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 one 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

The Correct Option is B

When both the wires vibrate simultaneously, beats per second,
$n_{1}\pm n_2=6$
or $\frac{1}{2l} \sqrt{\frac{T}{m}}+\frac{1}{2l} \sqrt{\frac{T '}{m}}=6$
$\Rightarrow \frac{1}{2l} \sqrt{\frac{T '}{m}}-\frac{1}{2l} \sqrt{\frac{T }{m}}=6$
$\Rightarrow \frac{1}{2l} \sqrt{\frac{T '}{m}}-600-6$
or $ \frac{1}{2l} \sqrt{\frac{T ' }{m}}=606\,\,\,\,\ldots\left(i\right)$
Given that fundamental frequency
$\Rightarrow \frac{1}{2l} \sqrt{\frac{T }{m}}=600\,\,\,\,\ldots\left(ii\right)$
Dividing E (i) by (ii),
$\frac{\frac{1}{2l}\sqrt{\frac{T'}{m}}}{\frac{1}{2l}\sqrt{\frac{T}{m}}} =\frac{606}{600}$
or $ \sqrt{\frac{T '}{T}}=\left(1.01\right) $
or $ \frac{T '}{T}=\left(1.02\right)\%$
or $ T '=T\left(1.02\right)$
Increase in tension
$\Delta T ' =T\times1.02-T $
$=\left(0.2T\right)$
Hence$\Delta T' =0.02$