A heavy but uniform rope of length L is suspended from a ceiling. A particle is dropped from the ceiling at the exact instant the bottom end of the rope is given a sudden sideways jerk. At what location will the particle meet the resulting pulse?

Step-by-step solution

Let x be the distance of the pulse from the bottom of the rope at time t . The tension in the rope at a distance x from the bottom is T = g x , where is the mass per unit length of the rope. The speed of the transverse pulse is v = T = g x . Since v = dx dt , we have dx dt = g x . Rearranging and integrating from t = 0 to t : ₀^ x x^ -1/2 dx = ₀^ t g dt 2 x = g t x = g t^2 4 At the same time t , the distance fallen by the particle dropped from the ceiling is y = 1 2 g t^2 . Comparing x and y , we get y = 2x . The p