Question

A uniform string of length l is fixed at both ends such that tension T is produced in it. The string is excited to vibrate with maximum displacement amplitude $a_0$.

The maximum kinetic energy of the string for its first overtone is

• A. $\frac{a_0^2{\pi }^{2}T}{4l}$
• B. $\frac{a_0^2{\pi }^{2}T}{l}$
• C. $\frac{a_0^2{\pi }^{2}T}{2l}$
• D. $\frac{a_0^2{\pi }^{2}T}{3l}$

Answer 1

The correct option is B

$\frac{a_0^2{\pi }^{2}T}{l}$

First overtone : Frequency, $n_1=\frac{v}{{\lambda }_1}=\frac{1}{l}\sqrt{\frac{T}{m}}$

Considering an elemental length dx of string at a distance from left end, its mass = m dx.

Its oscillating energy $=\frac{1}{2}\left(mdx\right)a_0^2.si{n}^2\left(\frac{2\pi x}{l}\right)(2\pi n_1{)}^2$

$=\frac{2{\pi }^2{a}_0^{2}T}{l^2}si{n}^2\left(\frac{2\pi x}{l}\right)dx$

$\therefore$ Total oscillation energy of the string

$=\frac{2{\pi }^2{a}_0^{2}T}{l^2}{\int }_0^{l}si{n}^2\left(\frac{2\pi x}{l}\right)dx$

or maximum kinetic energy of string in its overtone$=\frac{a_0^2{\pi }^{2}T}{l}$