• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Simple Harmonic motion

  • Thread starter Faiq
  • Start date
  • Tags
    shm
349
15
1. Homework Statement
The question is uploaded.
SHM.jpg

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?
 

TSny

Homework Helper
Gold Member
12,039
2,619
What value of ##\phi## does the pendulum oscillate about?
 
349
15
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
12,039
2,619
What value of ##\phi## corresponds to the equilibrium position of the pendulum. See equation [2].
 
349
15
##\theta##?
If yes? Can you tell why?
 
349
15
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
20,489
2,615
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...
 
349
15
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
20,489
2,615
Start with the expression:

##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##.
 
349
15
Start with the expression:

##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
12,039
2,619
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 [2] imply for the value of ##\phi## at equilibrium?
 
349
15
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?
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

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
31,667
4,667
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?
 
349
15
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

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
31,667
4,667
## \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?
 
349
15
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:

TSny

Homework Helper
Gold Member
12,039
2,619
upload_2017-4-11_18-34-47.png

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.
upload_2017-4-11_18-36-4.png

What angle is indicated by the blue angle?
 

Want to reply to this thread?

"Simple Harmonic motion" You must log in or register to reply here.

Related Threads for: Simple Harmonic motion

  • Posted
Replies
8
Views
3K
  • Posted
Replies
4
Views
2K
  • Posted
Replies
3
Views
10K
Replies
25
Views
419
Replies
1
Views
935
Replies
6
Views
10K
  • Posted
Replies
1
Views
807
  • Posted
Replies
17
Views
521

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top