First slide

Standing waves

Question

A wire having a linear density of 0.10kg/m is kept under a tension of 490N. It is observed that it resonates at a frequency of 400Hz and the next higher frequency is 450Hz. The length of the wire is

Moderate

A

14m
B

0.7m
C

1.6 m
D

0.49 m

Solution

$\frac{n}{2 l} \sqrt{\frac{T}{μ}} = 400 - - - - (1)$
and
$\frac{n + 1}{2 l} \sqrt{\frac{T}{μ}} = 450 - - - - (2) ∴ \frac{n + 1}{n} = \frac{450}{400} \Rightarrow n = 8 From - (1) \frac{8}{2 l} \times \sqrt{\frac{490}{0 .1}} = 400 \Rightarrow l = 0 .7 m$

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