A) 25 Hz
B) 200 Hz
C) 400 Hz
D) 1600 Hz
Correct Answer: D
Solution :
The frequency of transverse vibration in a stretched string is given by \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\] \[n\propto \frac{1}{l}\sqrt{T}\] Hence, \[\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}}\] (Given: \[{{l}_{2}}=\frac{{{l}_{1}}}{4},{{T}_{2}}=4{{T}_{1}},{{n}_{1}}=200\,Hz\]) \[\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{{{l}_{1}}}{\frac{{{l}_{1}}}{4}}\sqrt{\frac{4{{T}_{1}}}{{{T}_{1}}}}\] \[\frac{{{n}_{2}}}{{{n}_{1}}}=4\times 2=8\] \[{{n}_{2}}=8\times 200=1600\,Hz\]
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