A heavy but uniform rope of length L is suspended from a ceiling. A particle is dropped from the ceiling at the instant when the bottom end is given a transverse wave pulse. Where will the particle meet the pulse?

Question

Smart Achievers · Accepted Answer

(2)

For dropped body $L - x = \frac{1}{2} g t^{2}$ , for pulse $t = 2 \sqrt{\frac{x}{g}}$ , $x$ is the distance from bottom free end of row.

Solution

Q.
A heavy but uniform rope of length L is suspended from a ceiling. A particle is dropped from the ceiling at the instant when the bottom end is given a transverse wave pulse. Where will the particle meet the pulse

1 at a distance $\frac{2 L}{3}$ from the bottom

2 at a distance $\frac{L}{3}$ from the bottom

3 at a distance $\frac{3 L}{4}$ from the bottom

4 at a distance $\frac{L}{2}$ from the bottom

Ans.
(2)

For dropped body $L - x = \frac{1}{2} g t^{2}$ , for pulse $t = 2 \sqrt{\frac{x}{g}}$ , $x$ is the distance from bottom free end of row.

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Solution

Q. A heavy but uniform rope of length L is suspended from a ceiling. A particle is dropped from the ceiling at the instant when the bottom end is given a transverse wave pulse. Where will the particle meet the pulse

1 at a distance 2L3 from the bottom

2 at a distance L3 from the bottom

3 at a distance 3L4 from the bottom

4 at a distance L2 from the bottom

Ans. (2) For dropped body L-x=12gt2, for pulse t=2xg, x is the distance from bottom free end of row.