Chapter 14: Waves

Ans: The general equation of stationary wave is given by:

\[y(x,t)=Asin(kx)cos(\omega t)\]

This given equation is similar to the equation of stationary wave:

\[y(x,t)=0.06\sin \frac{2\pi x}{3}\cos (120\pi t)\]

Hence, the given function is a stationary wave. 

Ans: The transverse displacement of the string is given as:

\[y(x,t)=0.06\sin \left( \frac{2\pi }{3}x \right)\cos (120\pi t)\] 

From above equation, $k=\frac{2\pi }{3}$

\[\therefore \] Wavelength, $\lambda =\frac{2\pi }{k}=\frac{2\pi }{\frac{2\pi }{3}}=3m$

It is given that:

\[120\pi =2\pi \nu \]

Frequency, \[\nu =60\]Hz

Wave speed, \[v=\nu \lambda =60\times 3=180m{{s}^{-1}}\]

Ans: The transverse wave’s velocity travelling in a string is given by the relation:

\[v=\sqrt{\frac{T}{\mu }}\]    ……(i)

Where, Velocity of the transverse wave, \[v=180m{{s}^{-1}}\]

Mass of the string, \[m=3.0\times {{10}^{-2}}kg\]

Length of the string, \[l\text{ }=\text{ }1.5\text{ }m\]

Mass per unit length of the string, \[\mu =\frac{m}{l}=\frac{3.0}{1.5}\times {{10}^{-2}}=2\times {{10}^{-2}}kg{{m}^{-1}}\]

Tension in the string \[=\text{ }T\]

From equation (i), tension can be obtained as:

\[T={{v}^{2}}\mu \]

\[\Rightarrow T={{(180)}^{2}}\times 2\times {{10}^{-2}}\]

\[\Rightarrow T=648N\]

The tension in string is $648N$.

Ans: Yes, all the points on the string vibrate with the same frequency, except at the nodes which are having zero frequency.

Ans: Yes, all the points in any oscillating loop have the same phase, except at the nodes.

Ans: No, all the points in any oscillating loop have different vibration amplitudes.

Ans: The given equation is:

\[y(x,t)=0.06\sin \left( \frac{2\pi }{3}x \right)\cos \left( 120\pi t \right)\]

For x = 0.375 m and t = 0

Amplitude\[=0.06\sin \left( \frac{2\pi }{3}x \right)\cos 0\]

\[\Rightarrow a=0.06\sin \left( \frac{2\pi }{3}\times 0.375 \right)\times 1\]

\[\Rightarrow a=0.06\sin \left( 0.25\pi  \right)=0.06\sin \left( \frac{\pi }{4} \right)\]

\[\Rightarrow a=0.06\times \frac{1}{\sqrt{2}}=0.042m\]

The value of amplitude is \[0.042m\].

Ans: This equation demonstrates a stationary wave because the harmonic terms \[kx\] and \[wt\] seem separately in the equation.

Ans: This equation is not having any harmonic term. Therefore, it is not representing either a stationary wave or travelling wave.

Ans: This equation demonstrates a travelling wave as it is having harmonic terms \[kx\] and \[wt\] are in \[kx~wt\] combination.

Ans: This equation demonstrates a stationary wave because it is having harmonic terms \[kx\] and \[wt\] separately in the equation. It is the superposition of two stationary waves.

Ans: Provided that,

Mass of the wire, \[m=3.5\times {{10}^{-2}}kg\]

Linear mass density, \[\mu =\frac{m}{l}=4.0\times {{10}^{-2}}kg{{m}^{-1}}\]

Frequency of vibration, \[\nu =45Hz\]

 Length of the wire, \[l=\frac{m}{\mu }=\frac{3.5\times {{10}^{-2}}}{4.0\times {{10}^{-2}}}=0.875m\]

The wavelength of the stationary wave \[(\lambda )\] is given by: 

\[\lambda =\frac{2l}{n}\] where, \[n=\]number of nodes

 For fundamental node, \[n=1\]: 

\[\lambda =2l\]

\[\lambda =2\times 0.875=1.75m\]

The speed of the transverse wave in the string is given as: 

\[v=\nu \lambda =45\times 1.75=78.75m{{s}^{-1}}\]

The speed of transverse wave is \[78.75m{{s}^{-1}}\].

Ans: The tension produced in the string is given by the relation: 

\[T={{v}^{2}}\mu \]

\[\Rightarrow T={{(78.75)}^{2}}\times 4.0\times {{10}^{-2}}=248.06N\]

The tension in the string is $248.06N$.

Ans: It is provided that,

Frequency of the turning fork, \[\nu =340\text{ }Hz\]

Since the given pipe is attached with a movable piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure.

Such a system gives odd harmonics. The relation of fundamental note in a closed pipe is given by:

\[{{l}_{1}}=\frac{\lambda }{4}\]

Where, length of the pipe \[{{l}_{1}}=25.5cm=0.255m\]

\[\Rightarrow \lambda =4{{l}_{1}}=4\times 0.255=1.02m\]

The relation of sound speed is given by:

\[v=\nu \lambda =340\times 1.02=346.8m{{s}^{-1}}\]

Ans: We have,

Length of the steel rod, \[l\text{ }=\text{ }100\text{ }cm\text{ }=\text{ }1\text{ }m\]

Fundamental frequency of vibration, \[\nu =2.53kHz=2.53\times {{10}^{3}}Hz\]

An antinode (A) is formed at its centre, and nodes (N) are formed at its two ends when the rod is plucked at its middle, as shown in the given figure.

The distance between two successive nodes is \[\frac{\lambda }{2}\]

\[\therefore l=\frac{\lambda }{2}\]

\[\lambda =2l=2\times 1=2m\]

The sound speed in steel is given by:

\[v=\nu \lambda \]

\[\Rightarrow v=2.53\times {{10}^{3}}\times 2\]

\[\Rightarrow v=5.06\times {{10}^{3}}m{{s}^{-1}}\]

\[\Rightarrow v=5.06km{{s}^{-1}}\]

The speed of sound in steel is \[5.06km{{s}^{-1}}\].

Ans: Provided that,

Length of the pipe, \[l=20\text{ }cm=0.2\text{ }m\]

Source frequency \[=\text{ }{{n}^{th}}\] normal mode of frequency, \[{{\nu }_{n}}=\text{ }430\text{ }Hz\]

Speed of sound, \[v=340\text{ }m/s\]

In a closed pipe, the nth normal mode of frequency is given by the relation:\[{{\nu }_{n}}=(2n-1)\frac{v}{4l}\]

Where, n is an integer \[=\text{ }0,1,2,3\ldots ..\]

\[\Rightarrow 430=\frac{(2n-1)340}{4\times 0.2}\]

\[\Rightarrow 2n-1=\frac{430\times 4\times 0.2}{340}\]

\[\Rightarrow 2n-1=1.01\]

\[\Rightarrow n\approx 1\]

Clearly, the vibration frequency’s first mode is resonantly excited by the given source. In a pipe open at both ends, the \[{{n}^{th}}\] mode of vibration frequency is given by:

\[{{\nu }_{n}}=\frac{nv}{2l}\]

\[\Rightarrow n=\frac{2l{{v}_{n}}}{v}\]

\[\Rightarrow n=\frac{2\times 0.2\times 430}{340}=0.5\]

Since the number of the vibration mode has to be an integer, the given source does not generate a resonant vibration in an open pipe.

Ans: It is provided that,

Frequency of string A, \[{{f}_{A}}=324Hz\]

 Frequency of string B \[={{f}_{B}}\]

Beat’s frequency, \[n=6\text{ }Hz\]

Beat’s frequency is given as:

\[n=\left| {{f}_{A}}\pm {{f}_{B}} \right|\]

 \[\Rightarrow 6=324\pm {{f}_{B}}\]

\[\Rightarrow {{f}_{n}}=\text{ }330\text{ }Hz\text{ }or\text{ }318\text{ }Hz\]

Frequency decreases with a reduction in the tension in a string because frequency is directly proportional to the tension’s square root. It is given as: 

\[\nu \propto \sqrt{T}\]

 Hence, the beat frequency cannot be \[330\text{ }Hz\]

\[\therefore {{f}_{B}}=318Hz\]

Ans: A node is a point where the vibration amplitude is the minimum and pressure is the greatest. An antinode is a point where the maximum vibration amplitude and pressure are the lowest. Therefore, a displacement node is a pressure antinode and vice versa.

Ans: Bats emit powerful high-frequency ultrasonic sound waves. These waves get reflected toward them by obstructions. A bat takes a reflected wave and measures the distance, size, direction, and nature of an obstacle with the aid of its brain senses.

Ans: The overtones generated by a sitar and a violin, and the powers of these overtones, are different. Hence, one can differentiate between the notes produced by a sitar and a violin even if they have a similar vibration frequency.

Ans: Solids have a shear modulus. They can provide shearing stress. Since fluids do not hold any definite shape, they produce shearing stress. The transverse wave propagation is such that it provides shearing stress in a medium. The propagation of such a wave is probable only in solids and not in gases. Both solids and fluids have their respective bulk moduli. They can provide compressive stress. Hence, longitudinal waves can pass through solids and fluids.

(e) the shape of a pulse gets distorted during propagation in a dispersive medium. 

Ans: A pulse is a combination of waves having various wavelengths. These waves move in a dispersive medium with varying velocities, depending on the medium's nature. This results in the shape distortion of a wave pulse.

Ans: Provided that,

Frequency of the whistle, \[\nu =400\text{ }Hz\]

Speed of the train, \[{{v}_{T}}=10\text{ }m/s\]

Speed of sound, \[v=340\text{ }m/s\]

The whistle’s apparent frequency \[\left( \nu  \right)\] as the train approaches the platform is given by: 

\[\nu '=\left( \frac{v}{v-{{v}_{T}}} \right)\nu \]

\[\Rightarrow \nu '=\left( \frac{340}{340-10} \right)\times 400=412.12Hz\]

Ans: The apparent frequency (\[\nu ''\]) of the whistle as the train recedes from the platform is given by the relation: 

\[\nu ''=\left( \frac{v}{v+{{v}_{T}}} \right)\nu \]

\[\Rightarrow \nu ''=\left( \frac{340}{340+10} \right)\times 400=388.57Hz\]

The apparent frequency of the whistle is $388.57Hz$.

Ans: The apparent change in sound frequency is caused by the relative motions of the source and the observer. These relative motions generate no effect on the sound speed. Therefore, the sound speed in the air in both cases remains the same, i.e., \[340\text{ }m/s.\]