Question

A uniform string of length / is fixed at both ends such that tension T is produced in it. The string is excited to vibrate with maximum displacement amplitude $a_o$. The kinetic energy of the string for its f irst overtone is given as $\frac{a_0^2{\pi }^{2}T}{xl}$. Find $x$

Answer 1

Displacement equation of a stationary wave $y=2Asin\left(k_{n}x\right)sin\left(w_{n}t\right)$

First overtone means $n=2$

Kinetic energy of a small element $dE=\frac{1}{2}\left(\mu dx\right)(\frac{dy}{dt}{)}^2$

$⟹$ $dE=\frac{1}{2}\left(\mu dx\right)\left(2Asin\right(k_{n}x){)}^2\times w_n^{2}cos(w_{n}t)$

$\therefore$ $dE=2A^2{w}_n^2\mu si{n}^2\left(k_{n}x\right)dx$ .........(1)

Given : $A=a_o$

Also First overtone frequency ${\nu }_2=\frac{2}{2l}\sqrt{\frac{T}{\mu }}$ $⟹$ ${\nu }_2^2=\frac{T}{\mu l^2}$

As $w_n=2\pi {\nu }_2$

$\therefore$ $w_n^2=4{\pi }^2{\nu }_2^2=4{\pi }^2\times \frac{T}{\mu l^2}$

Also $k_n=\frac{n\pi x}{2l}=\frac{2\pi x}{2l}=\frac{\pi x}{l}$

Putting these values in (1) we get, $dE=2a_o^2\mu \times (4{\pi }^2\times \frac{T}{\mu l^2})si{n}^2\left(k_{n}x\right)dx$

$⟹$ $dE=\frac{8a_o^2{\pi }^{2}T}{l^2}si{n}^2(\frac{\pi x}{l})dx$

Total kinetic energy $E=\frac{8a_o^2{\pi }^{2}T}{l^2}\int si{n}^2(\frac{\pi x}{l})dx$

$E=\frac{8a_o^2{\pi }^{2}T}{l^2}{\int }_0^l\frac{1-cos\left(2\pi x/l\right)}{2}dx$

$E=\frac{8a_o^2{\pi }^{2}T}{l^2}\times \frac{1}{2}\times [x{|}_0^l-\frac{sin\left(2\pi x/l\right)}{\left(2\pi x/2\right)}{|}_0^l]$

$E=\frac{8a_o^2{\pi }^{2}T}{l^2}\times \frac{1}{2}\times [l-0]$ $⟹$ $E=\frac{4a_o^2{\pi }^{2}T}{l}$

$\therefore$ $x=4$