Question

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

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

Answer 1

Answer: (B)

The given equation is
$y\left(x, t\right) = 0.06\,sin \left(\frac{2\pi}{3}x\right)cos\left(120\,\pi t\right)$
Compare it with $y\left(x, t\right) = 2a\, sin\, kx\, cos \,\omega t$
we get, $k = \frac{2\pi}{3}$,
or $\frac{2\pi}{\lambda} = \frac{2\pi}{3}$
or $\lambda = 3\,m$
and $\omega = 120\pi$ or $2\pi\upsilon = 120\pi$ or $\upsilon = 60\,Hz = 60\,s^{-1}$
Velocity of wave, $v= \upsilon\lambda = \left(60\,s^{-1}\right)\,\left(3\,m\right) = 180\,ms^{-1}$
Mass per emit length of the string,
$\mu = \frac{3\times10^{-2}\,kg}{1.5\,m} = 2 \times 10^{-2}\,kg\,m^{-1}$
Velocity of transverse wave in the string,
$v = \sqrt{\frac{T}{\mu}}$ or $v^{2} = \frac{T}{\mu}$ or $T = v^{2}\mu$
$T = \left(180 \,m s^{- 1}\right)^{2 }\left(2 \times10^{-2}\, kg \,m ^{-1}\right) = 648 \,N$