Question

A string of length $l$ is fixed at both ends. It is vibrating in its $3^{rd}$ overtone with maximum amplitude $a$. The amplitude at a distance $\frac{l}{3}$ from one end is

• A. $a$
• B. $0$
• C. $\frac{\sqrt{3}a}{2}$
• D. $\frac{a}{2}$

Answer 1

The correct option is C $\frac{\sqrt{3}a}{2}$

General equation of wave in vibrating string when both ends fixed,

$y=2A\sin \left(kx\right)\cos \left(\omega t\right)$

For a string vibrating in its $n^{th}$ overtone or $(n+1{)}^{th}$ harmonic

$y=2A\sin \left(\frac{(n+1)\pi x}{l}\right)\cos \left(\omega t\right)...\left(1\right)$



For given problem,
$x=\frac{l}{3};2A=a$ and $n=3$

putting above value in eq. (1), we get

$y=\left[a\sin (\frac{4\pi }{l}.\frac{l}{3})\right]\cos \left(\omega t\right)$

$\Rightarrow y=a\sin (\frac{4\pi }{3})\cos \left(\omega t\right)$

$\Rightarrow y=-a\left(\frac{\sqrt{3}}{2}\right)\cos \left(\omega t\right)$

So, at $x=\frac{l}{3}$,
the amplitude of vibration is $\frac{\sqrt{3}a}{2}$.

Hence, option $\left(c\right)$ is the correct answer.