Question

Two vibrating tuning forks produce waves given by $ y_{1}=4\sin \,500\pi t, $ $ y_{1}=2\sin \,506\pi t $ If they are held near the ear of a person, the person will hear

• A. 3 beats/second with intensity ratio of maxima to minima equal to 9
• B. 3 beats/second with intensity ratio of maxima to minima equal to 2
• C. 6 beats/second with intensity ratio of maxima to minima to equal to 2
• D. 6 beats/second with intensity ratio of maxima to minima equal to 9

Answer 1

Answer: (A)

Comparing given equation with standard equations
$y=A \sin \omega t$
We get $A_{1}=4, \omega_{1}=500 \pi$
and $A_{2}=2, \omega_{2}=506 \pi$
Frequency $\quad n=\frac{\omega}{2 \pi}$
$\therefore \quad n_{1}=\frac{\omega_{1}}{2 \pi}=\frac{500 \pi}{2 \pi}=250$
$n_{2}=\frac{\omega_{2}}{2 \pi}=\frac{506 \pi}{2 \pi}=253$
$\therefore $ number of beats
$=n_{2}-n_{1}=253-250=3 $
$\frac{I_{\max }}{I_{\min }}=\frac{\left(A_{1}+A_{2}\right)^{2}}{\left(A_{1}-A_{2}\right)^{2}} $
$=\frac{(4+2)^{2}}{(4-2)^{2}}=\left(\frac{6}{2}\right)^{2}=\frac{9}{1}$