Question

Two identical wires are stretched by same tension of $100N$ and each emits a note of frequency $200cycles/sec$. The tension in one wire is increased by $1N$. Calculate the number of beats heard per second when the wires are plucked simultaneously.

Answer 1

Frequency of wave in the string is $f=\frac{v}{2l}=\frac{1}{2l}\sqrt{\frac{T}{\mu }}...\left(1\right)$ where $T=$ tension in the string and $\mu =$ mass per length of the string.
$df=\frac{1}{2l}\sqrt{\frac{1/2}{\mu T}}dT....\left(2\right)$
$\left(2\right)/\left(1\right),\frac{df}{f}=\frac{1}{2}.\frac{dT}{T}$
$\therefore df=\frac{f}{2}.\frac{dT}{T}=\frac{200}{2}\times \frac{1}{100}=1$
Thus number of beats heard per second $=1$