Q107 / 120 · JEE Advanced 2006 · Physics

Paragraph: Two plane harmonic sound waves are expressed by the equations. $$ \begin{aligned} & y_1(x, t)=A \cos (0.5 \pi x-100 \pi t) \\ & y_2(x, t)=A \cos (0.46 \pi x-92 \pi t) \end{aligned} $$ (All parameters are in MKS)Question: At $x=0$ how many times the amplitude of $y_1+y_2$ is zero in one second at $x=0$ ?

  • A
    192
  • B
    48
  • C
    100
  • Correct answer
    D
    96

Solution

At $x=0, y=y_1+y_2=2 A \cos 96 \pi t \cos 4 \pi t$ Frequency of $\cos (96 \pi t)$ function is $48 \mathrm{~Hz}$ and that of $\cos (4 \pi t)$ function is $2 \mathrm{~Hz}$. In one second cos function becomes zero at $2 f$ times, where $f$ is the frequency. Therefore, first function will become zero at 96 times and the second at 4 times. But second overlaps with first. Hence, net $y$ will become zero 96 times in 1 second.

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

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