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

The velocity of propagation of a transverse wave on a stretched string is given by
$$v = \sqrt {\left( {\frac{T}{\mu }} \right)} $$
where $$T$$ is tension in the string and $$\mu $$ is linear density of the string i.e. mass per unit length of the string.
$$\eqalign{ & {\text{Here,}}\,\,\mu = \frac{{0.035}}{{5.5}}kg/m \cr & T = 77\,N \cr & \therefore v = \sqrt {\frac{{77 \times 5.5}}{{0.035}}} \cr & = 110\,m/s \cr} $$

When both source and observer are moving towards each other, apparent frequency is given by
$${f_a} = {f_0}\left( {\frac{{v + {v_0}}}{{v - {v_s}}}} \right)$$
where,
$${{f_0}} = $$  original frequency of source
$${{v_s}} = $$  speed of source
$${{v_0}} = $$  speed of observer
$$v = $$  speed of sound
Frequency of the horn,
$${f_0} = 400\,Hz$$
Speed of observer in the second car,
$${v_0} = 16.5\,m/s$$

Speed of source,
$$\eqalign{ & {v_s} = {\text{speed of first car}} \cr & = 22\,m/s \cr} $$
Frequency heard by the driver in the second car
$$\eqalign{ & {f_a} = {f_0}\left( {\frac{{v + {v_0}}}{{v - {v_s}}}} \right) = 400\left( {\frac{{340 + 16.5}}{{340 - 22}}} \right) \cr & = 448\,Hz \cr} $$

$$\eqalign{ & y = A\sin \left( {\omega t - kx} \right) \cr & {\text{Particle velocity,}} \cr & {v_p} = \frac{{dy}}{{dt}} = A\omega \cos \left( {\omega t - kx} \right) \cr & \therefore {v_{p\,\max }} = A\,\omega \cr & {\text{wave}}\,{\text{velocity}} = \frac{\omega }{k} \cr & \therefore A\,\omega = \frac{\omega }{k} \cr & {\text{i}}{\text{.e}}{\text{.,}}\,\,A = \frac{1}{k}\,\,{\text{But}}\,\,k = \frac{{2\,\pi }}{\lambda } \cr & \therefore \lambda = 2\pi A \cr} $$

Maximum number of beats $$ = \left( {\nu + 1} \right) - \left( {\nu - 1} \right) = 2$$

Fundamental frequency,
$$\eqalign{ & f = \frac{v}{{2\,\ell }} \cr & = \frac{1}{{2\,\ell }}\sqrt {\frac{T}{\mu }} \cr & = \frac{1}{{2\,\ell }}\sqrt {\frac{T}{{A\,\rho }}} \left[ {\because \,\,v = \sqrt {\frac{T}{\mu }} \,{\text{and }}\mu = \frac{m}{\ell }} \right] \cr & {\text{Also, }}Y = \frac{{T\,\ell }}{{A\,\Delta \ell }} \cr & \Rightarrow \,\,\frac{T}{A} = \frac{{Y\,\Delta \ell }}{\ell } \cr & \Rightarrow \,\,f = \frac{1}{{2\,\ell }}\sqrt {\frac{{\gamma \,\Delta \ell }}{{\ell \rho }}} \,\,\,.....\left( {\text{i}} \right) \cr} $$
Putting the value of $$\ell ,\frac{{\Delta \ell }}{\ell },\rho $$   and $$\gamma $$ in eqn. (i) we get,
$$\eqalign{ & f = \sqrt {\frac{2}{7}} \times \frac{{{{10}^3}}}{3} \cr & {\text{or, }}f \approx 178.2\,Hz \cr} $$

$$\eqalign{ & {\text{Given,}}\,a = 1\,m \cr & {\text{As}}\,\,y = a\sin \left( {kx - \omega t} \right) \cr & = \sin \frac{{2\pi }}{{2\pi }}x - 2\pi \times \frac{1}{\pi }t \cr & = \sin \left( {x - 2t} \right) \cr} $$

The frequency $$\left( \nu \right)$$  produced by the air column is given by
$$\eqalign{ & \lambda \times \nu = \nu \cr & \Rightarrow \,\,\nu = \frac{\nu }{\lambda } \cr & {\text{Also, }}\frac{{3\lambda }}{4} = \ell = 75cm = 0.75m \cr & \therefore \,\,\lambda = \frac{{4 \times 0.75}}{3} \cr & \Rightarrow \,\nu = \frac{{340 \times 3}}{{4 \times 0.75}} \cr & = 340Hz \cr} $$
$$\therefore $$ The frequency of vibrating string = 340. Since this string produces 4 beats/sec with a tuning fork of frequency $$n$$ therefore $$n = 340 + 4$$   or $$n = 340 - 4.$$   With increase in tension, the frequency produced by string increases. As the beats/sec decreases therefore $$n = 340 + 4 = 344\,Hz.$$

$$\eqalign{ & y = A\sin \left( {kx - kct} \right) + A\sin \left( {kx + kct} \right) \cr & = 2A\sin \cr & \left( {\frac{{kx - kct + kx + kct}}{2}} \right) \cdot \cos \left( {\frac{{kx - kct - kx - kct}}{2}} \right) \cr & = 2A\sin \left( {kct} \right).\cos \,kx. \cr & {\text{Thus}}\,\,\frac{{2\pi }}{\lambda } = k,\,\,\therefore \lambda = \frac{{2\pi }}{k} \cr} $$
The distance between adjacent nodes
$$ = \frac{\lambda }{2} = \frac{\pi }{k}$$

No. of loops in longitudinal mode $$ = \frac{6}{2} = 3$$

For first resonant length $$\nu = \frac{v}{{4{\ell _1}}} = \frac{v}{{4 \times 18}}\,\left( {{\text{in winter}}} \right)$$
For second resonant length
$$\eqalign{ & \nu ' = \frac{{3v'}}{{4{\ell _2}}} = \frac{{3v'}}{{4x}}\,\left( {{\text{in summer}}} \right) \cr & \therefore \,\,\frac{v}{{4 \times 18}} = \frac{{3v'}}{{4 \times x}} \cr & \therefore \,\,x = 3 \times 18 \times \frac{{v'}}{v} \cr & \therefore \,\,x = 54 \times \frac{{v'}}{v}cm \cr} $$
$$v' > v$$  because velocity of light is greater in summer as compared to winter $$\left( {v\, \propto \,\sqrt T } \right)$$
$$\therefore $$ $$x > 54\,cm$$