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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 $x=0$ ?
Options:
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 $x=0$ ?
Solution:
1787 Upvotes
Verified Answer
The correct answer is:
96
96
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.
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.
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