Physics - Sound Wave Question with Solution | TestHub

When both the wires vibrate simultaneously, beats per second, $$\begin{array}{c}{n}_1\pm n_2=6\\ \text{ or }\frac{1}{2l}\sqrt{\frac{T}{m}}\pm \frac{1}{2l}\sqrt{\frac{T^{\prime }}{m}}=6\\ \Rightarrow \frac{1}{2l}\sqrt{\frac{T^{\prime }}{m}}-\frac{1}{2l}\sqrt{\frac{T}{m}}=6\\ \Rightarrow \frac{1}{2l}\sqrt{\frac{T^{\prime }}{m}}-600=6\Rightarrow \frac{1}{2l}\sqrt{\frac{T^{\prime }}{m}}=\mathrm{606...}(i)\end{array}$$ Given that fundamental frequency $$\frac{1}{2l}\sqrt{\frac{T}{m}}=\mathrm{600...}(ii)$$ Dividing Eq. (i) by(ii), $$\begin{array}{l}\frac{\frac{1}{2l}\sqrt{\frac{T^{\prime }}{m}}}{\frac{1}{2l}\sqrt{\frac{T}{m}}}=\frac{606}{600}\\ \Rightarrow \sqrt{\frac{T^{\prime }}{T}}=(1.01)\Rightarrow \frac{T^{\prime }}{T}=(1.02)\%\\ \Rightarrow T^{\prime }=T(1.02)\end{array}$$ Increase in tension $$\begin{array}{l}\Delta T^{\prime }=T\times 1.02-T=(0.02T)\\ \therefore \Delta T=0.02\end{array}$$