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:

  1. 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

$\therefore \quad T=\mu v^{2}=5 \times 10^{-3} \times 50 \times 50=12.5 N$

$$ \left(\because \text { given, } \mu=5 g / m=5 \times 10^{-3} kg / m\right) $$



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