A point particle of mass m is fixed to the bottom end of a thin wire suspended from a fixed point on the ceiling. The thin wire has total mass M and length L. The acceleration due to gravity is g. At time t = 0, the point m is given a very small tap.

(a) Find the tension in the wire and the speed of waves in the wire as a function of y, the distance from m.

(b) Find the total time needed for the perturbation to reach the top end of the wire (the ceiling).

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aristocles aristocles · Answer 1 · 2019-09-09T03:23:19+02:00

Answer:

[tex]T = (m + \frac{M}{L}y)g[/tex]

[tex]v = \sqrt{(\frac{mL}{M} + y)g}[/tex]

Part b)

[tex]t = 2(\frac{\sqrt{(\frac{mL}{M} + L)g}}{g} - \frac{\sqrt{(\frac{mL}{M})g}}{g})[/tex]

Explanation:

Part a)

tension in the wire at any distance "y" from the bottom end of the wire is due to the weight of the suspended part of the wire given by the equation

[tex]T = (m + \frac{M}{L}y)g[/tex]

So here we will have speed of the wave is given as

[tex]v = \sqrt{\frac{T}{M/L}}[/tex]

now we have

[tex]v = \sqrt{\frac{(m + \frac{M}{L}y)g}{M/L}}[/tex]

[tex]v = \sqrt{(\frac{mL}{M} + y)g}[/tex]

Part b)

now the time taken by the wave to reach the top is given as

[tex]t = \int \frac{dy}{v} [/tex]

[tex]t = \int_0^L \frac{dy}{\sqrt{(\frac{mL}{M} + y)g}}[/tex]

[tex]t = 2(\frac{\sqrt{(\frac{mL}{M} + L)g}}{g} - \frac{\sqrt{(\frac{mL}{M})g}}{g})[/tex]