JEE Advanced · Physics · 16. Waves & Sound
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
- D 96
Answer & Solution
Correct Answer
(D) 96
Step-by-step Solution
Detailed explanation
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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