Question

The equation of transverse displacement of a wire of mass \(20\,\mathrm{g}\) and length \(100\,\mathrm{cm}\) clamped at its ends is \[ y(x,t)=0.05\sin\!\left(\frac{2\pi}{5}x\right)\cos(80\pi t), \] where \(x\) is in metre and \(t\) is in second. The tension in the wire is

• A. \(200\,\mathrm{N}\)
• B. \(600\,\mathrm{N}\)
• C. \(400\,\mathrm{N}\)
• D. \(800\,\mathrm{N}\)

Hint:

For waves on a stretched string, \[ \boxed{ v=\frac{\omega}{k} =\sqrt{\frac{T}{\mu}} } \] where \[ \mu=\frac{\text{mass}}{\text{length}}. \]

Answer 1

The Correct Option is D

Step 1: Identify the wave parameters. Comparing with \[ y=A\sin(kx)\cos(\omega t), \] we get \[ k=\frac{2\pi}{5}\ \mathrm{m^{-1}}, \qquad \omega=80\pi\ \mathrm{rad\,s^{-1}}. \] Wave speed, \[ v=\frac{\omega}{k} =\frac{80\pi}{2\pi/5} =200~\mathrm{m\,s^{-1}}. \]

Step 2: Calculate the linear mass density. \[ \mu=\frac{m}{L} =\frac{0.02}{1} =0.02~\mathrm{kg\,m^{-1}}. \]

Step 3: Find the tension. For a stretched string, \[ v=\sqrt{\frac{T}{\mu}}. \] Therefore, \[ T=\mu v^2 =0.02\times(200)^2 =800~\mathrm{N}. \] Hence, \[ \boxed{T=800~\mathrm{N}} \] Therefore, \[ \boxed{(D)} \] is the correct answer.