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

The fundamental frequency of string
$$\eqalign{ & \nu = \frac{1}{{2l}}\sqrt {\frac{T}{m}} \cr & \therefore {\nu _1}{l_1} = {\nu _2}{l_2} = {\nu _3}{l_3} = k\,......\left( {\text{i}} \right) \cr & {\text{From}}\,{\text{Eq}}{\text{.}}\,\left( {\text{i}} \right) \cr & {l_1} = \frac{k}{{{\nu _1}}},{l_2} = \frac{k}{{{\nu _2}}},{l_3} = \frac{k}{{{\nu _3}}} \cr & {\text{Original length,}} \cr & l = \frac{k}{\nu } \cr & {\text{Here,}}\,l = {l_1} + {l_2} + {l_3} \cr & \frac{k}{\nu } = \frac{k}{{{\nu _1}}} + \frac{k}{{{\nu _2}}} + \frac{k}{{{\nu _3}}} \cr & \frac{1}{\nu } = \frac{1}{{{\nu _1}}} + \frac{1}{{{\nu _2}}} + \frac{1}{{{\nu _3}}} \cr} $$

$$\eqalign{ & {\text{Given,}}\,{y_1} = 4\sin 600\,\pi t \cr & {\text{and}}\,{y_2} = 5\sin 608\,\pi t \cr} $$
Comparing with general equation
$$\eqalign{ & y = a\sin 2\pi ft \cr & {\text{We}}\,{\text{get,}}\,\,{f_1} = 300\,Hz. \cr & {\text{and}}\,\,{f_2} = 304\,Hz \cr & {\text{So,number of beats}} = {f_2} - {f_1} = 4{s^{ - 1}} \cr & {\text{We know that,}} \cr & \frac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\frac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}}} \right)^2} \cr & = {\left( {\frac{{4 + 5}}{{4 - 5}}} \right)^2} = 81 \cr} $$

$$\eqalign{ & y\left( {x,t} \right) = f\left( {x - vt} \right) \cr & y = \left( {x,0} \right) = \frac{{4 \times {{10}^{ - 3}}}}{{8 - {x^2}}} \cr} $$
For a travelling wave in the $$x$$-direction
$$y\left( {x,t} \right) = \frac{{4 \times {{10}^{ - 3}}}}{{8 - {{\left( {x - 5t} \right)}^2}}}$$

KEY CONCEPT : The frequency of a tuning fork is given by the expression
$$f = \frac{{{m^2}k}}{{4\sqrt 3 \pi {\ell ^2}}}\sqrt {\frac{Y}{\rho }} $$
As temperature increases, $$\ell $$ increases and therefore $$f$$ decreases.

Total length of string $$\ell = {\ell _1} + {\ell _2} + {\ell _3}$$
(As string is divided into three segments)
$$\eqalign{ & {\text{But frequency}} \propto \frac{1}{{{\text{length}}}}\left( {\because f = \frac{1}{{2\ell }}\sqrt {\frac{T}{m}} } \right) \cr & {\text{so}}\,\frac{1}{n} = \frac{1}{{{n_1}}} + \frac{1}{{{n_2}}} + \frac{1}{{{n_3}}} \cr} $$

Fundamental frequency of closed organ pipe
$${V_c} = \frac{V}{{4{l_c}}}$$
Fundamental frequency of open organ pipe
$${V_0} = \frac{V}{{2{l_0}}}$$
Second overtone frequency of open organ pipe
$$\eqalign{ & = \frac{{3V}}{{2{l_0}}} \cr & {\text{From question,}} \cr & \frac{V}{{4{l_c}}} = \frac{{3V}}{{2{l_0}}} \cr & \Rightarrow {l_0} = 6{l_c} = 6 \times 20 = 120\,cm \cr} $$

Third overtone has a frequency $$7\,n,$$  which means $$L = \frac{{7\lambda }}{4} = $$   three full loops + one half loop, which would make four nodes and four antinodes.

As intensity is directly proportional to the square of amplitude
$${\text{i}}{\text{.e}}{\text{.}}\,\,I \propto {a^2}$$
So, maximum intensity is given by
$$\eqalign{ & {I_{\max }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2} \cr & \left[ {{I_1},{I_2}\,{\text{are intensities}}\,{\text{of two waves}}} \right] \cr & {\text{and}}\,{I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2} \cr & \therefore {I_{\max }} + {I_{\min }} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2} + {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2} \cr & = 2\left( {{I_1} + {I_2}} \right) \cr} $$

Sabine's formula for reverberation time is
$$\eqalign{ & T = \frac{{0.16\,V}}{{\Sigma as}} \cr & T \propto \frac{V}{s} \cr} $$
where, $$V$$ is volume of hall in $${m^3}$$
$$\Sigma as = {a_1}{s_1} + {a_2}{s_2} + ....$$
= total absorption of the hall (room)
Here, $${s_1},{s_2},{s_3}\,...$$   are surface areas of the absorbers and $${a_1},{a_2},{a_3}\,...$$   are their respective absorption coefficients.
So, for two different cases of reverberation.
$$\eqalign{ & \therefore \frac{{T'}}{T} = \frac{{V'}}{{s'}} \times \frac{s}{V} = \frac{{{{\left( 2 \right)}^3}}}{{{{\left( 2 \right)}^2}}} = \frac{8}{4} = 2 \cr & {\text{Hence,}}\,T' = 2T = 2 \times 1 = 2\,s \cr} $$

Distance between two successive nodes $$ = \frac{\lambda }{2}$$

$$\eqalign{ & {\text{but we know that, }}v = \nu \lambda \cr & \therefore \frac{\lambda }{2} = \frac{v}{{2\nu }} \cr & {\text{Given,}}\,v = 20\,m/s \cr & {\text{frequency, }}\nu {\text{ = n}} \cr & {\text{So,}}\,\frac{\lambda }{2} = \frac{{20}}{{2n}} = \frac{{10}}{n} \cr} $$