Question

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

• A. $324N$
• B. $684N$
• C. $832N$
• D. $972N$

Answer 1

Answer: (B)

The given equation is
$y\left(x,t\right)=0.06sin\left(\frac{2\pi }{3}x\right)cos\left(120\pi t\right)$
Compare it with $y\left(x,t\right)=2asinkxcos\omega t$
we get, $k=\frac{2\pi }{3}$,
or $\frac{2\pi }{\lambda }=\frac{2\pi }{3}$
or $\lambda =3m$
and $\omega =120\pi$ or $2\pi \upsilon =120\pi$ or $\upsilon =60Hz=60s^{-1}$
Velocity of wave, $v=\upsilon \lambda =\left(60s^{-1}\right)\left(3m\right)=180m{s}^{-1}$
Mass per emit length of the string,
$\mu =\frac{3\times 10^{-2}kg}{1.5m}=2\times 10^{-2}kg{m}^{-1}$
Velocity of transverse wave in the string,
$v=\sqrt{\frac{T}{\mu }}$ or $v^2=\frac{T}{\mu }$ or $T=v^2\mu$
$T={\left(180m{s}^{-1}\right)}^2\left(2\times 10^{-2}kg{m}^{-1}\right)=648N$