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A flexible steel cable of total length L and mass per unit length μ hangs vertically from a support at one end.(a) Show that the speed of a transverse wave down the cable is v = √(g(L - x)), where x is measured from the support. (b) How long will it take for a wave to travel down the cable ? |
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Answer» (a) Tension at a point at a distance x from the support is the due to the weight of the cable T = \(\frac{m(L-x)}{L}g\) \(\mu\) = \(\frac mL\) The speed of transverse wave v = \(\sqrt{\frac{T}{\mu}}\) v = \(\cfrac{\frac{m(L-x)g}{L}}{\frac mL}\) v = \(\sqrt{g(L-x)}\) (b) Time taken to transverse a distance dt = \(\frac{dx}{\sqrt{g(L-x)}}\) total time taken t = \(\int\limits_0^L\frac{dx}{\sqrt{g(L-x)}}\) t = 2\(\sqrt{\frac Lg}\) |
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