Question

The speed of a wave on a string is $150m{s}^{-1}$ when the tension is $120N$. 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 $\frac{T_1}{T_2}=(\frac{v_1}{v_2}{)}^2$
or $\frac{T_2-T_1}{T_1}=\frac{v_2^2-v_1^2}{v_1^2}$ ...(i)
Given, $T_1=120m$ and $v_1=150m{s}^{-1}$
$v_2=v_1+\frac{20}{100}{v}_1=\frac{120}{100}{v}_1=\frac{6}{5}{v}_1=\frac{6}{5}\times 150$
$=180m{s}^{-1}$
Substituting the values in Eq. (i), we get
$\frac{T_2-T_1}{T_1}=\frac{(180{)}^2-(150{)}^2}{(150{)}^2}$
$=\frac{30\times 330}{150\times 150}=0.44$
Percent increase in tension
$=\frac{T_2-T_1}{T_1}\times 100$
$=0.44\times 100=44\%$