Two identical piano wires kept under the same tension $T$ have a fundamental frequency of $600 \mathrm{~Hz}$…
Two identical piano wires kept under the same tension $T$ have a fundamental frequency of $600 \mathrm{~Hz}$. The fractional increase in the tension of one of the wires which will lead to occurrence of 6 beat/s when both the wires oscillate together would be
0.02
0.03
0.04
0.01
Solution
According to law of tension the frequency of the string varies directly as the square root of its tension
$\text {or } \quad \begin{aligned}
n & \propto \sqrt{T} \\
\frac{\Delta n}{n} & =\frac{1}{2} \cdot \frac{\Delta T}{T} \\
\frac{\Delta T}{T} & =2 \times \frac{\Delta n}{n} \\
\frac{\Delta T}{T} & =2 \times \frac{6}{600} \\
& =0.02
\end{aligned}$
.