Question

The tension of a stretched string is increases by $44\%$. In order to keep its frequency of vibration same its length must be increased by.

• A. $10\%$
• B. $20\%$
• C. $15\%$
• D. $25\%$

Answer 1

Answer: (B)

$20\%$

The frequency of vibration of a stretched string is given by
$f=\frac{p}{2l}\sqrt{\frac{T}{m}}$
where, $p=$ number of harmonics
$m=$ linear mass density
$\therefore f^2=\frac{p^2}{4l^2}\cdot \frac{T}{m}$
$l^2=\left(\frac{p^2}{4f^{2}m}\right)T$
Here, $\frac{p^2}{4f^{2}m}=$ constant
So, $l^2\propto t$
$\therefore \frac{l^{\prime }}{l}={\left(\frac{T+0.44T}{T}\right)}^{1/2}=(1.44{)}^{\frac{1}{2}}$
$\therefore l^{\prime }=1.2l=l+0.2l$
So, the length is increased by
$0.2\times 100=20\%$.