Question

The transverse displacement of a string fixed at both ends is given by $y=0.6 \sin \left(\frac{2 \pi x}{3}\right) \cos (120 \pi t)$ 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 $3.0 \times 10^{-2} kg$. The tension in the string is equal to

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

Answer 1

Answer: (A)

648 N

Answer 2

Given that $y=0.06 \sin \left(\frac{2 \pi x}{3}\right) \cos (120 \pi t)$
This is comparing with
$y=2 a \sin \left(\frac{2 \pi x}{\lambda}\right) \cos \left(\frac{2 \pi v t}{T}\right)$
we get $\lambda=3 \,m, v=60\, H z$
Now $v=v \lambda=60 \times 3=180\, m / s$
Now $v=\sqrt{\left(\frac{T}{\mu}\right)} 180$
$=\sqrt{\left(\frac{T}{\mu}\right)} T$
$=-180 \times 180 \times \mu$
$=180 \times 180 \times \frac{m}{l}$
$=180 \times 180 \times \frac{3 \times 10^{-2}}{1.5}$
$=648\, N$