Question

The transverse displacement of a string of a linear density $0.01kg{m}^{-1}$, clamped at its ends is given by $Y_{(x,t)}=0.03\sin \left(\frac{2\pi x}{3}\right)\cos (60\pi t)$, where $x$ and $y$ are in metres and time $t$ is in seconds. Tension in the string is

• A. 9 N
• B. 36 N
• C. 162 N
• D. 81 N

Answer 1

Answer: (D)

Transverse displacement
$y_{(X,t)}=0.03\sin \left(\frac{2\pi x}{3}\right)\cos 60\pi t$
Linear density $=0.01kg/m$
The standard equation of transverse displacement of a wave
$y=a\sin kx\cos \omega t$
Velocity, $V=\sqrt{\frac{T}{\mu }}=\frac{\omega }{k}$
$\therefore \sqrt{\frac{T}{\mu }}=\frac{\omega }{k}$ [Here, $\mu =$ mass per unit length $]$
Here, $\omega =60\pi ,k=2\pi /3$
$\sqrt{\frac{T}{0.01}}=\frac{60\pi }{2\pi /3}$
$=90$
$\frac{T}{0.01}=90\times 90$
$T=90\times 90\times 0.01$
$=81N$