Simple Harmonic motion

Faiq

1. Homework Statement
The question is uploaded. 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?

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TSny

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

Faiq

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

TSny

Homework Helper
Gold Member
What value of $\phi$ corresponds to the equilibrium position of the pendulum. See equation .

Faiq

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

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?

gneill

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...

Faiq

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

gneill

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, $\alpha$.

Faiq

$g~\left( sin(θ) - 7~cos(θ) \right)$

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, $\alpha$.
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 $\alpha$.

TSny

Homework Helper
Gold Member
Thus if I consider, the before acceleration state as equilibrium then $\theta$ is the value that corresponds to equilibrium.
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  imply for the value of $\phi$ at equilibrium?

Faiq

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  imply for the value of $\phi$ at equilibrium?
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:

haruspex

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Well if the pendulum is in equilibrium and hangs in free fall, then θ=0
Did you mean φ=0?

haruspex

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Well if the pendulum is in equilibrium and hangs in free fall, then $\theta=0$ and then $\small \frac{d^2\phi}{dt^2}$ should be equal to 0.
Your equation is $\ddot \phi=k\sin(\phi)$. For what value of φ will $\ddot \phi$ be zero?

Faiq

Your equation is $\ddot \phi=k\sin(\phi)$. For what value of φ will $\ddot \phi$ be zero?
$\phi = 2k\pi$ but since in this question $\phi$ is a small angle so $\phi = 0$

haruspex

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$\phi = 2k\pi$ but since in this question $\phi$ is a small angle so $\phi = 0$
Right. (I see you edited your post #14 after my post #15.)
So what does that give for θ at equilibrium?

Faiq

Right. (I see you edited your post #14 after my post #15.)
So what does that give for θ at equilibrium?
Yes it gives $\theta = \alpha$

Last edited:

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Faiq

Ok. All done?
Is this explanation correct? TSny

Homework Helper
Gold Member Your top right diagram indicates that $\theta$ is measured from the new equilibrium position. But, $\theta$ is defined to be the angle of the pendulum measured from the old equilibrium position (i.e., from the vertical position). $\phi$ is the angle of the pendulum as measured from the new equilibrium position. What angle is indicated by the blue angle?

"Simple Harmonic motion"

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