(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Help need with this problem.

A light spring AB of natural length 2a and of modulus of elasticity 2amn^{2}lies straight at its length and at rest on a smooth horizontal table. The end A is fixed to the table and a particle P of mass m is attached to the midpoint of the spring. The end B is then caused to move along the line AB so that after a time t the distance between A and B is a(2+sinnt). Denoting the distance of P from A by a+x, show that

d^{2}x/dt^{2}+ 4n^{2}x = 2an^{2}sinnt.

Find the value of t for which P first comes to rest.

3. The attempt at a solution

Here's my attempt so far, can't find the value of t for which P first comes to rest.

Am I right or wrong. Anyone has a better solution and a solution to the second part.

AB = 2(a+x) since AB at rest is twice a+x

At t, AB = 2a + asinnt

F = -k(x-e) where e is the extension and x the natural length

= k(e-x)

ma = 2amn^{2}/a (2a + asinnt -2a - 2x)

a = 2n^{2}(asinnt - 2a)

= 2n^{2}sinnt - 4n^{2}x

a + 4n^{2}x = 2n^{2}asinnt

=> d^{2}x/dt^{2}+ 4n^{2}x = 2n^{2}asinnt

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# Homework Help: Damped Oscillation problem.

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