The transverse displacement of a string of a linear density $0.01 \mathrm{~kg} \mathrm{~m}^{-1}$, clamped at…
The transverse displacement of a string of a linear density $0.01 \mathrm{~kg} \mathrm{~m}^{-1}$, 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
9 N
36 N
162 N
81 N
Solution
Transverse displacement
$
y_{(x, t)}=0.03 \sin \left(\frac{2 \pi x}{3}\right) \cos 60 \pi t \text {. }
$
This is equation of standing wave.
The standard equation of standing wave is $y=a \sin k x \cos \omega t$
Here, $\omega=60 \pi, k=2 \pi / 3$
So, $V=\frac{60 \pi}{2 \pi} \times 3=90$
So, $\sqrt{\frac{T}{\mu}}=90$
$
\Rightarrow \quad T=90^2 \mu \Rightarrow T=90^2 \times 10=81 \mathrm{~N}
$