Question

Two identical piano wires have a fundamental frequency of $60vib/sec$. when kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of six beats per second when both wires vibrate simultaneously.

Answer 1

Fundamental frequency $\alpha \frac{V}{l}$

where $V=$ Speed of wave in wire.

$l=$ Length of wire.

Fundamental frequency$=\frac{KV}{l}$ $[K=$ Constant$]$

According to the question

$\frac{KV}{l}=600$

$\because V=\sqrt{\frac{T}{\mu }}$

Where $T=$ Tension in wire.

$\mu =$ Linear mass density of wire.

$\therefore \frac{K\sqrt{T}}{l\sqrt{\mu }}=600⟶\left(i\right)$

$\sqrt{T}=\frac{600\sqrt{\mu }l}{K}$

Now, According to question

$f_1-f=6$$[$ Due to change in tension of one wire .Let that wire be $1]$

$\frac{K\sqrt{T_1}}{l\sqrt{\mu }}-\frac{K\sqrt{T}}{l\sqrt{\mu }}=6$

$\frac{K\sqrt{T_1}}{l\sqrt{\mu }}-600=6$

$\frac{\sqrt{T_1}.600}{\sqrt{T}}-600=6$ $[$ From equation $\left(i\right)]$

$600[\frac{\sqrt{T_1}}{\sqrt{T}}-1]=6$

$\sqrt{\frac{T_1}{T}}-1=\frac{1}{100}$

$\sqrt{\frac{T_1}{T}}=\frac{101}{100}$

$\frac{T_1}{T}=1.02$

$T_1=1.02T$

Fractional Increase in $T=\frac{1.02T-1T}{1T}$

$=0.02$