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

To calculate the time period of combined oscillation, calculate the beat produced from the given frequencies.
When both waves are combined, then beats are produced. Frequency of beats will be $$ = {v_1} - {v_2}$$
$$ = \frac{1}{{{T_1}}} - \frac{1}{{{T_2}}} = \frac{1}{3} - \frac{1}{4} = \frac{1}{{12}}$$
Hence, time period $$= 12\,s$$

$${\text{ }}{f_2} = \frac{{{f_0}v}}{{v + {v_0}}}$$
The wave which reaches wall $${f_1}$$ is reflected.
$${f_1} = \frac{{{f_0}v}}{{v - {v_0}}}$$
The reflected frequency is $${f_1}$$ as the wall is at rest.
$${\text{Beats}} = {f_1} - {f_2} = \frac{{{f_0}v}}{{v - {v_0}}} - \frac{{{f_0}v}}{{v + {v_0}}} = \frac{{2{f_0}v{v_0}}}{{{v^2} - v_0^2}}$$

From equation given,
$$\eqalign{ & \omega = 100\,\,{\text{and }}k = 20, \cr & v = \frac{\omega }{k} \cr & = \frac{{100}}{{20}} \cr & = 5\,m/s \cr} $$

Length of pipe $$= 85\,cm = 0.85\,m$$
Frequency of oscillations of air column in closed organ pipe is given by,
$$\eqalign{ & f = \frac{{\left( {2\,n - 1} \right)\upsilon }}{{4L}} \cr & f = \frac{{\left( {2\,n - 1} \right)\upsilon }}{{4L}} \leqslant 1250 \cr & \Rightarrow \,\,\frac{{\left( {2\,n - 1} \right) \times 340}}{{0.85 \times 4}} \leqslant 1250 \cr & \Rightarrow \,\,2n - 1 \leqslant 12.5 \approx 6 \cr} $$

The fundamental frequency in case (a) is $$f = \frac{v}{{2\,\ell }}$$
The fundamental frequency in case (b) is
$$f' = \frac{v}{{4\left( {\frac{\ell }{2}} \right)}} = \frac{u}{{2\,\ell }} = f$$

KEY CONCEPT : For a string vibrating between two rigid support, the fundamental frequency is given by
$$\eqalign{ & n = \frac{1}{{2\,\ell }}\sqrt {\frac{T}{\mu }} \cr & = \frac{1}{{2 \times 1}}\sqrt {\frac{{10 \times 9.8}}{{9.8 \times {{10}^{ - 3}}}}} \cr & = 50\,Hz \cr} $$
As the string is vibrating in resonance to a.c of frequency $$n,$$ therefore both the frequencies are same.

For both end open
$$\eqalign{ & \frac{{2{\lambda _1}}}{4} = \ell \cr & \Rightarrow \,\,{\lambda _1} = 2\ell \cr & {\nu _1} = \frac{c}{{{\lambda _1}}} = \frac{c}{{2\ell }}\,\,\,.....\left( {\text{i}} \right) \cr} $$
For one end closed
For third harmonic $$\frac{{3\,{\lambda _2}}}{4} = \ell $$
$$\eqalign{ & \Rightarrow \,\,{\lambda _2} = \frac{{4\ell }}{3} \cr & {\nu _2} = \frac{c}{{{\lambda _2}}} = \frac{{3c}}{{4\ell }}\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Given $${\nu _2} - {\nu _1} = 100$$
From (i) and (ii)
$$\frac{{{\nu _2}}}{{{\nu _1}}} = \frac{{\frac{3}{4}}}{{\frac{1}{2}}} = \frac{3}{2}$$
On solving, we get $${\nu _1} = 200Hz.$$

No. of beats heard when fork 2 is sounded with fork $$1 = \Delta n = 4$$
Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from 4 to 6 in this case) then the frequency of the unknown fork 2 is given by,
$$\eqalign{ & n = {n_0} - \Delta n \cr & = 200 - 4 \cr & = 196\,Hz \cr} $$

The sum of their $$KE$$  and $$PE$$  is a constant.

The general expression of travelling wave can be written as
$$y = a\sin \left( {\omega t \pm kx} \right)\,......\left( {\text{i}} \right)$$
For travelling wave along positive $$x$$-axis we should use minus $$\left( - \right)$$  sign only.
$$\eqalign{ & \therefore y = a\sin \left( {\omega t - kx} \right) \cr & {\text{but}}\,\,\omega = \frac{{2\,\pi v}}{\lambda }\,\,{\text{and}}\,\,k = \frac{{2\,\pi }}{\lambda } \cr & {\text{So,}}\,\,y = a\sin \frac{{2\,\pi }}{\lambda }\left( {vt - x} \right)\,.......\left( {{\text{ii}}} \right) \cr & {\text{Given,}}\,\,a = 0.2\,m,v = 360\,m/s,\lambda = 60\,m, \cr} $$
Substituting in Eq. (ii), we have
$$\eqalign{ & y = 0.2\,\sin \frac{{2\,\pi }}{{60}}\left( {360\,t - x} \right) \cr & {\text{or}}\,\,y = 0.2\,\sin 2\,\pi \left( {6\,t - \frac{x}{{60}}} \right) \cr} $$