A) 105 Hz
B) 1.05 Hz
C) 1050 Hz
D) 10.5 Hz
Correct Answer: A
Solution :
For string fixed at both the ends, resonant frequency are given by\[{{f}_{0}}=\frac{nv}{2L'}\]where symbols have their usual meanings, if is given that 315 Hz and 420 Hz are two consecutive resonant frequencies, Jet these be nth and \[(n+1)th\]harmonics. \[315=\frac{nv}{2L}\] ...(i) \[420=\frac{(n+1)v}{2L}\] ...(ii) Dividing Eq. (i) by Eq. (ii). we get\[\frac{315}{420}=\frac{n}{n+1}\] \[\Rightarrow \] \[n=3\] From Eq. (i), lowest resonant frequency, \[{{f}_{0}}=\frac{v}{2L}=\frac{315}{3}=105\,Hz,\,n=1\]You need to login to perform this action.
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