A) 4
B) 5
C) 7
D) 6
Correct Answer: D
Solution :
[d] In case of closed organ pipe frequency, \[{{f}_{n}}=(2n+1)\frac{v}{4l}\] For \[n=0,\,{{f}_{0}}=100Hz\] \[n=1,\,{{f}_{1}}=300Hz\] \[n=2,\,{{f}_{2}}=500Hz\] \[n=3,\,{{f}_{3}}=700Hz\] \[n=4,\,{{f}_{4}}=900Hz\] \[n=5,\,{{f}_{5}}=1100Hz\] \[n=6,\,{{f}_{6}}=1300Hz\] Hence possible natural oscillation whose frequencies < 1250 Hz=6(n= 0, 1, 2, 3, 4, 5)You need to login to perform this action.
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