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)

0.02

Beats per second when both the wires vibrate simultaneously.
$n_1\pm n_2=6$
or $\frac{1}{2l}\frac{\overline{T}}{m}\pm \frac{1}{2l}\frac{\overline{T^{\prime }}}{m}=6$
or $\frac{1}{2l}\frac{\overline{T^{\prime }}}{m}-\frac{1}{2l}\frac{\overline{T}}{m}=6$
or $\frac{1}{2l}\frac{\overline{T^{\prime }}}{m}-600=6$
$\frac{1}{2l}\frac{\overline{T^{\prime }}}{m}=606$
Given that fundamental frequency ,
$\frac{1}{2l}\frac{\overline{T^{\prime }}}{m}=606$
Dividing Eq. (i) by Eq. (ii), we gt
$\frac{\frac{1}{2l}\frac{\overline{T}}{m}}{\frac{1}{2l}\frac{\overline{T^{\prime }}}{m}}=\frac{606}{600}$
Or $\frac{\overline{T^{\prime }}}{T}=1.01$
Or $\frac{T^{\prime }}{T}=1.02$
Or $T^{\prime }=T1.02$
Increase in tension
$\Delta T^{\prime }=T\times 10.2-T$
$=0.02T$
Hence, $\Delta T^{\prime }=0.02$