A) 125 m/s
B) 250 m/s
C) 500 m/s
D) 1000 m/s
Correct Answer: C
Solution :
For string \[\lambda =\frac{2l}{p}\] where p = No. of loops = Order of vibration Hence for forth mode p = 4 Þ \[\lambda =\frac{l}{2}\] Hence v = nl \[=500\times \frac{2}{2}=500\,Hz\]You need to login to perform this action.
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