# Harmonic Motion of a Mass between two springs

See attached.

## Relevant Equations:

x(t)=Acos(wt+phi), v(t)=-wAsin(wt+phi)
So first I found the total energy of the system by calculating the potential Energy, Ep=0.5k(l^2+l^2) and get 2.0475 (this part is right).

Then I find w using the period T=2pi/w, so w=2pi/1.21=5.1927

I also found the amplitude using E=1/2kA^2, so A=sqrt(2E/k)=0.212132

Now this is the part I think Im messing up. I go

0.15=Acos(phi) to find phi

Then x(t)=Acos(wt+phi) to find location. But this gives me the wrong answer of -0.14 when the answer should be 0.010.

Another thing that doesn't make sense to me is that I solved the amplitude using E=1/2kA^2, but when I try and solve the amplitude using 0.15=Acos(phi) and 0=-wAsin(phi) I get an amplitude of 0.15, which is different.

#### Attachments

• Delta2

Delta2
Homework Helper
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Homework Statement:: See attached.
Relevant Equations:: x(t)=Acos(wt+phi), v(t)=-wAsin(wt+phi)

I also found the amplitude using E=1/2kA^2, so A=sqrt(2E/k)=0.212132
This is wrong, if you do write down the equation of motion for mass M you ll find out that it does SHM with constant ##k'=2k## (the angular frequency is ##w=\sqrt{\frac{M}{k'}}##). Therefore the correct equation to use is $$E=\frac{1}{2}k'A^2=kA^2$$ and you ll end up with A=0.15, same as when you using the other method which is correct.

• JoeyBob
BvU
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2.0475 (this part is right).
2.05 Joule, you mean !

I also found the amplitude using E=1/2kA^2
No. Two springs have energy !

0.15=Acos(phi) to find phi
At##t=0## that should give ##\phi = 0## !

(your notation is pretty awkward and easily confusing (##A\cos## versus ##\operatorname{acos}## for instance !)

Ah, ##\Delta## was quicker ...

• Delta2
In this case (and generally speaking for simple spring-mass systems starting at rest), there's no need to calculate the energy to find the amplitude. The initial displacement provides all of the energy in the system. Since there is no dissipative forces, you can write down immediately that the amplitude of oscillation should be the initial displacement. The system oscillates between entirely kinetic energy and entirely potential energy (at the point of maximum=initial displacement).

• Delta2, berkeman and BvU
This is wrong, if you do write down the equation of motion for mass M you ll find out that it does SHM with constant ##k'=2k## (the angular frequency is ##w=\sqrt{\frac{M}{k'}}##). Therefore the correct equation to use is $$E=\frac{1}{2}k'A^2=kA^2$$ and you ll end up with A=0.15, same as when you using the other method which is correct.
Okay I think I understand. Trying to derive the equation from the start I now get E=kA^2=m(wA)^2.

But testing this Im not so sure. I know E and k so I find that A=0.15. Now I can find m. But although these equations give me the right amplitude, the mass they give is wrong. The equation needs to be 2m(wA)^2, but doesn't

w^2=k/m and not w^2=k/2m?

Delta2
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Gold Member
You have two mistakes. First mistake is that you omit the coefficient 1/2 from the maximum kinetic energy. The second mistake is that it is ##w^2=\frac{k'}{m}=\frac{2k}{m}##. As I said before if you write the equation of motion for the mass m you ll be able to see that the constant of oscillation is ##k'=2k##, not ##k##. We have two springs there.

• JoeyBob
You have two mistakes. First mistake is that you omit the coefficient 1/2 from the maximum kinetic energy. The second mistake is that it is ##w^2=\frac{k'}{m}=\frac{2k}{m}##. As I said before if you write the equation of motion for the mass m you ll be able to see that the constant of oscillation is ##k'=2k##, not ##k##. We have two springs there.
Okay so I know E=kA^2 from deriving with 2 springs and like you said, theres two springs so ##w^2=\frac{k'}{m}=\frac{2k}{m}##.

So now E is also m/2(Aw)^2, which gives the right mass.

Now A=0.15, which lets me find out that phi is 0. Now solving x(t)=0.15cos(wt) gives 0.009733=0.01, the right answer.

Thanks for the help, I understand harmonic in two spring systems better now. When calculating forces/energies you're really just multiplying the equations you use for 1 spring systems by 2, and then w and E end up being different but correct.

• Delta2
Delta2
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When calculating forces/energies you're really just multiplying the equations you use for 1 spring systems by 2
Well yes but it is because in this problem both springs have the same constant k. If we had two springs with constants ##k_1,k_2## then to find the maximum potential energy you ll had to find ##k'=k_1+k_2## and the force would be ##F=-k'x=-(k_1+k_2)x## and similar for the maximum potential energy ##U=\frac{1}{2}k'A^2##.

• JoeyBob