M3 SHMhelp Watch

rbnphlp
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This is a screwed up question I cant get my head around.. If someone can help I will be grateful..

A elastic string of natural lebgth 2.5m and the modulous of elasticity 15N. A particle of mass is attached to the string a point K with the ratio 2:3.The ends of the string are attached to the points A and B which are 5m apart on a smooth horizontal surface.The particle is then pulled back aside and held at rest in contact at the point C where AC is 3m and ACB is a straight line.The particle is then released from rest.

a) Show that the particle is moving with SHM with a period of \frac{\pi\sqrt{2}}{5}

Im finding it quite difficult to atcually find the natural length of AP and BP.
Thanks again
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rbnphlp
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Unbounded
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When left unstretched, the natural length of the whole string is 2.5. We are told that the particle 'is attached to the string at a point K with ratio 2:3', so presumably that means AK:KB = 2:3 which implies that the natural length of AP is AK = 1m and the natural length of BP is KB = 1.5m.

The next step would 'normally' be:
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Find the equilibrium point of the system, i.e. when the tension in AP equals the tension in BP. The consider when the particle is a distance x from this equilibrium point - find the tensions of the two strings in terms of x and resolve.
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You should* find that  \ddot{x} = -w^2 x for some w, and hence the time period should be easy to find.
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The standard (quotable) result for SHM: time period = T = 2pi/w.
*I have not checked it myself, but I think it should work out.
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Pembo156
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I've read the question a couple of times, but I can't work out omega without the mass of the particle, unless I've missed something. This will then allow you to say T = 2(pi)/(omega) and find your answer.

If you read through your textbook there may be an example to help you derive the equation for SHM using Hooke's Law at equilibrium and then at a general position. I would expect an equation involving X (distance) and a (acceleration) which would allow you to compare this with
X (double dot) = -(omega^2) X
Unfortunately I can't help you much more with this question, but I agree it's difficult. Plus, I don't understand the part involving K with a ratio of 2:3. Is that your equilibrium position?
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rbnphlp
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(Original post by Unbounded)
When left unstretched, the natural length of the whole string is 2.5. We are told that the particle 'is attached to the string at a point K with ratio 2:3', so presumably that means AK:KB = 2:3 which implies that the natural length of AP is AK = 1m and the natural length of BP is KB = 1.5m.

The next step would 'normally' be:
Spoiler:
Show
Find the equilibrium point of the system, i.e. when the tension in AP equals the tension in BP. The consider when the particle is a distance x from this equilibrium point - find the tensions of the two strings in terms of x and resolve.
Spoiler:
Show
You should* find that  \ddot{x} = -w^2 x for some w, and hence the time period should be easy to find.
Spoiler:
Show
The standard (quotable) result for SHM: time period = T = 2pi/w.
*I have not checked it myself, but I think it should work out.
Thanks ..works out perfectly
(Original post by Pembo156)
I've read the question a couple of times, but I can't work out omega without the mass of the particle, unless I've missed something. This will then allow you to say T = 2(pi)/(omega) and find your answer.

If you read through your textbook there may be an example to help you derive the equation for SHM using Hooke's Law at equilibrium and then at a general position. I would expect an equation involving X (distance) and a (acceleration) which would allow you to compare this with
X (double dot) = -(omega^2) X
Unfortunately I can't help you much more with this question, but I agree it's difficult. Plus, I don't understand the part involving K with a ratio of 2:3. Is that your equilibrium position?
no thats not the equlibrium position ,its just where the particle is attached.
Funnily enough I could not use the 2:3 ratio, the rest of the bit is standard ..
Thanks for your input anyways
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