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 $v_2=v_2\lambda$
$\therefore \frac{v_1}{v_2}=\frac{v_1}{v_2}=\sqrt{n}$