Question

The transverse displacement of a string clamped at its both ends is given by
$y(x,t)=0.06$sin $\left(\frac{2\pi }{3}x\right)$ cos $\left(120\pi t\right)$ where $x$ and $y$
are in $m$ and $t$ in $s$. The length of the string is $1.5m$ and its mass is $3\times {12}^{-2}$ kg. The tension in the string is:

• A. $324N$
• B. $648N$
• C. $832N$
• D. $972N$

Answer 1

The correct option is B $648N$

The given equation is
$y(x,t)=0.06$sin $\left(\frac{2\pi }{3}x\right)$cos$\left(120\pi t\right)$

Copmpare it with $y(x,t)=2$a sin $kx$ cos $\omega t$

we get, $k=\frac{2\pi }{3}$, or $\frac{2\pi }{\lambda }=\frac{2\pi }{3}$ or $\lambda =3$
and $\omega =120\pi$ or $2\pi \upsilon$ or $\upsilon =60Hz=60s^{-1}$

$\mu =\sqrt{\frac{3\times {10}^{-2}kg}{1.5m}}=2\times {10}^{-2}kg{m}^{-1}$

Velocity of transverse wave in the string,

$\nu =\sqrt{\frac{T}{\mu }}$ or ${\nu }^2=\frac{T}{\mu }$ or $T={\nu }^2\mu$

$T=\left(180m{s}^{-1}{)}^2\right(2\times {10}^{-2}kg{m}^{-1})=648N$