The speed of transverse wave $v$ in a stretched string depends on tension $T$ in the string and linear mass density (mass per unit length ) $μ$ . Find the relation using method of dimensions.

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Solution

Given,

$S p e e d = v, T e n s i o n = t, D e n s i t y = \frac{m}{L}$

So, the velocity is given by,

$K L^{A} T^{B} M^{C}$

So. $[M^{0} L T^{- 1}] = [L]^{A} [M L T^{- 2}]^{B} [\frac{M}{L}]^{C}$

$[M^{0} L T^{- 1}] = [M^{B + C} L^{A + B - C} T^{- 2 B}]$

On computing the coefficient then we get,

$B + C = 0, A + B - C = 1 - 2 B = - 1$

Then we get $b = \frac{1}{2}, c = \frac{- 1}{2}, a = 0$

Thus, $V = k \sqrt{\frac{T}{M}}$

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Similar questions

A transverse wave described by

$y = (0.02 m) s i n [(1.0 m^{- 1}) x + (30 s^{- 1}) t]$

propagates on a stretched string having a linear mass density of $1.2 \times 10^{- 4} k g m^{- 1}$ . Find the tension in the string.

μ

y = 0.02 sin (x + 30 t)

12 g m^{- 1}

u

T

μ

V

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