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) n^{2} : 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}$