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Chapter 14: Problem 48

A uniform cable hangs vertically under its own weight. Show that the speed of waves on the cable is given by \(v=\sqrt{y g},\) where \(y\) is the distance from the bottom of the cable.

Short Answer

Expert verified
The speed of waves on a uniform cable hanging vertically under its own weight at a distance y from the bottom is given by \( v = \sqrt{y g} \).

Step by step solution

01

Determine the Linear Density of the Cable

Let \( \lambda \) denote the linear density of the cable, which is the mass per unit length. The mass of a section of cable of length dy at a distance y from the bottom is given by \( dm = \lambda dy \), where \( \lambda \) is constant as the cable is uniform.
02

Determine the Tension in the Cable

The tension T in the segment of the cable at a distance y from the bottom is the weight of the cable below that point. So, it can be expressed as \( T = \lambda g (L-y) \), where g is the gravitational acceleration and L is the total length of the cable. The term (L-y) is the length of the cable below the point at distance y.
03

Find the Wave Speed Based on Tension and Linear Density

The speed of waves on a string is given by the square root of the ratio of tension to linear density, v = \sqrt{T/位}. Substitute the tension T from step 2 into this equation: v = \sqrt{\lambda g (L-y) / \lambda}. Simplifying this equation gives: v = \sqrt{g (L-y)}. As we need the speed of waves at a distance y from the bottom, replace (L-y) with y. Therefore, the speed of the waves on the cable at distance y from the bottom is given by \( v = \sqrt{y g} \).

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