Why Is Answer A Correct for the Oscillation Problem?

  • Thread starter Thread starter magiclink
  • Start date Start date
  • Tags Tags
    Oscillation
AI Thread Summary
The discussion revolves around a physics question regarding oscillation, specifically why answer A is correct. The participant initially believed that since the ball was described as "stationary," all forces must be equal, leading them to choose answer C. However, upon further reflection, they realized that the forces must sum to zero, indicating that the vectors indeed add up correctly to support answer A. The confusion stemmed from the interpretation of the question and the nature of vector addition. Ultimately, the participant acknowledged their misunderstanding and clarified that the question pertains to IB Physics.
magiclink
Messages
23
Reaction score
0

Homework Statement


Hi! My question is about oscillation. The multiple choice answers and question is included in the picture attached! I also happen to know the answer is A by looking at the marscheme.


The Attempt at a Solution


I would've thought that since the ball is said to be held "stationary" in that position the forces would all have to be equal. Thus i would've thought C was the correct answer. If they had said that it's an oscillation where the ball is in position P and not mentioned it being held stationary, I would've answered A, however, knowing that the forces would've had to be unbalanced and acting towards the equilbrium point. Could anybody explain exactly why it's A? and not C? I think my problem is probably largely due to interpretation of the question. Thankyou for any replies!
 

Attachments

  • Oscillation question.jpg
    Oscillation question.jpg
    6.3 KB · Views: 484
Physics news on Phys.org
If the ball is stationary then the forces must sum to zero net force. If they were all equal, would the sum be zero? Remember, you're adding vectors here.
 
Hi, magiclink!
Is that IB Physics?
 
Hey! I just looked at the question and gave it a good look again. Turns out that the vectors did add vectorally to equal zero. A bit of an optical illusion to me. Thanks anyways! Sorry for wasting your time! :)


And yes, tis IB physics indeed. ;)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top