An audio transmitter (T) and a receiver (R) a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The angular amplitude is $	heta_0 = \cos^{-1}(0.9)$. Since $	heta_0$ is small,$\cos 	heta_0 \approx 1 - \frac{	heta_0^2}{2} = 0.9$,which gives

Vedclass · Accepted Answer

$(31.19$ to $32.27)$

Vedclass · Answer

(D) The angular amplitude is $	heta_0 = \cos^{-1}(0.9)$. Since $	heta_0$ is small,$\cos 	heta_0 \approx 1 - \frac{	heta_0^2}{2} = 0.9$,which gives $\frac{	heta_0^2}{2} = 0.1$,so $	heta_0^2 = 0.2$ and $	heta_0 = \sqrt{0.2} = \sqrt{\frac{1}{5}}$.For a simple pendulum,the maximum velocity is $v' = \ell \omega 	heta_0 = \ell \sqrt{\frac{g}{\ell}} 	heta_0 = 	heta_0 \sqrt{g\ell}$.Substituting the values: $v' = \sqrt{\frac{1}{5}} \sqrt{10 	imes 8} = \sqrt{\frac{80}{5}} = \sqrt{16} = 4 \ m/s$.The Doppler shift formula for source and observer moving towards each other is $f_{max} = \left( \frac{v + v'}{v - v'} ight) f$ and moving away is $f_{min} = \left( \frac{v - v'}{v + v'} ight) f$.The maximum variation in frequency is $\Delta f = f_{max} - f_{min} = f \left( \frac{v + v'}{v - v'} - \frac{v - v'}{v + v'} ight) = f \left( \frac{(v + v')^2 - (v - v')^2}{v^2 - v'^2} ight) = f \left( \frac{4vv'}{v^2 - v'^2} ight)$.Substituting $v = 330 \ m/s$,$v' = 4 \ m/s$,and $f = 660 \ Hz$:$\Delta f = 660 	imes \left( \frac{4 	imes 330 	imes 4}{330^2 - 4^2} ight) = 660 	imes \left( \frac{5280}{108900 - 16} ight) \approx 660 	imes \left( \frac{5280}{108884} ight) \approx 660 	imes 0.04849 \approx 32.003 \ Hz$.Thus,the value lies in the range $(31.19$ to $32.27)$. Therefore,the correct option is $D$.

Similar Questions

$A$ whistle of frequency $540 Hz$ rotates in a horizontal circle of radius $2 m$ at an angular speed of $15 rad/s$. What is the highest frequency heard by a listener at rest with respect to the center of the circle (in $Hz$)? (Velocity of sound in air $= 330 m/s$)

$A$ source and a listener are both moving towards each other with a speed of $v/10$,where $v$ is the speed of sound. If the frequency of the note emitted by the source is $f$,the frequency heard by the listener would be nearly:

Two factories are sounding their sirens at $800 \, Hz$. $A$ man walks from one factory to the other at a speed of $2 \, m/s$. The velocity of sound is $320 \, m/s$. The number of beats heard by the person in one second will be:

$A$ vehicle sounding a whistle of frequency $256 \,Hz$ is moving on a straight road towards a hill with a velocity of $10 \,ms^{-1}$. The number of beats per second observed by a person travelling in the vehicle is (velocity of sound $= 330 \,ms^{-1}$).

$Assertion :$ The Doppler formula for sound waves is symmetric with respect to the speed of the source and the speed of the observer.
$Reason :$ The motion of a source with respect to a stationary observer is not equivalent to the motion of an observer with respect to a stationary source.

Vedclass Test Series

Exam Paper Generator

Online Exam Module