Question

A wire under a tension of 144 N vibrating in its fundamental mode gives 5 beats per second with a tuning fork. When the tension applied to the wire is increased to 169 N, the number of beats heard per second remains the same. The frequency of the tuning fork is

• A. 125 Hz
• B. 60 Hz
• C. 65 Hz
• D. 55 Hz

Hint:

- Beat frequency = difference between frequencies. - Changing tension changes frequency: $f \propto \sqrt{T}$. - Always use ratios of tensions to find new frequency. - Consider both possibilities ($f_\text{t}>f$ or $f_\text{t}<f$) when solving beat problems.

Answer 1

The Correct Option is A

1. Fundamental frequency of stretched string: $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, $\mu$ = linear density.
2. Let wire frequency at $T_1 = 144~\text{N}$ be $f_1$, at $T_2 = 169~\text{N}$ be $f_2$.
3. Given beat frequency with tuning fork $f_\text{t} = |f_1 - f_\text{t}| = |f_2 - f_\text{t}| = 5~\text{Hz}$.
4. Using ratio: $f_2/f_1 = \sqrt{T_2/T_1} = \sqrt{169/144} = 13/12$.
5. Let $f_1 = f_\text{t} - 5$ or $f_\text{t} + 5$. Solving gives $f_\text{t} = 125~\text{Hz}$.