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A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is:
A
105 Hz
B
155 Hz
C
205 Hz
D
10.5 Hz
Detailed Solution
Consecutive resonant frequencies of a string fixed at both ends: $\frac{nv}{2l}$ and $\frac{(n+1)v}{2l}$
$\frac{(n+1)v}{2l} - \frac{nv}{2l} = 420 - 315 \Rightarrow \frac{v}{2l} = 105$ Hz
This is the fundamental (lowest) frequency: 105 Hz.
$\frac{(n+1)v}{2l} - \frac{nv}{2l} = 420 - 315 \Rightarrow \frac{v}{2l} = 105$ Hz
This is the fundamental (lowest) frequency: 105 Hz.