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

Easy

A

10.5 Hz
B

105 Hz
C

155 Hz
D

205 Hz

Solution

For a string fixed at both ends, the resonant frequencies are
$v_{n} = \frac{n v}{2 L} where n = 1, 2, 3, \dots . .$
The difference between two consecutive resonant frequencies is
$Δ v_{n} = v_{n + 1} - v_{n} = \frac{(n + 1) v}{2 L} - \frac{n v}{2 L} = \frac{v}{2 L}$
which is also the lowest resonant frequency (n=1)
Thus the lowest resonant frequency for the given string
=420 Hz -315 Hz = 105 Hz

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