Question

To increase the frequency of transverse oscillations of a stretched string by 40%, the tension must be increased by

• A. \(100\%\)
• B. \(40\%\)
• C. \(96\%\)
• D. \(140\%\)

Hint:

Frequency is proportional to the square root of tension.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a stretched string, \(f = \dfrac{1}{2\ell}\sqrt{T/\mu}\), so \(f \propto \sqrt T\).

Step 2: Compute
A 40% rise means \(f' = 1.4f\). Then:
\[ \frac{T'}{T} = \left(\frac{f'}{f}\right)^2 = 1.4^2 = 1.96 \]
The tension must increase by \(1.96 - 1 = 0.96\), that is 96%. Option (B) forgets the square, and (A) and (D) are not consistent with it.

Final Answer:
The tension must increase by 96%, option (C). \[ \boxed{96\%} \]