A) 155 Hz
B) 205 Hz
C) 10.5 Hz
D) 105 Hz
Correct Answer: D
Solution :
| Given \[I=75\,\,cm,\,\,{{f}_{1}}=420\,\,Hz\] and \[{{f}_{2}}=315\,\,Hz\] |
| As, two consecutive resonant frequencies for a |
| string fixed at both ends will be |
| \[{{f}_{1}}=\frac{nv}{2l}\] and \[{{f}_{2}}=\frac{(n+1)}{2R}\] |
| \[\Rightarrow \] \[{{f}_{2}}-{{f}_{1}}=420-315\] |
| \[\Rightarrow \] \[\frac{(n+1)v}{2}-\frac{nv}{2l}=105Hz\] |
| \[\Rightarrow \] \[\frac{v}{2l}=105Hz\] |
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