First slide

Speed of wave

Question

The string of length L is stretched by x when the speed of transverse wave along it is $V_{1}$ . The string is further stretched by $∆$ x when the speed of transverse wave along it is $V_{1} \sqrt{2}$ . Find $∆$ x:

Moderate

A

$(\sqrt{2} - 1) x$
B

$x \sqrt{2}$
C

x
D

$(\sqrt{2} + 1) x$

Solution

$V = \sqrt{\frac{T}{μ}} \Rightarrow V \propto \sqrt{T}$
and stress $\propto$ strain
hence, tension $\propto$ deformation
T $\propto$ $∆ l$
Hence $\frac{T_{1}}{T_{2}} = \frac{x}{x + ∆ x} = {[\frac{V_{1}}{V_{2}}]}^{2}$
$\frac{x}{x + ∆ x} = {[\frac{1}{\sqrt{2}}]}^{2} \Rightarrow 2 x = (x + ∆ x)$
$\Rightarrow x = ∆ x$

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