The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

This discussion is now closed.

Check out other Related discussions

Differential equations and simple harmonic motion

I'm struggling with the question below. I presume ω\omega has some form of differential part attached to it but I don't have a clue what it is.
I also don't know how to solve the d.e. which I presume comes from my lack of knowledge with the first bit. Any help is greatly appreciated.

Question
The equation for Simple Harmonic Motion with constant frequency ω\omega , is

d2xdt2=−ω2x\dfrac{d^2x}{dt^2} = -\omega ^2 x

Show that

d2xdt2=vdvdx\dfrac{d^2x}{dt^2}= v \dfrac{dv}{dx}

where u=dxdt u = \frac{dx}{dt} denotes velocity. Find and solve a differential equation in vv and xx given that x=a x=a when v=0 v=0

Hence show that

x(t)=asin⁡(ωt+ϵ)x(t) = a \sin(\omega t + \epsilon )

for some constant ϵ\epsilon
Reply 1
Avatar for DFranklin
DFranklin
18
w is a simple constant. I don't know what you mean by "I presume \omega has some form of differential part attached to it".

Were you able to do the "show that"? (I'll assume yes).

If d2xdt2=−ω2x\frac{d^2x}{dt^2} = -\omega^2 x and d2xdt2=vdvdx\frac{d^2x}{dt^2} = v\frac{dv}{dx}, it should be fairly obvious how to remove d2xdt2\frac{d^2x}{dt^2} from the 2nd equation using the first.
Reply 2
Avatar for The Muon
The Muon
OP
2
Thanks. I thought that as w was frequency perhaps there was some relation to time perhaps giving a dt somewhere were I didn't want one. Now I know its just a constant I should be able to give it a better stab.
The Muon
Thanks. I thought that as w was frequency perhaps there was some relation to time perhaps giving a dt somewhere were I didn't want one. Now I know its just a constant I should be able to give it a better stab.

The question does specify omega is constant. :p:
The Muon
Thanks. I thought that as w was frequency perhaps there was some relation to time perhaps giving a dt somewhere were I didn't want one. Now I know its just a constant I should be able to give it a better stab.


w is the angular displacement, and is equivalent to 2πf2\pi f, where f is the equivalent of frequency and the reciprocal of the period t.

EDIT: Yes, this information is not necessary to this particular question. I thought I'd tell it you for your own information. :smile:
Reply 5
Avatar for spex
spex
16
The cool way to solve this equation is to put x = Ae^kt and get complex values, which can also be written as a trig function in the form they want. A profound moment in my education, realising that.
Mathematician!
w is the angular displacement, and is equivalent to 2πf2\pi f, where f is the equivalent of frequency and the reciprocal of the period t.

EDIT: Yes, this information is not necessary to this particular question. I thought I'd tell it you for your own information. :smile:



w isn't the angular displacement.

The op has posted the differential equation for linear SHM; in which case w could be defined as the constant angular velocity of a particle moving on an associated circle.
Mrm.
w isn't the angular displacement.

The op has posted the differential equation for linear SHM; in which case w could be defined as the constant angular velocity of a particle moving on an associated circle.


Um, OK...
...My excuse? Haven't done Physics for the whole summer. :p:
The rest is right though, right?
Reply 8
Avatar for DFranklin
DFranklin
18
spex
The cool way to solve this equation is to put x = Ae^kt and get complex values, which can also be written as a trig function in the form they want. A profound moment in my education, realising that.
It's not actually an answer to the question here though. There's a difference between verifying x=Ae^kt is a solution (for certain values of k) and actually deriving the form that x has to take.
Mathematician!
Um, OK...
...My excuse? Haven't done Physics for the whole summer. :p:
The rest is right though, right?


not quite.
If you are referring to angular shm then theta will be the angular displacement at time t, and w will still be a constant.
Reply 10
Avatar for The Muon
The Muon
OP
2
Thanks everyone :yy:

Related discussions

Latest

Trending

Trending

Articles for you