Question

The transverse displacement of a string fixed at both ends is given by $y=0.06 \sin \left(\frac{2 \pi x}{3}\right) \cos (100 \pi t)$ where $x$ and $y$ are in metres and $t$ is in seconds. The length of the string is $1.5\, m$ and its mass is $3.0 \times 10^{-2} \,kg .$ What is the tension in the string?

• A. 225 N
• B. 300 N
• C. 450 N
• D. 675 N

Answer 1

Answer: (C)

On comparing $y=0.06 \sin \left(\frac{2 \pi x}{3}\right) \cos (100 \pi t)$
with $y=A \sin (k x) \cos (\omega t)$, we get
$k=\frac{2 \pi}{3}$
and $\omega=100 \pi rad / s$
Velocity of the wave, $v=\frac{\omega}{k}=\frac{(100 \pi rad / s )}{\left(\frac{2 \pi}{3} rad / m \right)}=150 \,m /\, s\,\,\,\,\,\,\,...(i)$
Here, $L=1.5 \,m , m=3.0 \times 10^{-2} \,kg$
Mass per unit length of the string,
$\mu=\frac{m}{L}=\frac{3.0 \times 10^{-2}\, kg }{1.5\, m }=2 \times 10^{-2} \,kg / m$
Velocity of a transverse wave on a stretched string is
$v=\sqrt{\frac{T}{\mu}}\,\,\,\,\,\,\,\,...(ii)$
where $T$ is the tension in the string and $\mu$ is the mass per unit length of the string.
From (i) and (ii), we get
$150=\sqrt{\frac{T}{2 \times 10^{-2}}}$ or $T=(150)^{2} \times 2 \times 10^{-2} N =450\, N$