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Simple Harmonic Motion questions
By Carl Evans (P2080) on Monday, July 3, 2000 - 01:46 am:
A particle of mass 0.5kg is hanging in equilibrium attached to the free end of an elastic spring of natural length 0.8m and modulus 60N which is hanging vertically. The particle is pulled downwards and released. Calculate the period of the resulting oscillations.

I have tried this for some time now, and I keep on thinking I'm lacking some information i.e. the amplitude (or subsequent extension). I know the extension of the particle under the influence of gravity whilst in equilibrium is 0.0653m, and I know I have to find w to work out the period. So I get to the following stage:

0.5g - (60×(x + 0.0653))/0.8 = 0.5a (where a is acceleration)

Then I'm stuck, because I can't compare the above equation (once its rearranged) to the standard SHM equation in order to find w.

Can anybody help?

Thanks
By David Loeffler (P865) on Monday, July 3, 2000 - 02:53 pm:

If you expand this out, it becomes
0.5a = 0.5g - 60×(0.0653)/0.8 - 60×x/0.8.
However, 60×(0.0653)/0.8 = 0.5g, so these two constant terms cancel and we have
0.5a = - 60×x/0.8
which is clearly a version of the standard SHM equation. Dividing through we have a=-150x, so w=sqrt(150).

By Carl Evans (P2080) on Monday, July 3, 2000 - 11:25 pm:

David, thanks for your help. I actually solved this myself after realising that the constants cancelled out if I took the initial extension to a greater number of decimal places. Stupid mistake!

However, I've been struggling with another SHM question today, and would really appreciate your help (or anybody else's) on the following if at all possible.

A small ball of mass 1.2kg is attached to an elastic string of unstretched length 0.75m and modulus 60N. The other end of the string is fixed to a point of the smooth horizontal floor on which the ball rests. The ball is pulled aside until the string measures 0.9m and released.

When the ball has travelled a distance 0.6m from its point of release, the floor becomes rough. Given that the coefficient of friction between the ball and the floor is 0.3, determine the time for which the ball is moving.

Now, I can get the times for the smooth harmonic motion bit (on smooth ground), together with the time for when the string is slack, but don't know how to calculate the time for when the ball is on rough ground, as I don't know how to incorparate the added frictional force.

Thanks very much for anybody's help.
By Andrew Smith (P2517) on Tuesday, July 4, 2000 - 07:57 pm:

Carl,
you can use the fact that F=coefficient of friction×R, i.e. F=0.3×1.2×g, to find the force acting on the particle on the rough ground. As this force slows the particle down -F=m.dv/dt so as m and F are constant you can solve this to get [-0.3×g×t]=[v] where the limits are 0 to t and u to 0. The particle remains at the speed it had when the string went slack until it reaches the rough surface as in the intervening time no force acts on it. You can find the speed when the string goes slack which is at the centre of the motion (v=wa), so u above is equal to wa where a is the amplitude. From this you can find the time until the particle comes to rest so combined with your earlier answers this should give you the total answer.