M4 oscillatory motion question!!! Watch

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#1
Report Thread starter 13 years ago
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hi...having big probs on this question - wonder whether anyone else gets it.

its M4- edexcel Pg 99 question 10 for those with books...otherwise:

a particle of mass 1kg which is free to slide horizontally inside a smooth cylindrical tuve. The particle is 0.5m and modulus of elasticity 2N. The system is initially at rest. The other end Q of the spring is then forced to oscillate with SHM so that at time t seconds its displacement from its initial position is 0.25sin 3t. The displacement of P from its initial position at time t sexonds is x metres, measured in the same direction as the displacement of Q;

a) show that d^2 x/dt^2 +4x = sin3t.

i've been stuck on it for ages and is the only one in the exercise i can't do. HELP!!!!

i'm basically getting stuck on the step of working out the length of the string in terms of acceleration or constant velocity, but dont know what particle P moves with, so i cant do any further calculations...would much appreciate your help

PK
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m:)ckel
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You mean Qu. 11, right? I've attached a diagram (sorry for the general crappiness):

m(d^2x/dt^2) = -T

d^2x/dt^2 = (-2/0.5)(x - 0.25sin3t)

d^2x/dt^2 = -4x + sin3t

(d^2x/dt^2) + 4x = sin3t


I put a 'v' at Q, since that's the part which is being made to oscillate. But the displacement is in fact measured from P- hence I use x (and similar) when referring to P.
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(Original post by mockel)
You mean Qu. 11, right? I've attached a diagram (sorry for the general crappiness):

m(d^2x/dt^2) = -T

d^2x/dt^2 = (-2/0.5)(x - 0.25sin3t)

d^2x/dt^2 = -4x + sin3t

(d^2x/dt^2) + 4x = sin3t


I put a 'v' at Q, since that's the part which is being made to oscillate. But the displacement is in fact measured from P- hence I use x (and similar) when referring to P.
i'm still pretty confused about directions of the tension. Say you consider P, then initiallly when the push on Q results in a correspoondinding push on P, isn't the force directed towards the particle P...i.e. in the direction of the acceleration and increase in displacement?

cheers

Pk, also, why can you assume that particle travels with the same speed v at t=0...did you assume there is no lag time/delay in the moving of P, after Q moves intially?
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Fermat
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Does this help ?

Edit: The assumptions about the displacements of P and Q are just so that you can decide whether the spring is in tension or compression, and therefore in which direction the force, T, is acting.
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(Original post by Fermat)
Does this help ?

Edit: The assumptions about the displacements of P and Q are just so that you can decide whether the spring is in tension or compression, and therefore in which direction the force, T, is acting.
still..from the quetion, as Q moves first, i've got this picture in my head of an initial compression, and i can't get rid of it!...it your method just as valid if you assume that the spring is initially compressed just after t=0, and if so, how would you do this question for this situation?

cheers

Pk
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Fermat
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(Original post by Phil23)
still..from the quetion, as Q moves first, i've got this picture in my head of an initial compression, and i can't get rid of it!...it your method just as valid if you assume that the spring is initially compressed just after t=0, and if so, how would you do this question for this situation?

cheers

Pk
Taking my pm into account, when the spring is initially compressed just after t=0, then s is +ve and x is +ve.

Either x > s or s > x (ignore x=s - it's not very fruitful)

If x > s, then spring in tension, T is -ve
If s > x, then spring in compression, T is +ve

...

develop eqn of motion
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(Original post by Fermat)
Taking my pm into account, when the spring is initially compressed just after t=0, then s is +ve and x is +ve.

Either x > s or s > x (ignore x=s - it's not very fruitful)

If x > s, then spring in tension, T is -ve
If s > x, then spring in compression, T is +ve

...

develop eqn of motion
Ah!!!!! so you do get the same thing...its all in the directions!

ignore my latter pm..i think i've sussed it now

Phil
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(Original post by Phil23)
Ah!!!!! so you do get the same thing...its all in the directions!

ignore my latter pm..i think i've sussed it now

Phil
I saw this post just after answering back to your pm
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