Oscillation Problem -- Why does my way not work?

AI Thread Summary
The discussion revolves around a mass-spring oscillation problem where the user incorrectly equates the weight of the object to the spring force at the lowest point of oscillation. The correct approach involves recognizing that the oscillating object moves beyond the equilibrium point, meaning the spring force cannot simply equal the weight at that position. The user initially calculated the frequency as 1.57 Hz, which is incorrect; the expected frequency is 2.2 Hz. The misunderstanding lies in the assumption that the displacement x represents the equilibrium position, which it does not during oscillation. Clarifying these concepts is essential for solving the problem accurately.
katha
Messages
3
Reaction score
0
1. Problem Description:
A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest at a position y. The object is then released from y and oscillates up and down, with its lowest position being 10cm below y.
What is the frequency of the oscillation?

Homework Equations


F=kx; w= sqroot k/m ; f= w/2pi ;3. I know that the right answer to the problem is 2.2 Hertz and I know how you could solve it using energy however I was wondering why the following does not give me the right answer. Is my math just wrong or is it a conceptual problem?

My Way:

F=kx
mg = kx ; x= .1m
mg = k (.1)
(9.8 /.1) = k/m
k/m = 98

w = sq-root k/m
w = sq-root (98)
w = 9.89

f = w/(2pi)
f= (9.89)/(2pi)
f= 1.57 Hz

This is obviously not the right answer. I hope you can give me an explanation as to why it's not.
Thanks in advance!
 
Physics news on Phys.org
katha said:
F=kx
mg = kx ; x= .1m

Why did you equate weight and spring force?
 
Mastermind01 said:
Why did you equate weight and spring force?

I don't no if this assumption is true, but I just thought that the mass/spring system's equilibrium point was at .1m and at that point the force of the spring and the force due to gravity should be equal (but in opposite direction).
 
katha said:
I don't no if this assumption is true, but I just thought that the mass/spring system's equilibrium point was at .1m and at that point the force of the spring and the force due to gravity should be equal (but in opposite direction).

Do you know the dynamics of oscillation? Think of a pendulum. What happens at its extreme point?
 
Mastermind01 said:
Do you know the dynamics of oscillation? Think of a pendulum. What happens at its extreme point?

Ok, I think I see where the problem is. The distance x can't be the equilibrium point because when smth oscillates, it goes beyond its equilibrium point. So F of the spring can't be equal to mg, right?
 
katha said:
Ok, I think I see where the problem is. The distance x can't be the equilibrium point because when smth oscillates, it goes beyond its equilibrium point. So F of the spring can't be equal to mg, right?

That is correct.
 
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