Wave Motion 5 Question 6
6. The extension in a string, obeying Hooke’s law, is $x$. The speed of transverse wave in the stretched string is $v$. If the extension in the string is increased to $1.5 x$, the speed of transverse wave will be
(1996, 2M)
(a) $1.22 v$
(b) $0.61 v$
(c) $1.50 v$
(d) $0.75 v$
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Answer:
Correct Answer: 6. (a) $\frac{2 \pi}{a}, \frac{b}{2 \pi}$
(b) $1.57 \times 10^{-3} s$
(d) $y=-1.6 A \sin a x \sin b t+0.2 A \cos (a x+b t)$. Antinodes are at $x=\left[n+\frac{(-1)^{n}}{2}\right] \frac{\pi}{a}$. Travelling wave is propagating in negative $x$-direction.
Solution:
- From Hooke’s law
Tension in a string $(T) \propto$ extension $(x)$
and speed of sound in string
$$ \begin{aligned} & v=\sqrt{T / \mu} \\ \text { or } \quad & \\ \text { v } & \propto \sqrt{T} \end{aligned} $$
Therefore, $\quad v \propto \sqrt{x}$
$x$ is increased to 1.5 times i.e. speed will increase by $\sqrt{1.5}$ times or 1.22 times. Therefore, speed of sound in new position will be $1.22 v$.
