Question

The transverse displacement of a string of a linear density $0.01\, kg\, m ^{-1}$, 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 $=0.01\, kg / m$
The standard equation of transverse displacement of a wave
$y=a \sin k x \cos \omega t$
Velocity, $V=\sqrt{\frac{T}{\mu}}=\frac{\omega}{k}$
$\therefore \sqrt{\frac{T}{\mu}}=\frac{\omega}{k}$ [Here, $\mu=$ mass per unit length $]$
Here, $\omega =60\, \pi, \,k=2 \pi / 3$
$\sqrt{\frac{T}{0.01}} =\frac{60 \pi}{2 \pi / 3}$
$=90$
$\frac{T}{0.01} =90 \times 90$
$T =90 \times 90 \times 0.01$
$=81\, N$