Question

Two wires are in unison. if the tension in one of the wires is increased by 2% , 5 beats are produced per second. the initial frequency of each wire is

• A. 200Hz
• B. 400Hz
• C. 500Hz
• D. 1000Hz

Answer 1

The correct option is C 500Hz

We know that,

$n\propto \sqrt{T}\Rightarrow \frac{\Delta n}{n}=\frac{\Delta T}{2T}$

If tension increases by $2\%,$ then frequency must increases by $1\%$

If initial frequency $n_1=n$ then final frequency $n-n=5$

$\Rightarrow \frac{101}{100}n-n=5\Rightarrow n=500Hz$

Short trick : If you can remember then apply following formula to solve such type of problems.Initial frequency of each wire $\left(n\right)$

$=\frac{\text{(Number of beats heard per}\sec )\times 200}{\text{(per centage change in tension of the wire})}$

Here $n=\frac{5\times 200}{2}=500Hz$