Question

The speed of a wave on a string is $150\, ms^{-1}$ when the tension is $120\, N$. The percentage increase in the tension in order to raise the wave speed by $20\%$ is

• A. 44%
• B. 40%
• C. 20%
• D. 10%

Answer 1

Answer: (A)

Speed of wave on a string
$ v= \sqrt{\frac{T}{m}} $
or $ v \propto \sqrt T $
$ \therefore \frac{v_1}{v_2}= \sqrt{\frac{T_1}{T_2}} $
or $ \frac{T_1}{T_2}=\bigg(\frac{v_1}{v_2}\bigg)^2 $
or $ \frac{T_2 - T_1}{T_1}=\frac{v_2^2 - v_1^2}{v_1^2}$ ...(i)
Given, $ T_1 = 120\, m $ and $v_1 = 150\, ms^{-1}$
$ v_2 = v_1 +\frac{20}{100}v_1=\frac{120}{100}v_1 = \frac{6}{5} v_1 = \frac{6}{5} \times 150 $
$ = 180\, ms^{-1}$
Substituting the values in Eq. (i), we get
$ \frac{T_2 - T_1}{T_1}=\frac{(180)^2 - (150)^2}{(150)^2}$
$ =\frac{30 \times 330 }{150 \times150}=0.44$
Percent increase in tension
$ = \frac{T_2 - T_1}{T_1} \times 100 $
$= 0.44 \times 100 = 44\% $