# SHM mass spring solutions

Hi community,
I've been looking at solutions for mass spring shm (undamped for now) ie that

x = Acoswt and x = Bcoswt work as solutions for dx2/dt2 = -(k/m)x

and that the general solution is the sum of these that with a trig identity can be written as

x = C Cos(wt - φ) where C is essentially the amplitude (and is given by √(A2 + B2)

My question is the physical significance of A and B in the two separate solutions above (before this youtube video) I've always gone for the solutions as either the Acoswt or Asinwt (with A being the amplitude) depending on where the mass is in its oscillating cycle at time t=0, i.e. would have gone with the coswt one if

x = +A at t=0.

If I let A and B both be A then my factor C (amplitude) comes out as √(2)A where I want it to represent the Amplitude A.

Would really appreciate help.
regards,
Glenn.

Last edited by a moderator:

berkeman
Mentor
Doesn't look like the video link is working...?

berkeman
Mentor
x = Acoswt and x = Bcoswt work as solutions
and is the 2nd term supposed to be sin()?

and is the 2nd term supposed to be sin()?
Sorry yes I did mean Bsinwt for the second solution...

In fact I think it is fine, because if you state that x = 0 after quarter of a cycle (so ∏/2 radians) then you get 0 = Bx1 therefore B = 0 (in x = Acoswt + Bsinwt) so the sine term disappears from the general solution anyway and you end up with C = √A^2 = A for the coefficient which is what I wanted.

thanks,
G.

• berkeman
berkeman
Mentor
Yeah, it just means that the spring doesn't necessarily have zero phase angle at t=0. It's a general way to express a sinusoid that has amplitude and phase information... Mister T
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