Question

The percentage increase in the speed of transverse waves produced in a stretched string if the tension is increased by 4 percent , will be ________ . 
 

Hint:

For small percentage changes, the concept of fractional error is very useful. If $y = k \cdot x^n$, then the fractional change is $\frac{\Delta y}{y} \approx n \frac{\Delta x}{x}$. Here, $v \propto T^{1/2}$, so the percentage change in v is half the percentage change in T.

Answer 1

Correct Answer: 2

The speed (v) of a transverse wave on a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$, where T is the tension and $\mu$ is the linear mass density.
From this formula, we can see that the speed is proportional to the square root of the tension: $v \propto \sqrt{T}$.
Let the initial tension be T and the initial speed be v.
The new tension, T', is the initial tension increased by 4%:
$T' = T + 0.04T = 1.04T$.
The new speed, v', will be proportional to the square root of the new tension:
$v' \propto \sqrt{T'} \implies \frac{v'}{v} = \sqrt{\frac{T'}{T}} = \sqrt{\frac{1.04T}{T}} = \sqrt{1.04}$.
We can use the binomial approximation for small changes: $(1+x)^n \approx 1+nx$.
$\sqrt{1.04} = (1+0.04)^{1/2} \approx 1 + \frac{1}{2}(0.04) = 1 + 0.02$.
So, $v' \approx 1.02v$.
The percentage increase in speed is calculated as:
$% \text{ increase} = \frac{v' - v}{v} \times 100 = \frac{1.02v - v}{v} \times 100 = 0.02 \times 100 = 2%$.