A) \[\frac{v}{2l}\]
B) \[\frac{v}{4l}\]
C) \[\frac{v\Delta l}{2{{l}^{2}}}\]
D) \[\frac{v\Delta l}{l}\] (here v is the speed of sound)
Correct Answer: C
Solution :
For open organ pipes, frequency \[f=\frac{v}{2l}\] So, \[{{f}_{1}}=\frac{v}{2l}\] and \[{{f}_{2}}=\frac{v}{2(l+\Delta l)}\] \[\therefore \]Number of beats produced per second \[=|{{f}_{1}}-{{f}_{2}}|\] \[=\left| \frac{v}{2l}-\frac{v}{2(l+\Delta l)} \right|\] \[=\frac{v}{2}\left| \frac{1}{l}-\frac{1}{l+\Delta t} \right|\] \[=\frac{v\Delta l}{2l(l+\Delta l)}\] \[=\frac{v\Delta l}{2{{l}^{2}}}\] \[\because \] \[\Delta l\]is negligible in comparison of I.
You need to login to perform this action.
You will be redirected in
3 sec
