# Simple Harmonic motion

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1. Apr 10, 2017

### Faiq

1. The problem statement, all variables and given/known data

3. The attempt at a solution
I have completed the whole question, however, stuck on the last part.
How to find the value about which $\rm \small \theta$ now oscillates?
A source stated that $\rm \small \alpha$ is the value about which $\rm \small \theta$ is oscillating. Can anybody tell me if that is correct? If yes, then why?

2. Apr 10, 2017

### TSny

What value of $\phi$ does the pendulum oscillate about?

3. Apr 10, 2017

### Faiq

I am not sure about the answer to your question. All I know is that the value of $\theta$ is technically the "amplitude" of the motion.

4. Apr 10, 2017

### TSny

What value of $\phi$ corresponds to the equilibrium position of the pendulum. See equation [2].

5. Apr 10, 2017

### Faiq

$\theta$?
If yes? Can you tell why?

6. Apr 10, 2017

### Faiq

Maybe I got this.
Before the acceleration, the $7/24 cos\theta$ term wasn't present. Thus if I consider, the before acceleration state as equilibrium then $\theta$ is the value that corresponds to equilibrium. But after applying the acceleration, the $7/24 cos\theta$ term emerged which gave rise to the $-\alpha$ value. As a result, the $-\alpha$ value must be the maximum amplitude or the value about which $\phi$ oscillates.
Rearranging the equation, we get that $\theta$ oscillates about the value of $\alpha$.
Correct?

7. Apr 10, 2017

### Staff: Mentor

Hint: Usually when you want to arrive at a result of the form trig(a - b) where "trig" is one of the trig functions, you are looking to employ one of the trig angle sum or difference identities in some fashion...

8. Apr 10, 2017

### Faiq

Apologies but I can't seem to relate your hint with my current confusion.

9. Apr 10, 2017

### Staff: Mentor

$g~\left( sin(θ) - 7~cos(θ) \right)$

You need to get it into a form where the sin(a - b) identity can be applied. That will mean massaging the expression until the coefficients of the existing sin(θ) and cos(θ) are the cosine and sine of some new angle, $\alpha$.

10. Apr 10, 2017

### Faiq

Yes, I did all of that and also obtained the correct expression $\rm \small -g\sin(\theta - 16.3)$
I am just stuck at "the value at which theta oscillates"? I mean why the value of alpha is the value at which theta oscillates?
Is this explanation correct?
Before the acceleration, the $7/24 cos\theta$ term wasn't present. Thus if I consider, the before acceleration state as equilibrium then $\theta$ is the value that corresponds to equilibrium. But after applying the acceleration, the $7/24 cos\theta$ term emerged which gave rise to the $-\alpha$ value. As a result, the $-\alpha$ value must be the maximum amplitude or the value about which $\phi$ oscillates.
Rearranging the equation, we get that $\theta$ oscillates about the value of $\alpha$.

11. Apr 10, 2017

### TSny

Try to be precise in how you state things. $\theta$ is a variable, so it cannot correspond to equilibrium. There is a specific value of $\theta$ that corresponds to equilibrium.

First, find the value of $\phi$ that corresponds to equilibrium. If the pendulum hangs at rest in the equilibrium position, what is the value of $\frac{d^2 \phi}{dt^2}$?

Then, what does equation [2] imply for the value of $\phi$ at equilibrium?

12. Apr 10, 2017

### Faiq

Well if the pendulum is in equilibrium and hangs in free fall, then $\phi=0$ and then $\small \frac{d^2\phi}{dt^2}$ should be equal to 0.
I get the meaning of your second question but unfortunately, I don't know the answer. Maybe $-\alpha$ because $\theta$ keeps changing.

Last edited: Apr 10, 2017
13. Apr 10, 2017

### haruspex

Did you mean φ=0?

14. Apr 10, 2017

### Faiq

Oh yes.

15. Apr 10, 2017

### haruspex

Your equation is $\ddot \phi=k\sin(\phi)$. For what value of φ will $\ddot \phi$ be zero?

16. Apr 10, 2017

### Faiq

$\phi = 2k\pi$ but since in this question $\phi$ is a small angle so $\phi = 0$

17. Apr 10, 2017

### haruspex

Right. (I see you edited your post #14 after my post #15.)
So what does that give for θ at equilibrium?

18. Apr 10, 2017

### Faiq

Yes it gives $\theta = \alpha$

Last edited: Apr 10, 2017
19. Apr 10, 2017

### haruspex

Ok. All done?

20. Apr 11, 2017

### Faiq

Is this explanation correct?