# A string is under tension so that its length is increased by 1/n times its original length…..

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

(a)   1 : n

(b) n2 : 1

(c) √n : 1

(d) n : 1

Ans: (c)

Sol:
Velocity of Longitudinal waves , $\displaystyle v_1 = \sqrt{\frac{Y}{\rho}}$

Velocity of Transverse waves , $\displaystyle v_2 = \sqrt{\frac{T}{\mu}}$

Where μ = Mass per unit length = (Mass/Volume)Area = ρ A

$\displaystyle v_2 = \sqrt{\frac{T}{\rho A}}$

$\displaystyle \frac{v_1}{v_2} = \sqrt{\frac{Y \rho A}{\rho T}}$

$\displaystyle \frac{v_1}{v_2} = \sqrt{\frac{Y A}{ T}}$ …(i)

As , $\displaystyle Y = \frac{T l}{A \Delta l}$

$\displaystyle \frac{Y A}{T} = \frac{l}{\Delta l}$

From(i)

$\displaystyle \frac{v_1}{v_2} = \sqrt{\frac{l}{\Delta l}}$

$\displaystyle \frac{v_1}{v_2} = (\frac{\Delta l}{l})^{-1/2}$

But , $\displaystyle \frac{\Delta l}{l}= \frac{1}{n}$

$\displaystyle \frac{v_1}{v_2} = (\frac{1}{n})^{-1/2} = \sqrt{n}$