First slide

Standing waves

Question

A string is under tension so that its length is increased by $\frac{1}{n}$ times its original length. The ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will be :

Moderate

A

1 : n
B

$n : 1$
C

$\sqrt{n} : 1$
D

n : 1.

Solution

Velocity of longitudinal wave $v_{1} = \sqrt{\frac{Y}{ρ}}$ and velocity of transverse wave
$V_{2} = \sqrt{\frac{T}{m}} = \sqrt{\frac{T}{ρs}}$
$\frac{v_{1}}{V_{2}} = \sqrt{\frac{Ys}{T}} = \sqrt{\frac{Y}{T / s}} = \sqrt{\frac{Y}{Y (\frac{Δl}{l})}} = \sqrt{n}$
$\begin{array}{l} Now, f \propto v \\ ∴ \frac{f_{1}}{f_{2}} = \frac{V_{1}}{V_{2}} = \sqrt{n} \end{array}$
In the above equation $ρ$ = density of string, s = area of cross–section of string, Y = Young’s modulus of elasticity.
$\frac{T}{s} = Stress = Y (strain) = Y \frac{Δl}{l}$

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