# M3: SHM - prove harmonic motion

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#1
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#2
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2 years ago
#3
(Original post by Maths&physics)
You need to consider when the particle is a distance x from the equilibrium position. Use F = ma and substitute Hooke's law into it for each string. This should simplify to (x double dot) = -kx where k should turn out to be positive. You have then proved that SHM is taking place and should say so.

In this question the strings are identical and so the equilibrium position is at the centre of AB.
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#4
(Original post by tiny hobbit)
You need to consider when the particle is a distance x from the equilibrium position. Use F = ma and substitute Hooke's law into it for each string. This should simplify to (x double dot) = -kx where k should turn out to be positive. You have then proved that SHM is taking place and should say so.

In this question the strings are identical and so the equilibrium position is at the centre of AB.
thanks

how can T - 4g = ma

and also T - 4g = T?
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2 years ago
#5
(Original post by Maths&physics)
thanks

how can T - 4g = ma

and also T - 4g = T?
The first is true.
For the second one, I don't think that you have spotted the bit that they are in the middle of writing, where it says -4 x 9. underneath.
They are in the process of subtracting 4g from each side.

I don't think that it is set out very well.

I would complete the T - 4g = -m(x double dot) and then substitute the expression from Hooke's law later. The reason for the minus is that x is increasing downwards and so x double dot is downwards and they are using F = ma upwards.
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#6
(Original post by tiny hobbit)
The first is true.
For the second one, I don't think that you have spotted the bit that they are in the middle of writing, where it says -4 x 9. underneath.
They are in the process of subtracting 4g from each side.

I don't think that it is set out very well.

I would complete the T - 4g = -m(x double dot) and then substitute the expression from Hooke's law later. The reason for the minus is that x is increasing downwards and so x double dot is downwards and they are using F = ma upwards.
yes, I now see what they did!

its quite confusing! 0
#7
(Original post by tiny hobbit)
The first is true.
For the second one, I don't think that you have spotted the bit that they are in the middle of writing, where it says -4 x 9. underneath.
They are in the process of subtracting 4g from each side.

I don't think that it is set out very well.

I would complete the T - 4g = -m(x double dot) and then substitute the expression from Hooke's law later. The reason for the minus is that x is increasing downwards and so x double dot is downwards and they are using F = ma upwards.
so, to prove SHM in this case: force/tension/acceleration has to be opposite to displacement?

leading to x(double dot) = - (omega^2)x ???

why is this the case? why should force/tension/acceleration have to be opposite to displacement?

can you please explain it because I dont get it? thanks
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2 years ago
#8
(Original post by Maths&amp;physics)
so, to prove SHM in this case: force/tension/acceleration has to be opposite to displacement?

leading to x(double dot) = - (omega^2)x ???

why is this the case? why should force/tension/acceleration has to be opposite to displacement?

can you please explain it because I dont get it? thanks
Are u familiar with the equation x=Acos(wt).

If u differentiate twice wrt to t u will get an equation for acceleration.

You should notice that this equation is similar to the initial equation for x. So u can sub in x where appropriate.

The resulting equation should look like the equation u have described. It is the characteristic equation of SHM, as it shows that the acceleration is in the opposite direction to the displacement, a condition for SHM.
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#9
(Original post by Shaanv)
Are u familiar with the equation x=Acos(wt).

If u differentiate twice wrt to t u will get an equation for acceleration.

You should notice that this equation is similar to the initial equation for x. So u can sub in x where appropriate.

The resulting equation should look like the equation u have described. It is the characteristic equation of SHM, as it shows that the acceleration is in the opposite direction to the displacement, a condition for SHM.
excellent explanation!
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#10
(Original post by Shaanv)
Are u familiar with the equation x=Acos(wt).

If u differentiate twice wrt to t u will get an equation for acceleration.

You should notice that this equation is similar to the initial equation for x. So u can sub in x where appropriate.

The resulting equation should look like the equation u have described. It is the characteristic equation of SHM, as it shows that the acceleration is in the opposite direction to the displacement, a condition for SHM.
whats the difference between the first period (part b), where he uses T = 2pi/w

and the second one (part b, ii), where he is finding the period but multiples it by the period found earlier?

thanks
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2 years ago
#11
(Original post by Maths&physics)
whats the difference between the first period (part b), where he uses T = 2pi/w

and the second one (part b, ii), where he is finding the period but multiples it by the period found earlier?

thanks
At the top of the motion in the second version, the string goes slack. What you shown is the proportion of the SHM oscillation that takes place. Now you need to calculate the time for the top part of the motion, when the particle just moves under gravity, using suvat with a=9.8
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#12
(Original post by tiny hobbit)
At the top of the motion in the second version, the string goes slack. What you shown is the proportion of the SHM oscillation that takes place. Now you need to calculate the time for the top part of the motion, when the particle just moves under gravity, using suvat with a=9.8
in the second screenshot, hes using a different equation to work out the period. what equation is that, whats the difference and whats the algebraic form? thanks
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2 years ago
#13
(Original post by Maths&physics)
in the second screenshot, hes using a different equation to work out the period. what equation is that, whats the difference and whats the algebraic form? thanks
He's finding what proportion of the whole period occurs when the string is stretched. The circle business is an alternative to using either x = a cos wt or x = a sin wt.

Do you get a running commentary with these videos that explain what is going on?
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#14
(Original post by tiny hobbit)
He's finding what proportion of the whole period occurs when the string is stretched. The circle business is an alternative to using either x = a cos wt or x = a sin wt.
whats the algebraic formula for that method?

Do you get a running commentary with these videos that explain what is going on?
yes, but he doesn't explain everything and he didn't explain that.
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2 years ago
#15
(Original post by Maths&physics)
whats the algebraic formula for that method?

yes, but he doesn't explain everything and he didn't explain that.
I suppose that it is something like (angle at centre of circle for the part that's required)/360 x period

It's not one I use, I go with the x=asin/cos wt
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#16
(Original post by tiny hobbit)
I suppose that it is something like (angle at centre of circle for the part that's required)/360 x period

It's not one I use, I go with the x=asin/cos wt
could you show me how you'd apply those in both instances, please? thanks
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