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

• A. $ 1: n$
• B. $n^2 : 1$
• C. $\sqrt n : 1$
• D. $n : 1$

Answer 1

Answer: (C)

Velocity of longitudinal waves,
$v_{1} = \sqrt{\frac{Y}{\rho}} $
Velocity of transverse waves
$v_{2} = \sqrt{\frac{T}{\mu}} $
If $A$ is area of cross-section of string, then
$\mu = \frac{\text{Mass}}{\text{Length}} = \frac{\text{Mass}}{\text{Volume}} \times$ Area $= \rho A$
$ \therefore v_{2} \sqrt{\frac{T}{\rho A}} $
$ \frac{v_{1}}{v_{2}} = \sqrt{\frac{y}{\rho} \frac{\rho A}{T}} = \sqrt{\frac{YA}{T}} $
As $Y = \frac{F}{A\left(\Delta l/l\right)} = \frac{T}{A\left(\Delta l/l\right)} $
$ \therefore \frac{v_{1}}{v_{2}} = \sqrt{\frac{T}{A\left(\frac{\Delta l}{l}\right)} \frac{A}{T}} = \left(\frac{\Delta l}{l}\right)^{\left(-1/2\right)}$
$\frac{\Delta l}{l} = \frac{1}{n}$ (Given)
$ \therefore \frac{v_{1}}{v_{2}} = \left(\frac{1}{n}\right)^{-1/2}=\sqrt{n} $
If $v_1, v_2$ are the corresponding fundamental frequencies of longitudinal and transverse vibrations, then
$v_{1} = v_{1}\lambda$ and $v_{2} = v_{2}\lambda $
$\therefore \frac{v_{1}}{v_{2}} = \frac{v_{1}}{v_{2}} = \sqrt{n}$