A) 256 Hz
B) 512 Hz
C) 1024 Hz
D) 128 Hz
Correct Answer: C
Solution :
Let\[l\]be the length of pipe,\[v\]the speed of sound, then the fundamental tone or first harmonic of closed tube \[{{n}_{1}}=\frac{v}{4l}\] For open tube \[{{n}_{2}}=\frac{v}{2l}\] \[\therefore \] \[{{n}_{2}}=2{{n}_{1}}\] Given, \[{{n}_{1}}=512\text{ }Hz,\] \[\therefore \] \[{{n}_{2}}=2\times 512=1024\text{ }Hz\]You need to login to perform this action.
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