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

Answer: (B)

Beats per second when both the wires vibrate simultaneously,
$ n_{1}\pm n_{2} = 6$
Or $ \frac{1}{2i} \frac{\overline{T}}{m} \pm \frac{1}{2l} \frac{ \overline{ T'}}{m} = 6 $
Or $\frac{1}{2l} \frac{ \overline{ T '}}{m} - \frac{1}{2l} \frac{ \overline{ T}}{m} = 6 $
Or $\frac{1}{2l} \frac{ \overline{ T '}}{m} - 600 = 6 $
$\frac{1}{2l} \frac{ \overline{ T '}}{m} = 606 $
Given that fundamental frequency
$\frac{1}{2l} \frac{ \overline{ T}}{m} = 600 $
Dividing Eq.(i) by Eq. (ii), we get
$\frac{\frac{1}{2l} \frac{ \overline{ T} '}{m}}{\frac{1}{2l} \frac{ \overline{ T}}{m}} = \frac{606}{600}$
Or $ \frac{ \overline{ T} '}{T} = 1.01$
Or $ \frac{T'}{T} = 1.02 \% $
Or $ T ' =T 1.02 $
Increase in tension
$\Delta T' = T \times 1.02 - T $
$= 0.02T $
Hence, $\Delta T' = 0.02 $