Question

The transverse displacement of a string fixed at both ends is given by
$y=0.06\sin \left(\frac{2\pi x}{3}\right)\cos \left(120\pi t\right)$ where $x$ and $y$ are in metres and $t$ is in seconds. The length of the string is $1.5m$ and its mass is $3.0\times {10}^{-2}kg$. The tension in the string is

• A. $648N$
• B. $724N$
• C. $832N$
• D. $980N$

Answer 1

The correct option is A $648N$

mass per unit length $=\frac{m}{l}=\frac{3\times {10}^{-2}}{1.5}=2\times {10}^{-2}$
Velocity of wave $=\frac{w}{K}=\frac{120\pi }{\frac{2\pi }{3}}=180$
$v=\sqrt{\frac{T}{\mu }}$
$180=\sqrt{\frac{T}{2\times {10}^{-2}}}$
$\Rightarrow T=(180{)}^2\times 2\times {10}^{-2}$
$=324\times {10}^2\times 2\times {10}^{-2}$
$=648N$