A) 1.47 Hz
B) 0.36 Hz
C) 6.73 Hz
D) 2.94 Hz
Correct Answer: A
Solution :
Given: Time required for maximum dis- placement (t) = 0.17 sec. We know that time period of sinosoidal wave \[\left( T \right)=4t=4\times 0.17=0.68\,sec.\] Therefore frequency\[(n)=\frac{1}{T}=\frac{1}{0.68}=1.47\,Hz.\]You need to login to perform this action.
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