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 $a_o$. The maximum kinetic energy of the string for its fundamental tone is given as $\frac{a_o^2{\pi }^{2}T}{xl}$. Find $x$.

Answer 1

Since tension on the string is $T$ so the velocity of the wave should be $v=\sqrt{\frac{T}{m}}$ where $m$ is mass per unit length.

Since frequency is $n=\frac{v}{\gamma }$ there for the frequency of the fundamental tone should be $n_1=\frac{1}{2l}\sqrt{\frac{T}{m}}$

Now consider an elemental length of a string at a distance $x$ of size $dx$ so its mass is $dm=m.dx$

So its oscillation energy is given by $\frac{1}{2}(m.dx)a^2({2\pi n_1)}^2$

$=\frac{a_o^2{\pi }^{2}T}{2l^2}{\sin }^2\left(\frac{2\pi x}{2l}\right)dx$

$[\because n_1=\frac{1}{2l}\sqrt{\frac{T}{m}}]$;

Integrating this we get, total energy as $\frac{a_o^2{\pi }^{2}T}{4l}$

This gives us $x=4$