First slide

Standing waves

Question

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

Moderate

A

105 Hz
B

155 Hz
C

205 Hz
D

10.5 Hz

Solution

Two consecutive resonant frequencies for a string fixed at both ends will be
$\frac{nv}{2 l} and \frac{(n + 1) v}{2 l}$
$\Rightarrow \frac{(n + 1) v}{2 l} - \frac{nv}{2 l} = 420 - 315$
$\Rightarrow \frac{v}{2 l} = 105 Hz .$ which is the minimum resonant frequency

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