# Simple Harmonic Motion (Total Energy)

1. Feb 4, 2014

### KiNGGeexD

I have a question about the derivation that I have attached! I understand that both KE and U are 1/2 kA^2

So how is it that the two combine is also equal to 1/2kA^2

Not sure if I'm missing something but I'm a little confused :(

2. Feb 4, 2014

### HallsofIvy

Staff Emeritus
Part of the problem is that what you "understand" is not true. It is NOT the case that "both KE and U are $1/2 kA^2$". What is true is that $E= 1/2 kA^2 sin^2(\omega t+ \phi)$ and $U= 1/2 kA^2 cos^2(\omega t+ \phi)$.

Adding those gives $KE+ U= 1/2 kA^2(sin^2(\theta)+ cos^2(\theta))$. Now use the fact that $sin^2(\theta)+ cos^2(\theta)= 1$ for all $\theta$!

3. Feb 4, 2014

### KiNGGeexD

So that leaves

1/2mv^2 + 1/2kA^2?

4. Feb 4, 2014

### Tanya Sharma

KE = (1/2)mv2 and v= ωAcos(ωt+ø)
So ,KE = (1/2)mω2A2cos2(ωt+ø)

But ,k = mω2

Hence KE = (1/2)kA2cos2(ωt+ø) (1)

PE = (1/2)kx2 and x= Asin(ωt+ø)
So ,PE = (1/2)kA2sin2(ωt+ø) (2)

E=KE+PE

Using the fact sin2θ+cos2θ =1 ,

E=(1/2)kA2

5. Feb 4, 2014

### KiNGGeexD

But there are two terms of kA^2?

I feel like I'm missing something obvious?

6. Feb 4, 2014

### Tanya Sharma

E = (1/2)kA2cos2(ωt+ø) + (1/2)kA2sin2(ωt+ø)

Take (1/2)kA2 common from both the terms ,

E = (1/2)kA2[cos2(ωt+ø) + sin2(ωt+ø)]

Hence E = (1/2)kA2

7. Feb 4, 2014

### KiNGGeexD

Ahh so you take 1/2kA^2 out as a common factor

8. Feb 4, 2014

### Tanya Sharma

Yes...just like if we have 2a+2b ,we can write it as 2(a+b) .

9. Feb 4, 2014

### KiNGGeexD

I knew it would be something trivial! That's a bunch mate!