Question

A string is clamped at both the ends and it is vibrating in its <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $4^{\text{th}}$ harmonic. The equation of the stationary wave is $Y=0.3\sin \left(0.157x\right)\cos \left(200\pi t\right).$ The length of the string is
(All quantities are in SI units.)

• A. $40\text{m}$
• B. $20\text{m}$
• C. $60\text{m}$
• D. $80\text{m}$

Answer 1

The correct option is D $80\text{m}$

Given,
$y=0.3\sin \left(0.157x\right)\cos \left(200\pi t\right)$

So,
$k=0.157\text{and}\omega =200\pi =2\pi f$

$\therefore f=100\text{Hz}\text{and}v=\frac{\omega }{k}=\frac{200\pi }{0.157}=4000\text{m}/\text{s}$

Now, using $f=\frac{nv}{2l}=\frac{4v}{2l}=\frac{2v}{l}$
Here $n=4$
$\therefore l=\frac{2v}{f}=\frac{2\times 4000}{100}=80\text{m}$