Question

Two identical piano wires have a fundamental frequency of $ \, 600$ cycle per second when kept under the same tension. What fractional increase in the tension of wires will lead to the occurrence of $6$ beats per second when both wires vibrate simultaneously?

• A. $0.01$
• B. $0.02$
• C. $0.03$
• D. $0.04$

Answer 1

Answer: (B)

Beats per second when both the wires vibrate simultaneously
n₁ ± n₂ = 6
or $\frac{1}{2 l }\sqrt{\frac{\textit{T}}{\textit{m}}}\pm\frac{1}{2 l ⁡}\sqrt{\frac{\textit{T}^{'}}{\textit{m}}}=6$
or $\frac{1}{2 l }\sqrt{\frac{\textit{T}^{'}}{\textit{m}}}-\frac{1}{2 l ⁡}\sqrt{\frac{\textit{T}}{\textit{m}}}=6$
or $\frac{1}{2 l }\sqrt{\frac{\textit{T}^{'}}{\textit{m}}}-600=6$ ...(i)
Given that fundamental frequency
$\frac{1}{2 l} \sqrt{\frac{T}{m}}=600$
Dividing Eq.(i) by Eq. (ii), we get
$\frac{\frac{1}{2 l } \sqrt{\frac{\textit{T}^{'}}{\textit{m}}}}{\frac{1}{2 l ⁡} \sqrt{\frac{\textit{T}}{\textit{m}}}}=\frac{6 0 6}{6 0 0}$
Or $\sqrt{\frac{\textit{T}^{'}}{\textit{T}}}=\text{1.01}$
Or $\frac{ T ^{'}}{ T ⁡} = 1 \text{.} 0 2 \text{\%}$
Or $T'=T\left(\right.1.02\left.\right)$
Increase in tension
$\Delta T'=\left(\right.T\times 1.02\left.\right)-T$
$=0.02T$
Hence, $\Delta T'=0.02$