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Original post by MathsAstronomy12
What method do you use for questions asking you to prove SHM? I use conservation of energy but am not sure if it's accepted..?!


If you've written out expression for the energy of a system (as a function of position), I think you can find the force by differentiating.
So the system is in SHM if the expression for the potential energy is a quadratic (positive quadratic, I.e. Has a minimum)
I'm not sure of any of this though.
(edited 8 years ago)
Original post by MathsAstronomy12
What method do you use for questions asking you to prove SHM? I use conservation of energy but am not sure if it's accepted..?!
Any method that ends up with you saying

x¨=Cx\ddot{x} = -C x (where C is a constant) should be fine.

[Also x¨=Cx+o(x)\ddot{x} = -Cx + o(x) if you're asked to show SHM for small oscillations].

So if you've shown:

E=Ax˙2+Bx2E = A\dot{x}^2 + B x^2 (A, B +ve constants) and you use con. energy to deduce

2Ax˙x¨=2Bx˙x2A \dot{x} \ddot{x} = -2B \dot{x} x

and then divide by 2Ax˙2A \dot{x} that's fine (and I don't believe you need to worry about the divide by zero when [latex]\dot{x} = 0, either).
(edited 8 years ago)
Original post by physicsmaths


I normally do but what about questions where the particle is attached between two strings horizontally?

http://www.madasmaths.com/archive/maths_booklets/mechanics/m3_m5_s.h.m.pdf

^q1a) on SHM dynamics for instance.
Original post by MathsAstronomy12
I normally do but what about questions where the particle is attached between two strings horizontally?

http://www.madasmaths.com/archive/maths_booklets/mechanics/m3_m5_s.h.m.pdf

^q1a) on SHM dynamics for instance.


Same. Take one direction positive x. Then the other extension in terms of the extension(x in most cases) for the other side.


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Original post by MathsAstronomy12
I normally do but what about questions where the particle is attached between two strings horizontally?

http://www.madasmaths.com/archive/maths_booklets/mechanics/m3_m5_s.h.m.pdf

^q1a) on SHM dynamics for instance.


If you're looking at the same question as me, there is only one string.

If there were two strings, then still use F = ma. It would be T1 - T2 = ma
If you sort the lengths properly, the constant bits will cancel out, leaving
x double dot = - C x
Original post by physicsmaths
Same. Take one direction positive x. Then the other extension in terms of the extension(x in most cases) for the other side.


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Wrong question sorry! I meant 7a) same section :P where there are two strings, would really appreciate the help
Original post by tiny hobbit
If you're looking at the same question as me, there is only one string.

If there were two strings, then still use F = ma. It would be T1 - T2 = ma
If you sort the lengths properly, the constant bits will cancel out, leaving
x double dot = - C x


I meant q7a) on the same section, where there are two strings!
Original post by MathsAstronomy12
Wrong question sorry! I meant 7a) same section :P where there are two strings, would really appreciate the help


Read tiny hobbits reply. There is an example explaining this method(displaying it really) in edexcel M3.


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Original post by physicsmaths
Read tiny hobbits reply. There is an example explaining this method(displaying it really) in edexcel M3.


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I'm afraid that I don't have an M3 textbook. Can you or anyone else upload the worked example in the textbook to which you're referring?
Original post by MathsAstronomy12
I'm afraid that I don't have an M3 textbook. Can you or anyone else upload the worked example in the textbook to which you're referring?


Have a look at example iv on page 18 inthe attached notes.
Original post by MathsAstronomy12
I'm afraid that I don't have an M3 textbook. Can you or anyone else upload the worked example in the textbook to which you're referring?


I am afraid i dont have the book anymore either.


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Original post by physicsmaths


hey, you know in some trig integrals, can you use a substitution in the argument of the trig function like x--> 90-x to change cos to sin and sin to cos...?
If you do this do you have to change all the functions?.. but then again wouldn't this make it pointless?

Im just wondering as im useless at substitutions lol
Original post by demigawdz
hey, you know in some trig integrals, can you use a substitution in the argument of the trig function like x--> 90-x to change cos to sin and sin to cos...?
If you do this do you have to change all the functions?.. but then again wouldn't this make it pointless?

Im just wondering as im useless at substitutions lol



You do need to apply the sub to all the functions, but it's useful. I'm not sure how to explain it exactly, best you work through an example and see for yourself.
Original post by tiny hobbit
Have a look at example iv on page 18 inthe attached notes.


Thank you mate :smile:
Original post by demigawdz
hey, you know in some trig integrals, can you use a substitution in the argument of the trig function like x--> 90-x to change cos to sin and sin to cos...?
If you do this do you have to change all the functions?.. but then again wouldn't this make it pointless?In isolation, such a transform would indeed be pointless (in general at any rate).

But sometimes you can then combine the integrals to simplify things.

e.g.

I=0π/2xsinxcosxdx\displaystyle I = \int_0^{\pi/2} x \sin x \cos x\, dx

The sub you describe lets us deduce that

I=0π/2(π/2x)cosxsinxdx\displaystyle I = \int_0^{\pi / 2} (\pi/2 - x) \cos x \sin x \, dx

So far, no progress made. But if we add these:

2I=0π/2π2cosxsinxdx\displaystyle 2I = \int_0^{\pi /2} \frac{\pi}{2} \cos x \sin x \, dx and we've got rid of that x. From there we can finish quickly:

8I=π0π/2sin2x=π\displaystyle 8I = \pi \int_0^{\pi/2} \sin 2x = \pi.

(Yes, you could do this example using integration by parts, but this isn't always the case).
Original post by DFranklin
In isolation, such a transform would indeed be pointless (in general at any rate).

But sometimes you can then combine the integrals to simplify things.

e.g.

I=0π/2xsinxcosxdx\displaystyle I = \int_0^{\pi/2} x \sin x \cos x\, dx

The sub you describe lets us deduce that

I=0π/2(π/2x)cosxsinxdx\displaystyle I = \int_0^{\pi / 2} (\pi/2 - x) \cos x \sin x \, dx

So far, no progress made. But if we add these:

2I=0π/2π2cosxsinxdx\displaystyle 2I = \int_0^{\pi /2} \frac{\pi}{2} \cos x \sin x \, dx and we've got rid of that x. From there we can finish quickly:

8I=π0π/2sin2x=π\displaystyle 8I = \pi \int_0^{\pi/2} \sin 2x = \pi.

(Yes, you could do this example using integration by parts, but this isn't always the case).

Thank you so much!! :smile:
Does anybody have a copy of the spreadsheet to print out and keep track of which questions you have done that was available on the Prep Thread last year? The link doesn't work, or else it has been taken down.
Original post by Jordan\
Does anybody have a copy of the spreadsheet to print out and keep track of which questions you have done that was available on the Prep Thread last year? The link doesn't work, or else it has been taken down.


This?
Original post by rcmehta
This?

That's it! Thankyou so much 😆
https://3110b3b48cdc993f8fb622042e2ffa4f00a024fe.googledrive.com/host/0B1ZiqBksUHNYMTZlRm1iQnJPTWs/Papers/2006%20STEP%202.pdf

q11

does northerly direction mean due north (directly north)? or only a bit north (like at an angle theta)? I dont see why they cannot be clear lol.

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