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. 0.44
• B. 0.4
• C. 0.2
• D. 0.1

Answer 1

Answer: (A)

0.44

Answer 2

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 E (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
$= \frac{T_2 - T_1}{T_1} \times 100$
$= 0.44 \times 100 = 44\%$