A) \[\frac{50}{\pi }Hz\]
B) \[\frac{100}{\pi }Hz\]
C) 100 Hz
D) 50 Hz
Correct Answer: A
Solution :
Compare the given equation with \[y=a\sin (\omega t+kx)\]. We get \[\omega =2\pi n=100\] Þ \[n=\frac{50}{\pi }Hz\]
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