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 fundamental tone 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 A

$\frac{a_0^2{\pi }^{2}T}{4l}$

When a string oscillates, nodes are produced at its ends. In case of fundamental tone, it vibrates in single loop. Hence, wavelength of fundamental tone, ${\lambda }_0=2l$ and in case of first overtone it vibrates in two loops as shown in figure. Hence wavelength of first overtone is ${\lambda }_1=l$.

When string particles performs SHM, displacement amplitude at a point x away from one end is given by

$a=a_{0}sin\left(\frac{2\pi x}{\lambda }\right)$ . . . .(1)

where $a_0$ is maximum displacement amplitude which occurs at antinode. Since, tension in string is T, velocity of transverse wave is given by $v=\sqrt{\frac{T}{m}}$ where m is mass per unit length of string.

Fundamental tone : Since, frequency is $n=\frac{v}{\lambda }$ , therefore, frequency of fundamental tone.

$n_0=\frac{1}{2l}\sqrt{\frac{T}{m}}$

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

Its oscillation energy $=\frac{1}{2}\left(mdx\right)a^2(2\pi n_0{)}^2$

$=\frac{a_0^2{\pi }^{2}T}{2l^2}si{n}^2\left(\frac{2\pi x}{2l}\right)$ dx

$\therefore$ Total oscillation energy of the string

$=\frac{a_0^2{\pi }^{2}T}{2l^2}{\int }_0^{l}si{n}^2\left(\frac{\pi x}{2l}\right)dx=\frac{a_0^2{\pi }^{2}T}{4l}$

Since, maximum kinetic energy of a particle performing SHM is equal to its oscillation energy, therefore, maximum kinetic energy of the string in its fundamental tone $=\frac{a_0^2{\pi }^{2}T}{4l}$