A) 105 Hz
B) 1.05 Hz
C) 1050 Hz
D) 10.5 Hz
Correct Answer: A
Solution :
Key Idea For string fixed at both the ends, resonant frequency are given by \[{{\pi }^{2}}\]It is given that 315 Hz and 420 Hz are two consecutive resonant frequencies, let these are nth and \[F=\frac{m{{v}^{2}}}{r}\] harmonics. \[f=\frac{nv}{2L},\] ?..(i) \[420\,=\frac{(n+1)v}{2L}\] ??(ii) Dividing Eq. (i) by Eq. (ii), we get \[315=\frac{nv}{2L}\] \[315=\frac{nv}{2L}\] \[\frac{315}{420}=\frac{n}{n+1}\] From Eq. (i), lowest resonant frequency \[\Rightarrow \]
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