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 $\frac{1}{2l}\sqrt{\frac{\text{T}}{\text{m}}}\pm \frac{1}{2l}\sqrt{\frac{{\text{T}}^{{}^{\prime }}}{\text{m}}}=6$
or $\frac{1}{2l}\sqrt{\frac{{\text{T}}^{{}^{\prime }}}{\text{m}}}-\frac{1}{2l}\sqrt{\frac{\text{T}}{\text{m}}}=6$
or $\frac{1}{2l}\sqrt{\frac{{\text{T}}^{{}^{\prime }}}{\text{m}}}-600=6$ ...(i)
Given that fundamental frequency
$\frac{1}{2l}\sqrt{\frac{T}{m}}=600$
Dividing Eq.(i) by Eq. (ii), we get
$\frac{\frac{1}{2l}\sqrt{\frac{{\text{T}}^{{}^{\prime }}}{\text{m}}}}{\frac{1}{2l}\sqrt{\frac{\text{T}}{\text{m}}}}=\frac{606}{600}$
Or $\sqrt{\frac{{\text{T}}^{{}^{\prime }}}{\text{T}}}=\text{1.01}$
Or $\frac{T^{{}^{\prime }}}{T}=1\text{.}02\text{\%}$
Or $T^{\prime }=T(1.02)$
Increase in tension
$\Delta T^{\prime }=(T\times 1.02)-T$
$=0.02T$
Hence, $\Delta T^{\prime }=0.02$