A) 155 Hz
B) 205 Hz
C) 10.5 Hz
D) 105 Hz
Correct Answer: D
Solution :
Given \[I=75cm,{{f}_{1}}=420Hz\]and \[{{f}_{2}}=315Hz.\] As, two consecutive resonant frequencies for a string fixed at both ends will be \[{{f}_{1}}=\frac{nv}{2l}\]and \[{{f}_{2}}=\frac{(n+1)}{2R}\] Þ \[{{f}_{2}}-{{f}_{1}}=420-315\] \[\Rightarrow \,\,\,\,\,\,\,\frac{(n+1)v}{2}-\frac{nv}{2l}=105Hz\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\frac{v}{2l}=105Hz\] Thus, lowest resonant frequency of a string is 105 Hz.You need to login to perform this action.
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