Question

A string of length $l$ along $x-$ axis is fixed at both ends and is vibrating in second harmonic. If amplitude of incident wave is $2.5\text{mm}$, the equation of standing wave is (T is tension and $\mu$ is linear density)

• A. $\left(2.5\text{mm}\right)\text{sin}\left(\frac{2\pi }{l}x\right)\text{cos}\left(2\pi \sqrt{\left(\frac{T}{\mu l^2}\right)}t\right)$
• B. $\left(5\text{mm}\right)\text{sin}\left(\frac{\pi }{l}x\right)\text{cos}2\pi t$
• C. $\left(5\text{mm}\right)\text{sin}\left(\frac{2\pi }{l}x\right)\text{cos}\left(2\pi \sqrt{\left(\frac{T}{\mu l^2}\right)}t\right)$
• D. $\left(7.5\text{mm}\right)\text{cos}\left(\frac{2\pi }{l}x\right)\text{cos}\left(2\pi \sqrt{\left(\frac{T}{\mu l^2}\right)t}\right)$

Answer 1

The correct option is C $\left(5\text{mm}\right)\text{sin}\left(\frac{2\pi }{l}x\right)\text{cos}\left(2\pi \sqrt{\left(\frac{T}{\mu l^2}\right)}t\right)$

Equation of standing wave fixed at both ends is given as,
$y=2\text{a sin(kx) cos}\left(\omega t\right)$
$\Rightarrow$ Putting $a=2.5\text{mm}$ (amplitude of incident wave)

For second harmonic , $n=2$
$\Rightarrow l=n\times \left(\frac{\lambda }{2}\right)$
$\therefore l=2\times \frac{\lambda }{2}=\lambda$
Using $k=\frac{2\pi }{\lambda }$
$k=\frac{2\pi }{l}$

Angular frequency
$\Rightarrow \omega =vk=\frac{2\pi }{l}\sqrt{\frac{T}{\mu }}$
Substituting value of $\lambda$, we get
$\omega =2\pi \sqrt{\frac{T}{\mu l^2}}$
Hence equation of standing wave will be,
$y=2\text{a sin}\left(\frac{2\pi x}{l}\right)\text{cos}\left[2\pi \sqrt{\frac{T}{\mu l^2}}t\right]$
$\therefore y=5\text{mm}\text{sin}\left(\frac{2\pi }{l}x\right)\text{cos}\left(2\pi \sqrt{\frac{T}{\mu l^2}}t\right)$