Waves MCQ Questions & Answers in Oscillation and Mechanical Waves | Physics

Given,
$$\eqalign{ & {y_1} = 4\sin 500\,\pi t\,......\left( {\text{i}} \right) \cr & {y_2} = 2\sin 506\,\pi t\,......\left( {{\text{ii}}} \right) \cr} $$
Comparing Eqs. (i) and (ii), we get
$$y = a\sin \omega t\,.......\left( {{\text{iii}}} \right)$$
$$\eqalign{ & {\text{We}}\,{\text{have,}}\,\,{\omega _1} = 500\,\pi \cr & \Rightarrow {n_1} = \frac{{500\,\pi }}{{2\pi }} = 250\,beats/s\,\left[ {\therefore n = \frac{\omega }{{2\pi }}} \right] \cr & {\text{and}}\,\,{\omega _2} = 506\,\pi \cr & \Rightarrow {n_2} = \frac{{506\,\pi }}{{2\,\pi }} = 253\,beats/s \cr} $$
Thus, number of beats produced
$$\eqalign{ & = {n_2} - {n_1} = 253 - 250 = 3\,beats/s \cr & = 3 \times 60\,beats/\min = 180\,beats/\min \cr} $$
NOTE
If equation of wave is given and to find physical quantities like amplitude, wavelength, time period, frequency, just compare the given equation with standard equation of wave.

The standard equation of standing wave is
$$y = a\sin \left( {\omega t} \right)\cos \left( {kx} \right)\,......\left( {\text{i}} \right)$$
Given equation is
$$y = a\sin \left( {100\,t} \right)\cos \left( {0.01x} \right)\,......\left( {{\text{ii}}} \right)$$
Comparing Eqs. (i) and (ii)
$$\omega = 100\,\,{\text{and}}\,\,k = 0.01$$
∴ Velocity of wave is
$$\eqalign{ & v = \frac{\lambda }{T} = \frac{\omega }{k} = \frac{{100}}{{0.01}}\,\left[ {{\text{As,}}\,\omega = \frac{{2\pi }}{T}\,\,{\text{and}}\,\,k = \frac{{2\pi }}{\lambda }} \right] \cr & = {10^4}\,m/s \cr} $$

$$y = 0.25\sin \left( {10\pi x - 2\pi t} \right)$$
Comparing this equation with the standard wave equation
$$\eqalign{ & y = a\sin \left( {kx - \omega t} \right) \cr & {\text{We}}\,{\text{get,}}\,\,k = 10\pi \Rightarrow \frac{{2\pi }}{\lambda } = 10\pi \Rightarrow \lambda = 0.2\,m \cr & {\text{And}}\,\omega = 2\pi \,\,{\text{or,}}\,\,2\pi v = 2\pi \Rightarrow v = 1\,Hz. \cr} $$
The sign inside the bracket is negative, hence the wave travels in $$+ ve$$  $$x$$-direction.

The frequency of vibrations of string is
$$\eqalign{ & n = \frac{1}{2}\sqrt {\frac{g}{l}} \,......\left( {\text{i}} \right) \cr & {\text{Given,}}\,\,{n_A} = 2{n_B} \cr & \therefore \frac{1}{2}\sqrt {\frac{g}{{{l_A}}}} = 2 \cdot \frac{1}{2}\sqrt {\frac{g}{{{l_B}}}} \cr & {\text{or}}\,\,\frac{1}{{{l_A}}} = \frac{4}{{{l_B}}}\,\,or\,\,{l_B} = 4{l_A} \cr} $$
It is obvious from Eq. (i), the frequency of vibrations of strings does not depend on their masses.

$$\eqalign{ & y = 0.02\left( m \right)\sin \left[ {2\pi \left( {\frac{t}{{0.04\left( s \right)}}} \right) - \frac{x}{{0.05\left( m \right)}}} \right] \cr & {\text{But}}\,y = a\sin \left( {\omega t - kx} \right) \cr & \therefore \omega = \frac{{2\pi }}{{0.50}} \Rightarrow \nu = \frac{1}{{0.04}} = 25\,Hz \cr & k = \frac{{2\pi }}{{0.50}} \Rightarrow \lambda = 0.5\,m \cr & \therefore {\text{velocity,}}\,\nu = \nu \lambda = 25 \times 0.5\,m/s = 12.5\,m/s \cr} $$
Velocity on a string is given by
$$\nu = \sqrt {\frac{T}{\mu }} \,\,\therefore T = {\nu ^2} \times \mu = {\left( {1.25} \right)^2} \times 0.04 = 6.25\,N$$

The position of such a wave changes in two dimensional plane with time.

Velocity of a wave is given by
\[v = n\lambda \,\,\left[ {\begin{array}{*{20}{c}} {n = {\text{frequency of wave}}} \\ {\lambda = {\text{wavelength of wave}}} \end{array}} \right]\]
So, for two different cases,
$$\eqalign{ & {v_1} = {n_1}{\lambda _1} \cr & {v_2} = {n_2}{\lambda _2} \cr & {\lambda _2} = {\lambda _1}\frac{{{v_2}}}{{{v_1}}} = {\lambda _1} \times \frac{{3500}}{{350}} = {\lambda _1} \times 10\,\left[ {\because {n_1} = {n_2}} \right] \cr & {\lambda _2} = 10\,{\lambda _1} \cr} $$

In longitudinal waves, energy is propagated along with the wave motion without any net transport of the mass of the medium.

A tuning fork of frequency $$256\,Hz$$   makes 5 beats/second with the vibrating string of a piano. Therefore the frequency of the vibrating string of piano is $$\left( {256 \,\pm 5} \right)\,Hz$$    i.e., either $$261\,Hz$$  or $$251\,Hz.$$  When the tension in the piano string increases, its frequency will increases. Now since the beat frequency decreases, we can conclude that the frequency of piano string is $$251\,Hz$$

Statement (D) is not true, because at the open ends pressure change will be zero.