Calculating Amplitude in Simple Harmonic Motion

Click For Summary
The discussion centers on calculating the amplitude in simple harmonic motion, specifically addressing a problem from a past paper. The user is confused about how to derive the amplitude from the given equation, A = √(c₁² + c₂²), where x(t) is expressed as a combination of cosine and sine functions. Clarification is provided regarding the equilibrium point of a block, indicating that the amplitude is the excess distance from this point. The user expresses gratitude for the explanation, indicating that the concept is becoming clearer. Understanding the relationship between equilibrium position and amplitude is crucial for solving such problems in simple harmonic motion.
affirmative
Messages
13
Reaction score
0
My question is with part c, more specifically the calculating of the amplitude part.
[PLAIN]http://img691.imageshack.us/img691/1750/shmquestion.jpg

The answer to the question is:
[PLAIN]http://img80.imageshack.us/img80/8921/shmanswer.jpg

I do not understand how to arrive at this conclusion in order to calculate the amplitude; it baffles me. Any poking or prodding in the right direction (or even an outright answer) would be greatly appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
Isn't the amplitude

A = \sqrt{c^2_1 + c^2_2}

Where

x(t) = c_1cos(wt) + c_2sin(wt)
 
I have no idea unfortunately. My question is how did the examiner arrive at the answer above (it is the verbatim answer for the past paper I'm currently working on).
 
see the equilibrium point of smaller block is .05g/k = .01 m but initially it is at .03 m thus this excess distance is its amplitude . as after each oscillation it wll come back to this point .
 
Thank you very much, that actually kind of makes sense now!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 13 ·
Replies
13
Views
1K
Replies
3
Views
815
  • · Replies 1 ·
Replies
1
Views
1K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K