A) 320 Hz
B) 160 Hz
C) 480 Hz
D) 640 Hz
Correct Answer: B
Solution :
If a wire, vibrates in p segments, then its frequency is given by \[n=\frac{p}{2\,l}\sqrt{\frac{T}{m}}\] where \[l\] is length of wire, T is tension and m is mass per unit length. The frequency of first harmonic is \[n=\frac{1}{2\,l}\sqrt{\frac{T}{m}}\] ?. (i) The frequency of second harmonic or first overtone is \[n=\frac{2}{2\,l}\sqrt{\frac{T}{m}}\] ?. (ii) \[\Rightarrow \] \[n=2\,n\] \[\Rightarrow \] \[n=\frac{n}{2}=\frac{320}{2}=160\,Hz\]
You need to login to perform this action.
You will be redirected in
3 sec
