Question

The speed of a wave on a string is $150\mathrm{m}/\mathrm{s}$ when the tension is $120\mathrm{N}$. The percentage increase in the tension in order to raise the wave speed by $20\%$ is

• A. $10\%$
• B. $20\%$
• C. $40\%$
• D. $44\%$

Answer 1

The correct option is D

$44\%$

Step 1: Given parameters

Speed of a wave on the string $\left(v_1\right)$ = $150\mathrm{m}/\mathrm{s}$.

Tension in the string $\left(T_1\right)$ = $120\mathrm{N}$.

Step 2: Parameter to be calculated

Percentage increase in tension in order to raise the wave speed by $20\%$.

Step 3: Formula used

Speed of wave on a string $\left(v\right)$,

$v=\sqrt{\frac{T}{m}}$

Where, $T$ is the tension, and $m$ is the mass of the string.

$v\propto \sqrt{T}$

$\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}$

$\frac{T_1}{T_2}={\left(\frac{v_1}{v_2}\right)}^2$

$$\begin{gathered} \frac{T_2}{T_1}=\frac{v_2^2}{v_1^2} \\ Subtracting1fromboththesides \\ \frac{T_2}{T_1}-1=\frac{v_2^2}{v_1^2}-1 \\ \frac{T_2-T_1}{T_1}=\left(\frac{v_2^2-v_1^2}{v_1^2}\right) \end{gathered}$$

From the given statement, the wave speed $\left(v_2\right)$ is increased by $20\%$ ,

$$\begin{gathered} v_2=v_1+\frac{20}{100}{v}_1=\frac{120}{100}{v}_1 \\ {v}_2=\frac{6}{5}{v}_1=\frac{6}{5}\times 150 \\ {v}_2=180\mathrm{m}/\mathrm{s} \end{gathered}$$

Now,

$\frac{T_2-T_1}{T_1}=\frac{{v_2}^2-{v_1}^2}{{v_1}^2}$

After substituting $v_2=180$ and $v_1=150$, we get

$$\begin{gathered} \frac{T_2-T_1}{T_1}=\frac{{180}^2-{150}^2}{{150}^2} \\ =\frac{32400-22500}{22500} \\ =0.44 \end{gathered}$$

Percentage decrease,

$$\begin{gathered} \frac{T_2-T_1}{T_1}\times 100=0.44\times 100 \\ =44\% \end{gathered}$$

Hence the percentage increase in the tension in order to raise the wave speed by $20\%$ is $44\%$.

Therefore option (D) is correct.