One end of a nylon rope is tied to a stationary support at the top of a vertical

Question

One end of a nylon rope is tied to a stationary support at the top of a vertical mine shaft of depth 84.0m . The rope is stretched taut by a box of mineral samples with mass 16.0kg attached at the lower end. The mass of the rope is 2.50kg . The geologist at the bottom of the mine signals to his colleague at the top by jerking the rope sideways. (Do not neglect the weight of the rope.)

1. What is the wave speed at the bottom of the rope?

2. What is the wave speed at the middle of the rope?

3. What is the wave speed at the top of the rope?

Explanation / Answer

linear density u = mass of the rope / lengthn of the rope = 2.5 kg/ 84 m = 0.02976 kg /m

a) at bottom Tension T1 = M g = 16*9.8 N = 156.8 N

speed = sqrt (T1/u) = sqrt (16*9.8/0.02976 ) = 72.5866 m/s

b)at middle T2 =( M+u *(L/2)*)g = [16 + 0.02976 *(84/2)]*9.8 = 169.049 N

speed = sqrt (T2 / u) = 75.36m/s

c) at top T3 = (M+m)g = 181.3 N

speed = sqrt(T3/u ) = 78.05 m/s

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