrm{~kg} \mathrm{~m}^{-1}}}\)
or \(v=\sqrt{1600} \mathrm{~ms}^{-1}=40 \mathrm{~ms}^{-1}\)
Time,
\(
t=\frac{l}{v}=\frac{20.0}{40} s=\frac{1}{2} s=0.5 \mathrm{~s} .
\)

Hint

Here, \(h=300 \mathrm{~m}, \mathrm{~g}=9.8 \mathrm{~ms}^{-2}\) and velocity of sound, \(\mathrm{v}=340 \mathrm{~ms}^{-1}\) Let \(\mathrm{t}_1\) be the time taken by the stone to reach at the surface of pond.
Then, using
\(
s=u t+\frac{1}{2} a t^2 \frac{1}{2} a t^2 \Rightarrow h=0 \times t+\frac{1}{2} g t_1^2
\)
\(
\therefore \quad t_1=\sqrt{\frac{2 \times 300}{9.8}}=7.82 \mathrm{~s}
\)
Also, if \(t_2\) is the time taken by the sound to reach at a height \(h\), then
\(
t_2=\frac{h}{v}=\frac{300}{340}=0.88 \mathrm{~s}
\)
\(\therefore\) Total time after which sound of splash is heard \(=t_1+t_2\)
\(
=7.82+0.88=8.7 \mathrm{~s} \text {. }
\)

Hint

Here,
\(
\begin{aligned}
l & =12.0 \mathrm{~m}, \mathrm{M}=2.10 \mathrm{~kg} \\
v & =343 \mathrm{~ms}^{-1}
\end{aligned}
\)
\(
\text { Mass per unit length }=\frac{M}{l}=\frac{2.10}{12.0}=0.175 \mathrm{~kg} \mathrm{~m}^{-1}
\)
\(
\begin{array}{ll}
\text { As } & v=\sqrt{\frac{T}{m}} \\
\therefore & T=v^2 {m}=(343)^2 \times 0.175=2.06 \times 10^4 \mathrm{~N} .
\end{array}
\)

Hint

We are given that \(\quad v=\sqrt{\frac{\gamma P}{\rho}}\)
We know
We know \(\quad P V=n \mathrm{R} T\)
(for \(n\) moles of ideal gas)
\(
\Rightarrow \quad P V=\frac{m}{M} R T
\)
where \(m\) is the total mass and \(\mathrm{M}\) is the molecular mass of the gas.
\(
\therefore \quad P=\frac{m}{M} \cdot \frac{R T}{M}=\frac{\rho R T}{M} \Rightarrow \frac{P}{\rho}=\frac{\mathrm{R} T}{M}
\)
(a) For a gas at constant temperature, \(\frac{P}{\rho}=\) constant
\(\therefore\) As \(P\) increase, \(\rho\) also increases and vice versa. This implies that \(v=\sqrt{\frac{\gamma P}{\rho}}=\) constant, i.e., velocity is independent of pressure of the gas.
(b) Since \(\frac{P}{\rho}=\frac{R T}{M}\), therfore, \(v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{\gamma R T}{M}}\)
Clearly \(v \propto \sqrt{T}\) i.e., speed of sound in air increases with increase in temperature.
(c) Increase in humidity decreases the effective density of air. Therefore the velocity \(\left(v \propto \frac{1}{\sqrt{\rho}}\right)\) increases.

Hint

(a) Does not represent a wave
(b) Represents a wave
(c) Does not represent a wave
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 \(t\).
Explanation:
(a) For \(\mathrm{x}=0\) and \(t=0\), the function \((x-v t)^2\) becomes 0 .
Hence, for \(x=0\) and \(t=0\), the function represents a point and not a wave.
(b) For \(x=0\) and \(t=0\), the function
\(
\log \left(\frac{x+v t}{x_0}\right)=\log 0=\infty
\)
Since the function does not converge to a finite value for \(x=0\) and \(t=0\), it represents a travelling wave.
(c) For \(x=0\) and \(t=0\), the function
\(
\frac{1}{x+v t}=\frac{1}{0}=\infty
\)
Since the function does not converge to a finite value for \(x=0\) and \(t=0\), it does not represent a travelling wave.

Hint

Here,
\(
\begin{aligned}
v & =1000 \times 10^3 \mathrm{~Hz}=10^6 \mathrm{~Hz}, v_a=340 \mathrm{~ms}^{-1}, \\
v_w & =1486 \mathrm{~ms}^{-1}
\end{aligned}
\)
Wavelength of reflected sound, \(\lambda_a\)
\(
=\frac{v_a}{v}=\frac{340}{10^6} \mathrm{~m}=3.4 \times 10^{-4} \mathrm{~m}
\)
Wavelength of transmitted sound, \(\lambda_\omega\)
\(
=\frac{v_w}{v}=\frac{1486}{10^6} \mathrm{~m}=1.49 \times 10^{-3} \mathrm{~m}
\)

Hint

Frequency \((f)=4.2 \mathrm{MHz}=4.2 \times 10^6 \mathrm{~Hz}\) and speed of sound
\(
(v)=1.7 \mathrm{~km} / \mathrm{s}=1.7 \times 10^3 \mathrm{~m} / \mathrm{s} \text {. }
\)
Wavelength of sound in tissue
\(
\begin{aligned}
(\lambda) & =\frac{v}{f}=\frac{1.7 \times 10^3}{4.2 \times 10^6} \\
& =4.1 \times 10^{-4} \mathrm{~m}
\end{aligned}
\)

Hint

The given equation is \(y(x, t)=3.0 \sin \left(36 t+0.018 x+\frac{\pi}{4}\right)\), where \(x\) and \(y\) are in \(\mathrm{cm}\) and \(t\) in \(\mathrm{s}\).
The equation is the equation of a travelling wave, travelling from right to left (i.e., along \(-v e\) direction of \(x\) because it is an equation of the type
\(
y(x, t)=A \sin (\omega t+k x+\phi)
\)
Here, \(\quad A=3.0 \mathrm{~cm}, \omega=36 \mathrm{rad} \mathrm{s}^{-1}, k=0.018 \mathrm{~cm}\) and \(\phi=\frac{\pi}{4}\).
\(\therefore\) Speed of wave propagation,
\(
v=\frac{\omega}{k}=\frac{36 \mathrm{rad} \mathrm{s}^{-1}}{0.018 \mathrm{~cm}^{-1}}=\frac{36 \mathrm{rad} \mathrm{s}^{-1}}{0.018 \times 10^2 \mathrm{~ms}^{-1}}=20 \mathrm{~ms}^{-1}
\)

Hint

(a) Amplitude of wave, \(A=3.0 \mathrm{~cm}=0.03 \mathrm{~m}\)
Frequency of wave \(v=\frac{\omega}{2 \pi}=\frac{36}{2 \pi}=5.7 \mathrm{~Hz}\)
(b) Initial phase at the origin, \(\phi=\frac{\pi}{4}\)
(c) Least distance between two successive crests in the wave
\(
\begin{aligned}
& =\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{0.018} \\
& =349 \mathrm{~cm}=3.5 \mathrm{~m}
\end{aligned}
\)

Hint

The transverse harmonic wave is
\(
y(x, t)=3.0 \sin \left(36 t+0.018 x+\frac{\pi}{4}\right)
\)
For
\(
\begin{aligned}
x & =0, \\
y(0, t) & =3 \sin \left(36 t+0+\frac{\pi}{4}\right)=3 \sin \left(36 t+\frac{\pi}{4}\right) \dots(1)
\end{aligned}
\)
Here
\(
\omega=\frac{2 \pi}{T}=36 \Rightarrow T=\frac{2 \pi}{36}
\)
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 eqn. (1)).
\(
\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline \mathrm{t} & 0 & \frac{T}{8} & \frac{2 T}{8} & \frac{3 T}{8} & \frac{4 T}{8} & \frac{5 T}{8} & \frac{6 T}{8} & \frac{7 T}{8} & \mathrm{~T} \\
\hline \mathrm{y} & \frac{3}{\sqrt{2}} & 3 & \frac{3}{\sqrt{2}} & 0 & \frac{-3}{\sqrt{2}} & -3 & \frac{-3}{\sqrt{2}} & 0 & \frac{3}{\sqrt{2}} \\
\hline
\end{array}
\)
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 \(y=2 \mathrm{~cm}\) and \(x=4 \mathrm{~cm}\). The oscillatory motion in the travelling wave only differs in respect of phase. Amplitude and frequency of oscillatory motion remains the same in all the cases.

Hint

The given equation can be rewritten as under:
or
\(
\begin{aligned}
& y(x, t)=2.0 \cos [2 \pi(10 t-0.0080 x)+2 \pi \times 0.35] \\
& y(x, t)=2.0 \cos \left[2 \pi \times 0.0080\left(\frac{10 t}{0.0080}-x\right)+0.7 \pi\right]
\end{aligned}
\)
Comparing this equation with the standard equation of a travelling harmonic wave,
\(
\frac{2 \pi}{\lambda}=2 \pi \times 0.0080 \text { or } \lambda=\frac{1}{0.0080} \mathrm{~cm}=125 \mathrm{~cm}
\)
The phase difference between oscillatory motion of two points separated by a distance \(\Delta x\) is given by
\(
\Delta \phi=\frac{2 \pi}{\lambda} \Delta x
\)
(a) When
\(
\begin{aligned}
& \Delta x=4 \mathrm{~m}=400 \mathrm{~cm}, \text { then } \\
& \Delta \phi=\frac{2 \pi}{125} \times 400=6.4 \pi \mathrm{rad}
\end{aligned}
\)
(b) When \(\Delta x=0.5 \mathrm{~m}=50 \mathrm{~cm}\), then \(\Delta \phi=\frac{2 \pi}{125} \times 50=0.8 \pi \mathrm{rad}\)
(c) When \(\Delta x=\frac{\lambda}{2}=\frac{125}{2} \mathrm{~cm}\), then \(\Delta \phi=\frac{2 \pi}{125} \times \frac{125}{2}=\pi \mathrm{rad}\)
(d) When \(\Delta x=\frac{3 \lambda}{4}=\frac{3 \times 125}{4} \mathrm{~cm}\), then \(\Delta \phi=\frac{2 \pi}{125} \times \frac{3 \times 125}{4}=\frac{3 \pi}{2} \mathrm{rad}\).

Hint

The general equation representing a stationary wave is given by the displacement function:
\(y(x, t)=2 a \sin k x \cos \omega t\)
This equation is similar to the given equation:
\(
y(x, \mathrm{t})=0.06 \sin \left(\frac{2}{3} x\right) x \cos (120 \pi \mathrm{t})
\)
Hence, the given function represents a stationary wave.

Hint

The given equation is \(y(x, t)=0.06 \sin \frac{2 \pi}{3} \times \cos 120 \pi t \dots(1)\)
We know that when a wave pulse
\(
\mathrm{y}_1=r \sin \frac{2 \pi}{\lambda}(v t-x)
\)
travelling along + direction of \(x\)-axis is superimposed by the reflected wave
\(
y_2=-r \sin \frac{2 \pi}{\lambda}(v t+x)
\)
travelling in opposite direction, a stationary wave
\(y=y_1+y_2=-2 r \sin \frac{2 \pi}{\lambda} x \cos \frac{2 \pi}{\lambda} v t \dots(2)\) is formed
Comparing equation (1) and (2) we find that
\(
\frac{2 \pi}{\lambda}=\frac{2 \pi}{3} \Rightarrow \lambda=3 m
\)
Also \(\frac{2 \pi}{\lambda} v=120 \pi\) or \(v=60 \lambda=60 \times 3=180 \mathrm{~ms}^{-1}\)
Frequency, \(f=\frac{v}{\lambda}=\frac{180}{3}=60 \mathrm{~Hz}\)
Note that both the waves have same wavelength same frequency and same speed.

Hint

Velocity of transverse waves is
\(
\begin{aligned}
v & =\sqrt{\frac{T}{m}} \quad \text { or } \quad v^2=\frac{T}{m} \\
\therefore \quad T & =m v^2, \text { where } \quad m=\frac{3 \times 10^{-2}}{1.5}=2 \times 10^{-2} \mathrm{~kg} / \mathrm{m} \\
\therefore \quad T & =(180)^2 \times 2 \times 10^{-2}=648 \mathrm{~N} .
\end{aligned}
\)

Hint

(i) For the wave on the string described in questions we have seen that \({l}=1.5 \mathrm{~m}\) and \(\lambda=3 \mathrm{~m}\). So, it is clear that \(\lambda=\lambda / 2\) 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 sring particles, except nodes, vibrate with the same frequency \({f}=60 \mathrm{~Hz}\).
(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 \(=2 A=0.06 \mathrm{~m}\). It gradually falls on going towards nodes and at nodes, and at nodes, it is zero.
(ii) Amplitude at a point \(x=0.375 \mathrm{~m}\) will be obtained by putting \(\cos (120 \pi \mathrm{t})\) as +1 in the wave equation.
\(
\therefore A(x)=0.06 \sin \left(\frac{2 \pi}{3} \times 0.375\right) \times 1=0.06 \sin \frac{\pi}{4}=0.042 m
\)

Hint

(a) The given equation represents a stationary wave because the harmonic terms \(\mathrm{kx}\) and \(\omega \mathrm{t}\) appear separately in the equation.
(b) The given equation does not contain any harmonic term. Therefore, it does not represent either a travelling wave or a stationary wave.
(c) The given equation represents a travelling wave as the harmonic terms \(\mathrm{kx}\) and \(\omega t\) are in the combination of \(\mathrm{kx}\) – \(\omega \mathrm{t}\).
(d) The given equation represents a stationary wave because the harmonic terms kx and \(\omega\) t appear separately in the equation. This equation actually represents the superposition of two stationary waves.

Hint

Here, \(\quad f=45 \mathrm{~Hz}, \quad M=3.5 \times 10^{-2} \mathrm{~kg}\)
Mass per unit length \(=m=4.0 \times 10^{-2} \mathrm{~kg} \mathrm{~m}^{-1}\)
\(\therefore \quad l=\frac{M}{m}=\frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}}=\frac{7}{8}\)
As \(\quad \frac{l}{2}=\lambda=\frac{7}{8} \quad \therefore \quad \lambda=\frac{7}{4} \mathrm{~m}=1.75 \mathrm{~m}\).
(a) The speed of the transverse wave, \(v=f \lambda=45 \times 1.75=78.75 \mathrm{~m} / \mathrm{s}\)
(b) As
\(
\therefore \quad T=v^2 \times m=(78.75)^2 \times 4.0 \times 10^{-2}=248.06 \mathrm{~N} .
\)

Hint

Frequency of \(n^{\text {th }}\) mode of vibration of the closed organ pipe of length
\(
l_1=(2 n-1) \frac{v}{4 l_1}
\)
Frequency of \((n+1)^{\text {th }}\) mode of vibration of closed pipe of length
\(
l_2=[2(n+1)-1] \frac{v}{4 l_2}=(2 n+1) \frac{v}{4 l_2}
\)
Both the modes are given to resonate with a frequency of \(340 \mathrm{~Hz}\).
\(
\begin{aligned}
& \therefore \quad(2 n-1) \frac{v}{4 l_1}=(2 n+1) \frac{v}{4 l_2} \\
& \text { or } \quad \frac{2 n-1}{2 n+2}=\frac{l_1}{l_2}=\frac{25.5}{79.3}=\frac{1}{3} \\
&
\end{aligned}
\)
[Approximation has been used because 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 simplification, \(n=1\)
\(
\text { Now, }(2 n-1) \frac{v}{4 l_1}=340 \text {. Substituting values }
\)
\(
(2 \times 1-1) v \times 100 / 4 \times 25.5=340 \text { or } v=346.8 \mathrm{~ms}^{-1}
\)

Hint

Here, \(\mathrm{L}=100 \mathrm{~cm}=1 \mathrm{~m}, {f}=2.53 \mathrm{k} \mathrm{Hz}=2.53 \times 10^3 \mathrm{~Hz}\)
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 ant mode is formed at each end.
Therefore, as is clear from Fig.
\(
\begin{aligned}
& L=\frac{\lambda}{4}+\frac{\lambda}{4}=\frac{\lambda}{2} \\
& \lambda=2 L=2 \mathrm{~m} \\
& \text { As } \quad v=f \lambda \\
& \therefore \quad v=2.53 \times 10^3 \times 2 \\
& =5.06 \times 10^3 \mathrm{~ms}^{-1} \\
&
\end{aligned}
\)

Hint

Answer: Here length of pipe, \(l=20 \mathrm{~cm}=0.20 \mathrm{~m}\), frequency \({f}=430 \mathrm{~Hz}\) and speed of sound in air \({v}=340\) \(\mathrm{ms}^{-1}\)
For closed end pipe, \(f=\frac{(2 n-1) v}{4 l}\), where \(n=1,2,3 \ldots \ldots \ldots\).
\(
\begin{aligned}
& \therefore \quad(2 n-1)=\frac{4 f l}{v}=\frac{4 \times 430 \times 0.20}{340}=1.02 \\
& \Rightarrow \quad 2 n=1.02+1=2.02 \Rightarrow n=\frac{0.20}{2}=1.01
\end{aligned}
\)
Hence, resonance can occur only for first (or fundamental) mode of vibration.
As for an open pipe \(f=\frac{n v}{2 l}\), where \(n=1,2,3 \ldots \ldots \ldots\).
\(
\therefore \quad n=\frac{2 l f}{v}=\frac{2 \times 430 \times 0.20}{340}=0.51 \text {. }
\)
As \(n<1\), hence, in this case resonance position cannot be obtained.

Hint

Frequency of string \(A_1 f_A=324 \mathrm{~Hz}\)
Frequency of string \(B=f_B\)
Beat’s frequency, \(n=6 \mathrm{~Hz}\)
Beat’s Frequency is given as:
\(
\begin{aligned}
& n=\left|f_A \pm f_B\right| \\
& 6=324 \pm f_B \\
& F_B=330 \mathrm{~Hz} \text { or } 318 \mathrm{~Hz}
\end{aligned}
\)
The frequency decreases with a decrease in the tension in a string. This is because the frequency is directly proportional to the square root of the tension. It is given as:
\(
f \propto \sqrt{T}
\)
Hence the beat frequency cannot be \(330 \mathrm{~Hz}\)
\(
\therefore f_n=318 \mathrm{~Hz}
\)

Hint

(a) In a sound wave, a decrease in displacement i.e., displacement node causes an increase in the pressure there i.e., a pressure antinode is formed. Also, an increase in displacement is due to the decrease in pressure.
(b) Bats emit ultrasonic waves of high frequency from their mouths. These waves after being reflected back from the obstacles on their path are observed by the bats. These waves give them an idea of distance, direction, nature and size of the obstacles.
(c) The quality of a violin note is different from the quality of sitar. Therefore, they emit different harmonics which can be observed by human ear and used to differentiate between the two notes.
(d) This is due to the fact that gases have only the bulk modulus of elasticity whereas solids have both, the shear modulus as well as the bulk modulus of elasticity.
(e) A pulse of sound consists of a combination of waves of different wavelength. In a dispersive medium, these waves travel with different velocities giving rise to the distortion in the wave.

Hint

Frequency of whistle, \(f=400 \mathrm{~Hz}\); speed of sound, \(v=340 \mathrm{~ms}^{-1}\), speed of train, \(v_{\mathrm{s}}=10 \mathrm{~ms}^1\)
(i) (a) When the train approaches the platform (i.e., the observer at rest),
\(
\mathrm{f}^{\prime}=\frac{v}{v-v_s} \times f=\frac{340}{340-10} \times 400=412 \mathrm{~Hz} \text {. }
\)
\(
f^{\prime}=\frac{v}{v+v_s} \times f=\frac{340}{340+10} \times 400=389 \mathrm{~Hz}
\)
\(
\text { (ii) The speed of sound in each case does not change.It is } 340 \mathrm{~ms}^{-1} \text { in each case. }
\)

Hint

Here actual frequency of whistle of train \(f=400 \mathrm{~Hz}\), speed of sound in still air \(v=340 \mathrm{~ms}^{-1}\).
As wind is blowing in the direction from the yard to the station with a speed of \(v_m=10 \mathrm{~ms}^{-1}\)
\(\therefore\) For an observer standing on the platform, the effective speed of sound
\(
v^{\prime}=v+v_{\mathrm{m}}=340+10=350 \mathrm{~ms}^{-1}
\)
As there is no relative motion between the sound source (rail engine) and the observer, the frequency of sound for the observer, \(f=400 \mathrm{~Hz}\)
\(
\begin{aligned}
\therefore \text { Wavelength of sound heard by the observer } \lambda^{\prime} & =\frac{v^{\prime}}{f}=\frac{350}{400} \\
& =0.875 \mathrm{~m} .
\end{aligned}
\)
The situation is not identical to the case when the air is still and the observer runs towards the yard at a speed of \(v_0=10 \mathrm{~ms}^{-1}\). In this situation as the medium is at rest. Hence \(v^{\prime}=v\) \(=340 \mathrm{~ms}^{-1}\).
\(
\begin{aligned}
& f^{\prime}=\frac{v+v_0}{v} f=\frac{340+10}{340} \times 400=412 \mathrm{~Hz} \\
& \lambda^{\prime}=\lambda=\frac{v}{f}=\frac{340}{400}=0.85 \mathrm{~m}
\end{aligned}
\)

No, because in this case, with respect to the medium, both the observer and the source are in motion.

Hint

Here, frequency of SONAR (source) \(=40.0 \mathrm{kHz}=40 \times 10^3 \mathrm{~Hz}\)
Speed of sound waves, \(v=1450 \mathrm{~ms}^{-1}\)
Speed of observers, \(v_0=360 \mathrm{~km} / \mathrm{h}=360 \times \frac{5}{18}=100 \mathrm{~ms}^{-1}\).
Since the source is at rest and observer moves towards the source (SONAR),
\(
\therefore \quad f^{\prime}=\frac{v+v_0}{v} \cdot f=\frac{1450+100}{1450} \times 40 \times 10^3=4.276 \times 10^4 \mathrm{~Hz}
\)
This frequency \(\left(f^{\prime}\right)\) is reflected by the enemy ship and is observed by the SONAR (which now acts as observer). Therefore, in this case, \(v_{\mathrm{s}}=360 \mathrm{~km} / \mathrm{h}=100 \mathrm{~ms}^{-1}\).
\(
\therefore \text { Apparent frequency, } \begin{aligned}
f^{\prime \prime} & =\frac{v}{v-v_s} f^{\prime}=\frac{1450}{1450-100} \times 4.276 \times 10^4 \\
& =4.59 \times 10^4 \mathrm{~Hz}=45.9 \mathrm{kHz} .
\end{aligned}
\)

Hint

Here speed of \(S\) wave, \(v_s=4.0 \mathrm{~km} \mathrm{~s}^{-1}\) and speed of \(P\) wave, \(v_p=8.0 \mathrm{~km} \mathrm{~s}^{-1}\). Time gap between \(P\) and \(S\) waves reaching the resimograph, \(t=40 \mathrm{~min}=240 \mathrm{~s}\).
Let distance of earthquake centre \(=\mathrm{sKm}\)
\(\therefore \quad t=t_s-t_p=\frac{S}{v_s}-\frac{S}{v_p}=\frac{S}{4.0}-\frac{S}{8.0}=\frac{S}{8.0}=240 \mathrm{~s}\)
or
\(s=240 \times 8.0=1920 \mathrm{~km}\).

Hint

Here, the frequency of sound emitted by the bat, \(f=40 \mathrm{kHz}\). Velocity of bat, \(v_s=0.03 \mathrm{v}\), where \(v\) is velocity of sound. Apparent frequency of sound striking the wall
\(
\begin{aligned}
f^{\prime} & =\frac{v}{v-v_s} \times f=\frac{v}{v-0.03 v} \times 40 \mathrm{kHz} \\
& =\frac{40}{0.97} \mathrm{kHz}
\end{aligned}
\)
This frequency is reflected by the wall and is received by the bat moving towards the wall. So \(v_s=0\),
\(
\begin{aligned}
v_L & =0.03 v . \\
f^{\prime \prime} & =\frac{\left(v+v_L\right)}{v} \times f^{\prime}=\frac{(v+0.03 v)}{v}\left(\frac{40}{0.97}\right) \\
& =\frac{1.03}{0.97} \times 40 \mathrm{kHz}=42.47 \mathrm{kHz} .
\end{aligned}
\)

Hint

Given,
Speed of the wave pulse passing on a string in the negative \(x\)-direction \(=40 \mathrm{cms}^{-1}\) As the speed of the wave is constant, the location of the maximum after \(5 \mathrm{~s}\) will be
\(
\begin{aligned}
s & =v \times t \\
& =40 \times 5 \\
& =200 \mathrm{~cm} \text { (along the negative } x \text {-axis) }
\end{aligned}
\)
Therefore, the required maximum will be located after \(x=-2 \mathrm{~m}\).

Hint

\(
\text { At } t=t_0, f\left(x, t_0\right)=A \sin (x / a) \dots(1)
\)
For a wave traveling in the positive \(x\)-direction, the general equation is given by
\(
y=A sin\left(\frac{x}{a}-\frac{t}{T}\right)
\)
Putting \(t=-t_0\) and comparing with equation (1), we get
\(
\begin{aligned}
& \Rightarrow f(x, 0)=A \sin \left\{(x / a)+\left(t_0 / T\right)\right\} \\
& \Rightarrow f(x, t)=A \sin \left\{(x / a)+\left(t_0 / T\right)-(t / T)\right\}
\end{aligned}
\)
As \(\mathrm{T}={a} / {v} (\mathrm{a}=\) wave length, \(v=\) speed of the wave \()\)
\(
\begin{aligned}
& \Rightarrow \mathrm{y}=A \sin \left(\frac{x}{a}+\frac{t_0}{(a / v)}-\frac{t}{(a / v)}\right) \\
& =A \sin \left(\frac{x+v\left(t_0-t\right)}{a}\right) \\
& \Rightarrow y=A \sin \left[\frac{x-v\left(t-t_0\right)}{a}\right]
\end{aligned}
\)

Hint

Given, Wave speed \((v)=10 \mathrm{~ms}^{-1}\)
Time period \((T)=20 \mathrm{~ms}\) \(=20 \times 10^{-3}=2 \times 10^{-2} s\)
(a) Wavelength of the wave:
\(
\begin{aligned}
& \lambda=v t=10 \times 2 \times 10^{-2} \\
& =0.02 \mathrm{~m}=20 \mathrm{~cm}
\end{aligned}
\)
(b) Displacement of the particle at a certain instant:
\(
\begin{aligned}
& y=a \sin (\omega t-k x) \\
& \Rightarrow 1.5=a \sin (\omega t-k x)
\end{aligned}
\)
Phase difference of the particle at a distance \(x=10 \mathrm{~cm}\) :
\(
\phi=\frac{2 \pi x}{\lambda}=2 \pi \times \frac{10}{20}=\pi
\)
The displacement is given by
\(
\begin{aligned}
& y^{\prime}=a \sin (\omega t-k x+\pi) \\
& =-a \sin (\omega t-k x)=-1.5 \mathrm{~mm} \\
& \therefore \text { Displacement }=-1.5 \mathrm{~mm}
\end{aligned}
\)

Hint

mass \(=5 \mathrm{~g}\), length \(l=64 \mathrm{~cm}\)
\(\therefore\) mass per unit length \(=m=5 / 64 \mathrm{~g} / \mathrm{cm}\)
\(\therefore\) Tension, \(\mathrm{T}=8 \mathrm{~N}=8 \times 10^5\) dyne
\(
\begin{aligned}
& \text { Speed, } v=\sqrt{\left(\frac{T}{m}\right)} \\
& =\sqrt{\frac{\left(8 \times 10^5 \times 64\right)}{5}} \\
& =3200 \mathrm{~cm} / \mathrm{s}=32 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

Given, Length of the string \(=20 \mathrm{~cm}\)
Linear mass density of the string \(=0.40 \mathrm{~g} \mathrm{~cm}^{-1}\)
Applied tension \(=16 \mathrm{~N}=16 \times 10^5 \mathrm{dyne}\)
The velocity of the wave:
\(
\begin{aligned}
& v=\sqrt{\left(\frac{T}{m}\right)} \\
& =\sqrt{\frac{\left(16 \times 10^5\right)}{0.4}} \\
& =2000 \mathrm{~cm} / \mathrm{s} \\
& \therefore \text { Time taken to reach the other end } \\
& =\frac{20}{2000}=0.01 \mathrm{~s}
\end{aligned}
\)
Time taken to see the pulse again in the original position
\(
=0.01 \times 2=0.02 \mathrm{~s}
\)
(b) At \(\mathrm{t}=0.01 \mathrm{~s}\), there will be a trough at the right end as it is reflected.

Hint

Let the velocity of the 1st string will be \(V_1\)
\(
\Rightarrow \mathrm{v}_1=\sqrt{\frac{\mathrm{T}}{\mathrm{m}_1}}
\)
because \(\mathrm{m}_1=\) mass per unit lengh \(=\frac{\rho_1 a_1 l_1}{l_1}=\rho_1 a_1\)
where \(a_1=\) area pf the cross section
\(
\Rightarrow \mathrm{v}_1=\sqrt{\frac{\mathrm{T}}{\rho_1 \mathrm{a}_1}}
\)
Let the velocity of the 2nd string will be \(\mathrm{V}_2\)
\(
\begin{aligned}
& \Rightarrow \mathrm{v}_2=\sqrt{\frac{\mathrm{T}}{\mathrm{m}_2}} \\
& \Rightarrow \mathrm{v}_2=\sqrt{\frac{\mathrm{T}}{\rho_2 \mathrm{a}_2}}
\end{aligned}
\)
given that \(V_1=V_2\)
\(
\begin{aligned}
& \Rightarrow \sqrt{\frac{T}{\rho_1 a_1}}=2 \sqrt{\frac{T}{\rho_2 a_2}} \\
& \Rightarrow \frac{\rho_1}{\rho_2}=\frac{1}{4}
\end{aligned}
\)
\(
\Rightarrow \rho_1: \rho_2=1: 4\left(\text { because } a_1=a_2\right)
\)

Hint

\(\mathrm{m}=\) mass per unit length \(=1.2 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\) \(Y=(0.02 \mathrm{~m}) \sin \left[\left(1.0 \mathrm{~m}^{-1}\right) \mathrm{x}+\left(30 \mathrm{~s}^{-1}\right) \mathrm{t}\right]\)
Here, \(\mathrm{k}=1 \mathrm{~m}^{-1}=2 \pi / \lambda\)
\(\omega=30 \mathrm{~s}^{-1}=2 \pi \mathrm{f}\)
\(\therefore\) velocity of the wave in the stretched string
\(v=\lambda f=\omega / \mathrm{k}=30 / {1}=30 \mathrm{~m} / \mathrm{s}\)
\(
\Rightarrow v=\sqrt{\mathrm{T} / \mathrm{m}} \Rightarrow 30 \sqrt{(\mathrm{T} / 1.2) \times 10^{-4} \mathrm{~N}}
\)
\(\Rightarrow \mathrm{T}=10.8 \times 10^{-2} \mathrm{~N} \Rightarrow \mathrm{T}=1.08 \times 10^{-1}\) Newton.

Hint

Given, Length of the string, \(L=40 \mathrm{~cm}\)
Mass of the string \(=10 \mathrm{gm}\)
Mass per unit length
\(
=\frac{10}{40}=\frac{1}{4}(\mathrm{gm} / \mathrm{cm})
\)
Spring constant, \(k=160 \mathrm{~N} / \mathrm{m}\)
Deflection, \(x=1 \mathrm{~cm}\)
\(=0.01 \mathrm{~m}\)
Tension, \(T=k x=160 \times 0.01\)
\(\Rightarrow T=1.6 N=16 \times 10^4 d y n e\)
Now,
\(
\begin{aligned}
& v=\sqrt{\left(\frac{T}{m}\right)}=\sqrt{\left(\frac{16 \times 10^4}{\frac{1}{4}}\right)} \\
& \Rightarrow v=8 \times 10^2 \mathrm{~cm} / \mathrm{s}=800 \mathrm{~cm} / \mathrm{s}
\end{aligned}
\)
\(\therefore\) Time taken by the pulse to reach the spring,
\(
t=\frac{40}{800}=\frac{1}{20}=0.05 \mathrm{~s}
\)

Hint

\(
m_1=m_2=3.2 \mathrm{~kg}
\)
mass per unit length of \(A B=10 \mathrm{~g} / \mathrm{mt}=0.01 \mathrm{~kg} \cdot \mathrm{mt}\)
mass per unit length of \(C D=8 \mathrm{~g} / \mathrm{mt}=0.008 \mathrm{~kg} / \mathrm{mt}\)
for the string \(\mathrm{CD}, \mathrm{T}=3.2 \times \mathrm{g}\)
\(
\Rightarrow {v}=\sqrt{(\mathrm{T} / \mathrm{m})}=\sqrt{(3.2 \times 10) / 0.008}=\sqrt{\left(32 \times 10^3\right) / 8}=2 \times 10 \sqrt{10}=20 \times 3.14=63 \mathrm{~m} / \mathrm{s}
\)
for the string \(A B, T=2 \times 3.2 \mathrm{~g}=6.4 \times \mathrm{g}=64 \mathrm{~N}\)
\(
\Rightarrow v=\sqrt{(T / \mathrm{m})}=\sqrt{(64 / 0.01)}=\sqrt{6400}=80 \mathrm{~m} / \mathrm{s}
\)

Hint

Given,
Mass of the block \(=2 \mathrm{~kg}\)
Total length of the string \(=2+0.25=2.25 \mathrm{m}\)
Mass per unit length of the string:
\(
\begin{aligned}
& m=\frac{4.5 \times 10^{-3}}{2.25} \\
& =2 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \\
& T=2 g=20 \mathrm{~N}
\end{aligned}
\)
Wave speed,\(v=\sqrt{\left(\frac{T}{m}\right)}\)
\(
\begin{aligned}
& =\sqrt{\frac{20}{\left(2 \times 10^{-3}\right)}} \\
& =\sqrt{10^4} \\
& =10^2 \mathrm{~m} / \mathrm{s}=100 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Time taken by the disturbance to reach the pulley:
\(
\begin{aligned}
& t=\left(\frac{s}{v}\right) \\
& =\frac{2}{100}=0.02 \mathrm{~s}
\end{aligned}
\)

Hint

Given, Mass of the block \(=4.0 \mathrm{~kg}\)
Linear mass density,
\(
m=19.2 \times 10^{-3} \mathrm{~kg} / \mathrm{m}
\)
From the free body diagram,
\(
\begin{aligned}
& T-4 g-4 a=0 \\
& \Rightarrow T=4(a+g) \\
& =4(2+10)=48 \mathrm{~N}
\end{aligned}
\)
Wavespeed,\(v=\sqrt{\left(\frac{T}{m}\right)}\)
\(
\begin{aligned}
& =\sqrt{\frac{48}{19.2 \times 10^3}} \\
& =\sqrt{\left(2.5 \times 10^{-3}\right)}=50 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

Given, Speed of the transverse pulse when the car is at rest, \(v_1=60 \mathrm{~cm} \mathrm{~s}^{-1}\)
Speed of the transverse pulse when the car accelerates, \(v_2=62 \mathrm{~cm} \mathrm{~s}^{-1}\)
Let the Mass of the heavy ball suspended from the ceiling \(=M\)
Mass per unit length \(=m\)
Now, Wave speed, \(v=\sqrt{\left(\frac{T}{m}\right)}=\sqrt{\left(\frac{M g}{m}\right)}\)
When car is at rest: Tension in the string, \(T=M g\)
\(
\begin{aligned}
& \Rightarrow v_1=\sqrt{\left(\frac{M g}{m}\right)} \\
& \Rightarrow \frac{M g}{m}=(60)^2 \ldots(i)
\end{aligned}
\)
When car is having acceleration: Tension, and from the free body diagram above
\(
\begin{aligned}
& T=\sqrt{(M a)^2+(M g)^2} \\
& \text { Again, } v_2=\sqrt{\left(\frac{T}{m}\right)} \\
& \Rightarrow 62=\frac{\left[(M a)^2+(M g)^2\right]^{1 / 4}}{m^{1 / 2}} \\
& \Rightarrow \sqrt{\frac{\left[(M a)^2+(M g)^2\right]}{m}}=(62)^2 \ldots(ii)
\end{aligned}
\)
From equations (i) and (ii), we get:
\(
\begin{aligned}
& \left(\frac{M g}{m}\right) \times \frac{m}{\sqrt{(M a)^2+(M g)^2}}=\left(\frac{60}{62}\right)^2 \\
& \Rightarrow a=3.76 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)

Hint

Let, \({V}=\) Linear velocity of the string
\({m}=\) Mass per unit length of the the string.
\(R=\) Radius of the loop
\(\omega=\) Angular velocity
Consider one half of the string, as shown in the figure.
The half loop experiences centrifugal force at every point (away from the centre) balanced by tension 2T. Consider an element of angular part \(d \theta\) at angle \(\theta\).
So, the Length of the element \(=R d \theta\), \(\text { mass }=m R d \theta\)
Centrifugal force experienced by the element \(=(m R d \theta) \omega^2 R\)
Resolving the centrifugal force into rectangular components, Since the horizontal components cancel each other, the net force on the two symmetric elements is given as
\(
d F=2 m R^2 d \theta \omega^2 \sin \theta [\text { horizontal components cancels each other }]
\)
Now we will integrate the force from 0 to \(\frac{\pi}{2}\)
\(
\int d F=\int_{0 \pi}^{\frac{\pi}{2}} 2 m R^2 d \theta \omega^2 \sin \theta
\)
Total force we will get by integrating
\(
\begin{aligned}
& F=2 m R^2 \omega^2[-\cos \theta] \\
& \Rightarrow 2 m R^2 \omega^2
\end{aligned}
\)
As we can see in the diagram force is equal to twice tension
\(
\begin{aligned}
& 2 T=2 m R^2 \omega^2 \\
& \Rightarrow T=m R^2 \omega^2
\end{aligned}
\)
Velocity of the transverse vibration is given as
\(
\begin{aligned}
V^{\prime} & =\sqrt{\left(\frac{T}{m}\right)} \\
V^{\prime} & =\sqrt{\left(\frac{m R^2 \omega^2}{m}\right)}=\omega R
\end{aligned}
\)
Linear velocity of the string, \(V=\omega R\)
\(\therefore\) Speed of the disturbance, \(V^{\prime}={V}\)

Hint

Given,
Linear density of each of two long strings \(A\) and \(B, m=1.2 \times 10^{-2} \mathrm{~kg} / \mathrm{m}\)
String \(A\) is stretched by tension \(T_a=4.8 \mathrm{~N}\).
String \(B\) is stretched by tension \(T_b=7.5 \mathrm{~N}\).
Let \(v_a\) and \(v_b\) be the speeds of the waves in strings \(A\) and \(B\).
Now,
\(
\begin{aligned}
& v_a=\sqrt{\frac{T_a}{m}} \\
& \Rightarrow v_a=\sqrt{\frac{(4.8)}{\left(1.2 \times 10^{-2}\right)}}=20 \mathrm{~m} / \mathrm{s} \\
& v_b=\sqrt{\frac{T_b}{m}} \\
& \Rightarrow v_b=\sqrt{\frac{7.5}{\left(1.2 \times 10^{-2}\right)}}=25 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\(
\text { The pulse produced on } A \text { and } B \text { meets when, } x_A=x_B
\)
Let ‘ \(t\) ‘ be the time at which both pulses meet
So,
\(
\begin{aligned}
& \mathrm{x}_A=\mathrm{x}_B \Rightarrow 20 \mathrm{t}=25(\mathrm{t}-0.02) \\
& \Rightarrow 5 \mathrm{t}=0.5=0.1 \mathrm{~s}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& r=0.5 \mathrm{~mm} \\
& =0.5 \times 10^{-3} \mathrm{~m}, \\
& f=100 \mathrm{~Hz} \\
& T=100 \mathrm{~N}, \mathrm{V}=100 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\(
\begin{aligned}
& v=\sqrt{\left(\frac{T}{m}\right)} \\
& \Rightarrow v^2=\left(\frac{T}{m}\right) \\
& \Rightarrow m=\frac{T}{v^2}=\frac{100}{(100)^2} \\
& =0.01=0.01 \mathrm{~kg} / \mathrm{m}
\end{aligned}
\)
\(
\begin{aligned}
& P_{\text {ave }}=2 \pi^2 \mathrm{mvr}^2 \mathrm{f}^2 \\
& =2(3.14)^2(0.01) \times 100 \times\left(0.5 \times 10^{-3}\right)^2 \times(100) \\
& =2 \times 9.86 \times 0.25 \times 10^{-6} \times 10^4 \\
& =19.7 \times 0.0025=0.049 \text { watt } \\
& =49 \times 10^{-3} \text { watt }=49 \mathrm{~mW}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& A=1 \mathrm{~mm}=10^{-3} \mathrm{~m}, \mathrm{~m}=6 \mathrm{~g} / \mathrm{m}=6 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \\
& \mathrm{T}=60 \mathrm{~N}, \mathrm{f}=200 \mathrm{~Hz}
\end{aligned}
\)
\(
\therefore V=\sqrt{T / \mathrm{m}}=100 \mathrm{~m} / \mathrm{s}
\)
\(
\text { (a) } P_{\text {average }}=2 \pi^2 \mathrm{mv} \mathrm{A}^2 \mathrm{f}^2=0.47 \mathrm{~W}
\)
(b) Length of the string is \(2 \mathrm{~m}\). So, \(\mathrm{t}=2 / 100=0.02 \mathrm{sec}\). Energy \(=2 \pi^2 \mathrm{mvf}^2 \mathrm{~A}^2 \mathrm{t}=9.46 \mathrm{~mJ}\).

Hint

Phase difference \(\phi=\pi / 2\)
\(\mathrm{f}\) and \(\lambda\) are same. So, \(\omega\) is same.
\(y_1=r \sin w t, y_2=r \sin (w t+\pi / 2)\)
From the principle of superposition
\(
\begin{aligned}
& y=y_1+y_2 \rightarrow=r \sin w t+r \sin (w t+\pi / 2) \\
& =r[\sin w t+\sin (w t+\pi / 2)] \\
& =r[2 \sin \{(w t+w t+\pi / 2) / 2\} \cos \{(w t-w t-\pi / 2) / 2\}] \\
& \Rightarrow \mathrm{y}=2 \mathrm{r} \sin (\mathrm{wt}+\pi / 4) \cos (-\pi / 4) \\
& \text { Resultant amplitude }=\sqrt{2} r=4 \sqrt{2} \mathrm{~mm} \quad(\text { because } r=4 \mathrm{~mm}) \\
&
\end{aligned}
\)

Hint

Length of a stretched string \((L)=1 \mathrm{~m}\)
Wave speed \((v)=60 \mathrm{~m} / \mathrm{s}\)
Fundamental frequency \(\left(f_0\right)\) of vibration is given as follows:
\(
\begin{aligned}
& f_0=\frac{v}{2 L} \\
& =\frac{60}{2} \times 1 \\
& =30 \mathrm{~s}^{-1}=30 \mathrm{~Hz}
\end{aligned}
\)

Hint

Given Length of the wire \((L)=2.00 \mathrm{~m}\)
Fundamental frequency of the vibration \(\left(f_0\right)=100 \mathrm{~Hz}\)
Applied tension \((T)=160 \mathrm{~N}\)
Fundamentalfrequency, \(f_0=\frac{1}{2 L} \sqrt{\left(\frac{T}{m}\right)}\)
\(
\begin{aligned}
& \Rightarrow 10=\frac{1}{4} \sqrt{\frac{160}{m}} \\
& \Rightarrow m=1 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \\
& \Rightarrow m=1 \mathrm{~g} / \mathrm{m}
\end{aligned}
\)
So, the linear mass density of the wire is \(1 \mathrm{~g} / \mathrm{m}\).

Hint

Given mass of the steel wire \(=4.0 \mathrm{~g}\)
Length of the steel wire \(=80 \mathrm{~cm}=0.80 \mathrm{~m}\)
Tension in the wire \(=50 \mathrm{~N}\)
Linear mass density \((m)\)
\(
=\left(\frac{4}{80}\right) \mathrm{g} / \mathrm{cm}=0.005 \mathrm{~kg} / \mathrm{m}
\)
Wave speed, \(v=\sqrt{\left(\frac{T}{m}\right)}\)
\(
=\sqrt{\left(\frac{50}{0.005}\right)}=100 \mathrm{~m} / \mathrm{s}
\)
Fundamental frequency,\(f_o=\frac{1}{2 L} \sqrt{\left(\frac{T}{m}\right)}\)
\(
\begin{aligned}
& =\frac{1}{2 \times 0.8} \times \sqrt{\left(\frac{50}{0.005}\right)} \\
& =\frac{100}{2 \times 0.8}=62.5 \mathrm{~Hz}
\end{aligned}
\)
First harmonic \(=62.5 \mathrm{~Hz}\)
If \(f_4\) =frequency of the fourth harmonic:
\(
\begin{aligned}
& \Rightarrow f_4=4 f_0=62.5 \times 4 \\
& \Rightarrow f_4=250 \mathrm{~Hz}
\end{aligned}
\)
Wavelength of the fourth harmonic,\(\lambda_4=\frac{v}{f_4}=\frac{100}{250}\)
\(
\Rightarrow \lambda_4=0.4 \mathrm{~m}=40 \mathrm{~cm}
\)

Hint

\(
\begin{aligned}
& l=90 \mathrm{~cm}=0.9 \mathrm{~m} \\
& \mathrm{~m}=(6 / 90) \mathrm{g} / \mathrm{cm}=(6 / 900) \mathrm{kg} / \mathrm{m} \\
& \mathrm{f}=261.63 \mathrm{~Hz} \\
& \mathrm{f}=1 / 2l \sqrt{(\mathrm{T} / \mathrm{m})} \Rightarrow T=1478.52 \mathrm{~N}\approx 1480 N.
\end{aligned}
\)

Hint

First harmonic be \(f_0\), second harmonic be \(f_1\)
\(
\begin{aligned}
& \therefore \mathrm{f}_1=2 \mathrm{f}_0 \\
& \Rightarrow \mathrm{f}_0=\mathrm{f}_1 / 2 \\
& \mathrm{f}_1=256 \mathrm{~Hz} \\
& \therefore 1^{\text {st }} \text { harmonic or fundamental frequency } \\
& \mathrm{f}_0=\mathrm{f}_1 / 2=256 / 2=128 \mathrm{~Hz}
\end{aligned}
\)
When the fundamental wave is produced, we have:
\(
\begin{aligned}
& \frac{\lambda}{2}=1.5 \mathrm{~m} \\
& \Rightarrow \lambda=3 \mathrm{~m} \\
& \text { Wave speed, } v=f_0 \lambda \\
& \Rightarrow v=128 \times 3=384 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

Given Length of the wire between two pulleys \((L)=1.5 \mathrm{~m}\)
Mass of the wire \(=12 \mathrm{~gm}\)
Mass per unit length, \(m=\frac{12}{1.5} \mathrm{~g} / \mathrm{m}\)
\(=8 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\)
Tensioninthewire, \(T=9 \times g\)
\(=90 \mathrm{~N}\)
Fundamental frequency is given by:
\(
f_0=\frac{1}{2 L} \sqrt{\left(\frac{T}{m}\right)}
\)
For second harmonic (when two loops are produced):
\(
\begin{aligned}
& f_1=2 f_0=\frac{1}{1.5} \sqrt{\left(\frac{90}{8} \times 10^{-3}\right)} \\
& =\frac{(106.06)}{1.5} \\
& =70.7 H z \approx 70 \mathrm{~Hz}
\end{aligned}
\)

Hint

A string of mass \(40 \mathrm{~g}\) is attached to the tuning fork \(\mathrm{m}=\left(40 \times 10^{-3}\right) \mathrm{kg} / \mathrm{m}\)
The fork vibrates with \(\mathrm{f}=128 \mathrm{~Hz}\)
No. of loops formed, \((n)=4\)
\(
\begin{aligned}
& L=\frac{n \lambda}{2} \\
& \Rightarrow \lambda=\frac{2 L}{n}=\frac{2 \times 1}{4} \\
& \Rightarrow \lambda=0.5 m
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{v}=\mathrm{f} \lambda=128 \times 0.5=64 \mathrm{~m} / \mathrm{s} \\
& v=\sqrt{T / m} \Rightarrow T=v^2 \mathrm{~m}=163.84 \mathrm{~N} \Rightarrow 164 \mathrm{~N} \\
&
\end{aligned}
\)

Hint

Given Wire makes a resonant frequency of \(240 \mathrm{~Hz}\) and \(320 \mathrm{~Hz}\) when its both ends are fixed. Therefore, fundamental frequency \(\left(f_0\right)\) of the wire must be the factor of \(240 \mathrm{~Hz}\) and \(320 \mathrm{~Hz}\).
(a) Maximum value of fundamental frequency, \(f_0=80 \mathrm{~Hz}\)
(b) Wave speed \((v)=40 \mathrm{~m} / \mathrm{s}\)
And if \(\lambda\) is the wave length:
\(
\begin{aligned}
& \frac{\lambda}{2}=L \\
& \therefore v=\lambda \times f_0 \\
& \Rightarrow v=2 \times L \times f_0 \\
& \Rightarrow L=\frac{40}{2 \times 80} \\
& \Rightarrow L=\frac{1}{4} m=0.25 m = 25 cm
\end{aligned}
\)

Hint

Given separation between two consecutive nodes when the string vibrates in resonant mode \(=2.0 \mathrm{~cm}\),  Let there be \(n\) loops and \(\lambda\) be the wavelength.
\(\therefore\) \(\lambda=2 \times\) Separation between the consecutive nodes
\(
\lambda_1=2 \times 2=4 \mathrm{~cm}
\)
\(
\lambda_2=2 \times 1.6=3.2 \mathrm{~cm}
\)
The length of the wire is \(L\).
In the first case:
\(
L=\left(\frac{n \lambda_1}{2}\right)
\)
In the second case:
\(
\begin{aligned}
& L=(n+1) \frac{\lambda_2}{2} \\
& \Rightarrow \frac{n \lambda_1}{2}=(n+1) \frac{\lambda_2}{2} \\
& \Rightarrow n \times 4=(n+1)(3.2) \\
& \Rightarrow 4 n-3.2 n=3.2 \\
& \Rightarrow 0.8 n=3.2 \\
& \Rightarrow n=4 \therefore \text { length of the string } \\
& L=\frac{\left(n \lambda_1\right)}{2}=\frac{(4 \times 4)}{2}=8 \mathrm{~cm}
\end{aligned}
\)

Hint

Given Fundamental frequency \(\left(f_0\right)\) of the steel wire \(=200 \mathrm{~Hz}\)
Let the highest harmonic audible to the person be \(n\).
Frequency of the highest harmonic, \(f^{\prime}=14000 \mathrm{~Hz}\)
\(
\begin{aligned}
& \therefore f^{\prime}=n f_0 \quad \ldots(1) \\
& \frac{f^{\prime}}{f_0}=\frac{14000}{200} \\
& \Rightarrow \frac{n f_0}{f_0}=70 \\
& \Rightarrow n=70
\end{aligned}
\)
Thus, the highest harmonic audible to man is the \(70^{\text {th }}\) harmonic.

Hint

Given the tensions in the two wires are in the ratio of 2:1.
\(
\Rightarrow \frac{T_1}{T_2}=2
\)
Ratio of the radii is \(3: 1\).
\(
\Rightarrow \frac{r_1}{r_2}=3=\frac{D_1}{D_2}
\)
Density in the ratios of 1:2.
\(
\Rightarrow \frac{\rho_1}{\rho_2}=\frac{1}{2}
\)
Let the length of the wire be \(L\).
\(
\begin{aligned}
& \text { Frequency, } f=\frac{1}{L D} \sqrt{\frac{T}{\pi \rho}} \\
& \Rightarrow f_1=\frac{1}{L D_1} \sqrt{\frac{T_1}{\pi \rho_1}} \\
& \Rightarrow f_2=\frac{1}{L D_2} \sqrt{\frac{T_2}{\pi \rho_2}} v \\
& \therefore \frac{f_1}{f_2}=\frac{L D_2}{L D_1} \sqrt{\frac{T_1}{T_2}} \sqrt{\frac{\pi \rho_2}{\pi \rho_1}} \\
& \Rightarrow \frac{f_1}{f_2}=\frac{1}{3} \sqrt{\frac{2}{1} \times \frac{2}{1}} \\
& \Rightarrow f_1: f_2=2: 3
\end{aligned}
\)

Hint

Given Length of the \(\operatorname{rod}(L)=40 \mathrm{~cm}=0.40 \mathrm{~m}\)
Mass of the \(\operatorname{rod}(m)=1.2 \mathrm{~kg}\)
Let the mass of \(4.8 \mathrm{~kg}\) be placed at \(x\) distance from the left.
As per the question, frequency on the left side \(=f_0\)
Frequency on the right side \(=2 f_0\)
Let tension be \(T_l\) and \(T_r\) on the left and the right side, respectively.
\(
\begin{aligned}
& \therefore \frac{1}{2 L} \sqrt{\frac{T_l}{m}}=\frac{2}{2 L} \sqrt{\frac{T_r}{m}} \\
& \Rightarrow \sqrt{\frac{T_l}{T_r}}=2 \\
& \Rightarrow \frac{T_l}{T_r}=4 \ldots(1)
\end{aligned}
\)
From the free body diagram:

\(
\begin{aligned}
& T_l+T_r=48+12=60 N \\
& \Rightarrow 4 T_r+T_r=5 T_r=60 N[\text { using equation }(1)] \\
& \therefore T_r=12 N \\
& \text { and } T_l=48 N
\end{aligned}
\)
Now, taking moment about point A:

\(
\begin{aligned}
& T_r \times(0.4)=48 x+12 \times 0.2 \\
& \Rightarrow 4.8=48 x-2.4 \\
& \Rightarrow 4.8 x=2.4 \\
& \Rightarrow x=\frac{2.4}{4.8}=\frac{1}{20} m=5 \mathrm{~cm}
\end{aligned}
\)
Therefore, the mass should be placed at a distance of \(5 \mathrm{~cm}\) from the left end.

Hint

Given Length of the aluminium wire \(\left(L_a\right)=60 \mathrm{~cm}=0.60 \mathrm{~m}\)
Length of the steel wire \(\left(L_s\right)=80 \mathrm{~cm}=0.80 \mathrm{~m}\)
Tension produced \((T)=40 \mathrm{~N}\)
Area of cross-section of the aluminium wire \(\left(A_a\right)=1.0 \mathrm{~mm}^2\)
Area of cross-section of the steel wire \(\left(A_s\right)=3.0 \mathrm{~mm}^2\)
Density of aluminium \(\left(\rho_a\right)=2 \cdot 6 \mathrm{~g} \mathrm{~cm}^{-3}\)
Density of steel \(\left(\rho_{\mathrm{s}}\right)=7.8 \mathrm{~g} \mathrm{~cm}^{-3}\)
Mass per unit length of the steel, \(m_s=\rho_s \times A_s\)
\(
\begin{aligned}
& =7.8 \times 10^{-2} \mathrm{gm} / \mathrm{cm} \\
& =7.8 \times 10^{-3} \mathrm{~kg} / \mathrm{m}
\end{aligned}
\)
Mass per unit length of the aluminium, \(\quad \mathrm{m}_A=\rho_A A_A\)
\(
\begin{aligned}
& =2.6 \times 10^{-2} \times 3 \mathrm{gm} / \mathrm{cm} \\
& =7.8 \times 10^{-2} \mathrm{gm} / \mathrm{cm} \\
& =7.8 \times 10^{-3} \mathrm{~kg} / \mathrm{m}
\end{aligned}
\)
A node is always placed at the joint. Since aluminium and steel rod has same mass per unit length, the velocity of wave (\(v\)) in both of them is same.
Let \(v\) be the velocity of wave.
\(
\begin{aligned}
& \Rightarrow v=\left(\frac{T}{m}\right) \\
& =\sqrt{\left\{\frac{40}{7.8 \times 10^{-3}}\right\}} \\
& =\sqrt{\left(\frac{4 \times 10^4}{7.8}\right)} \\
& \Rightarrow v=71.6 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
For minimum frequency, there would be maximum wavelength.
For maximum wavelength, the minimum number of loops are to be produced.
\(\therefore\) Maximum distance of a loop \(=20 \mathrm{~cm}\)
\(\Rightarrow\) Wavelength, \(\lambda=2 \times 20=40 \mathrm{~cm}\)
or \(\lambda=0.4 \mathrm{~m}\)
\(
\therefore \text { Frequency, } f=\frac{v}{\lambda}=\frac{71.6}{0.4}=180 \mathrm{~Hz}
\)

Hint

Given the Equation of the standing wave:
\(
\begin{aligned}
& y=(0.4 \mathrm{~cm}) \sin \left[\left(0.314 \mathrm{~cm}^{-1}\right) x\right] \cos \left[\left(600 \pi \mathrm{s}^{-1}\right) t\right] \\
& \Rightarrow k=0.314=\frac{\pi}{10} \\
& \text { Also, } k=\frac{2 \pi}{\lambda} \\
& \Rightarrow \lambda=20 \mathrm{~cm}
\end{aligned}
\)
We know:
\(
L=\frac{n \lambda}{2}
\)
For the smallest length of teh string,as wavelength remains constant, the strng should vibrate in the fundamental frequency. So \(n=1\):
\(
\Rightarrow L=\frac{\lambda}{2}=\frac{20 \mathrm{~cm}}{2}=10 \mathrm{~cm}
\)
Therefore, the required length of the string is \(10 \mathrm{~cm}\).

Hint

Given Length of the wire \((L)=40 \mathrm{~cm}=0.40 \mathrm{~m}\)
Mass of the wire \(=3.2 \mathrm{~g}=0.003 \mathrm{~kg}\)
Distance between the two fixed supports of the wire \(=40.05 \mathrm{~cm}\)
Fundamental mode frequency \(=220 \mathrm{~Hz}\)
Therefore, linear mass density of the wire \((m)\) is given by:
\(
m=\frac{0.0032}{0.4}=8 \times 10^{-3} \mathrm{~kg} / \mathrm{m}
\)
Change in length, \(\Delta L=40.05-40=0.05 \mathrm{~cm}\)
\(
\begin{aligned}
& =0.05 \times 10^{-2} \mathrm{~m} \\
& \text { Strain }=\frac{\Delta L}{L}=\frac{0.05 \times 10^{-2}}{0.4} \\
& =0.125 \times 10^{-2}
\end{aligned}
\)
\(
\begin{aligned}
& f_0=\frac{1}{2 L} \sqrt{\frac{T}{m}} \\
& =\frac{1}{2 \times(0.4005)} \sqrt{\frac{T}{8 \times 10^{-3}}} \\
& \Rightarrow 220 \times 220=\left[\frac{1}{(0.801)^2}\right] \times T \times\left(\frac{10^3}{8}\right) \\
& \Rightarrow T \times 1000=220 \times 220 \times 0.641 \times 0.8 \\
& \Rightarrow T=248.19 \mathrm{~N}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Young’s modulus, } Y=\frac{\text { stress }}{\text { strain }} \\
& =\frac{248.19 \times 10^6}{0.125 \times 10^{-2}} \\
& \Rightarrow Y=19852 \times 10^8 \\
& =1.98 \times 10^{11} \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)
Hence, the required Young’s modulus of the wire is
\(
1.98 \times 10^{11} \mathrm{~N} / \mathrm{m}^2
\)

Hint

Density of the block \(=\rho\)
Volume of block \(=\mathrm{V}\)
\(\therefore\) Weight of the block is, \(W=\rho V g\)
\(\therefore\) Tension in the string, \(T=W\)
The tuning fork resonates with different frequencies in the two cases.
Let the tenth harmonic be \(f_{10}\).
\(
\begin{aligned}
& f_{11}=\frac{11}{2 L} \sqrt{\frac{T^{\prime}}{m}} \\
& =\frac{11}{2 L} \sqrt{\frac{\left(\rho-\rho_w\right) V g}{m}}
\end{aligned}
\)
The frequency ( \(f\) ) of the tuning fork is the same.
\(
\begin{aligned}
& \therefore f_{10}=f_{11} \\
& \Rightarrow \frac{10}{2 L} \sqrt{\frac{\rho V g}{m}}=\frac{11}{2 L} \sqrt{\frac{\left(\rho-\rho_\omega\right) V g}{m}} \\
& \Rightarrow \frac{100}{121}=\frac{\rho-1}{\rho}\left(\text { because }\rho_\omega=1 \mathrm{gm} / \mathrm{cc}\right) \\
& \Rightarrow 100 \rho=121 \rho-121 \\
& \Rightarrow \rho=\frac{121}{21}=5.8 \mathrm{gm} / \mathrm{cc} \\
& =5.8 \times 10^3 \mathrm{~kg} / \mathrm{m}^3
\end{aligned}
\)
Therefore, the required density is \(5.8 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\)

Hint

Let \(T\) be the tension in the string and \(m\) be the mass per unit length of the heavy string.
In the first part of the question, the heavy string is fixed at only one end.
So, the lowest frequency is given by:
\(f_0=\frac{1}{4 L} \sqrt{\frac{T}{m}} \ldots(1)\)
When the movable support is pushed by \(10 \mathrm{~cm}\) to the right, the joint is placed on the pulley and the heavy string becomes fixed at both ends (keeping \(T\) and \(m\) same).
Now, the lowest frequency is given by:
\(
f_0^{\prime}=\frac{1}{2 L} \sqrt{\frac{T}{m}} \ldots(2)
\)
Dividing equation (2) by equation (1), we get:
\(
f_0^{\prime}=2 f_0=240 \mathrm{~Hz}
\)

Alternate explanation:

Initially because the end \(A\) is free, an antinode will be formed.
So, \(l=\lambda_1 / 4\)
Again, if the movable support is pushed to the right by \(10 \mathrm{~m}\) so that the joint is placed on the pulley, a node will be formed there.
So, \(l=\lambda_2 / 2\)
Since the tension remains the same in both cases, velocity remains the same.
As the wavelength is reduced by half, the frequency will become twice as that of \(120 \mathrm{~Hz}\) i.e. \(240 \mathrm{~Hz}\).