Wave Motion 2 Question 6
6. Equation of travelling wave on a stretched string of linear density $5 g / m$ is $y=0.03 \sin (450 t-9 x)$, where distance and time are measured in SI units. The tension in the string is
(Main 2019, 11 Jan I)
(a) $5 N$
(b) $12.5 N$
(c) $7.5 N$
(d) $10 N$
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Answer:
Correct Answer: 6. (b)
Solution:
- Given, equation can be rewritten as,
$$ y=0.03 \sin 450\left(t-\frac{9 x}{450}\right) $$
We know that the general equation of a travelling wave is given as,
$$ y=A \sin \omega(t-x / v) $$
Comparing Eqs. (i) and (ii), we get
$$ \text { velocity, } v=\frac{450}{9}=50 m / s $$
and angular velocity, $\omega=450 rad / s$
As, the velocity of wave on stretched string with tension $(T)$ is given as $v=\sqrt{T / \mu}$
where, $\mu$ is linear density
$\therefore \quad T=\mu v^{2}=5 \times 10^{-3} \times 50 \times 50=12.5 N$
$$ \left(\because \text { given, } \mu=5 g / m=5 \times 10^{-3} kg / m\right) $$
