Q120 / 200 · NEET 2012 (Screening) · Physics

Two sources of sound placed close to each other, are emitting progressive waves given by $y_1=4 \sin 600 \pi t \text { and } y_2=5 \sin 608 \pi t$ An observer located near these two sources of sound will hear

  • A
    4 beats per second with intensity ratio $25: 16$ between waxing and waning
  • B
    8 beats per second with intensity ratio $25: 16$ between waxing and waning
  • C
    8 beats per second with intensity ratio $81: 1$ between waxing and waning
  • Correct answer
    D
    4 beats per second with intensity ratio $81: 1$ between waxing and waning

Solution

Given, $y_1=4 \sin 600 \pi t$ and $y_2=5 \sin 608 \pi t$ Comparing with general equation $y=a \sin 2 \pi f t$ we get, $f_2=300 \mathrm{~Hz}$ and $f_2=304 \mathrm{~Hz}$ Number of beats $=f_2-f_1=4 \mathrm{~s}^{-1}$ $\begin{aligned} \frac{I_{\max }}{I_{\min }} & =\left(\frac{a_1+a_2}{a_1-a_2}\right)^2 \\ & =\left(\frac{4+5}{4-5}\right)^2=\frac{81}{1} \end{aligned}$

Hence, the correct answer is option: 'D'.

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