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प्रश्न
A heavy but uniform rope of length L is suspended from a ceiling. (a) Write the velocity of a transverse wave travelling on the string as a function of the distance from the lower end. (b) If the rope is given a sudden sideways jerk at the bottom, how long will it take for the pulse to reach the ceiling? (c) A particle is dropped from the ceiling at the instant the bottom end is given the jerk. Where will the particle meet the pulse?
उत्तर
(a) Let m be the mass per unit length of the string.
Consider an element at a distance x from the lower end.
Here,
Weight acting downwards = (mx)g
∴ Tension in the string at the upper part = mgx
The velocity of transverse vibration is given as
\[v = \sqrt{\left( \frac{T}{m} \right)} = \sqrt{\left( \frac{mgx}{m} \right)}\]
\[ \Rightarrow v = \sqrt{\left( gx \right)}\]
(b) Let the time taken be dt for the small displacement dx.
Thus, we have:
\[dt = \frac{dx}{v} = \frac{dx}{\sqrt{\left( gx \right)}}\]
\[\therefore \text{ Total time, } T = \int\limits_0^L \frac{dx}{\sqrt{\left( gx \right)}} = \sqrt{\left( \frac{4L}{g} \right)}\]
(c) Suppose after time t, the pulse meets the particle at a distance y from the lower end of the rope.
Now,
\[t = \int\limits_0^y \frac{dx}{\sqrt{\left( gx \right)}}\]
\[ = \sqrt{\left( \frac{4y}{g} \right)}\]
∴ Distance travelled by the particle in this time, S = \[L - y\]
Using the equation of motion, we get:
\[S = ut + \frac{1}{2} g t^2 \]
\[ \Rightarrow L - y = \left( \frac{1}{2} \right) g \times \left\{ \left( \sqrt{\frac{4y}{g}} \right)^2 \right\}\]
\[ \Rightarrow L - y = 2y\]
\[ \Rightarrow 3y = L\]
\[ \Rightarrow y = \frac{L}{3}\]
Thus, the particle will meet the pulse at a distance
\[\frac{L}{3}\] from the lower end.
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