What is the amplitude of the resulting oscillations?

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Homework Help Overview

The discussion revolves around a physics problem involving a mass attached to a spring, subjected to an oscillating force. The participants are trying to determine the amplitude of the resulting oscillations while neglecting damping effects. The problem falls under the subject area of oscillations and simple harmonic motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the applied force and the spring force, questioning the calculation of amplitude based on the given parameters. Some express confusion about the role of gravity and how it affects the equilibrium position without altering the amplitude. Others inquire about the correct application of Newton's second law and the net forces acting on the mass.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the problem. Some have offered insights into the equations of motion and the nature of driven oscillations, while others are still seeking clarity on their calculations and the underlying principles.

Contextual Notes

There is a noted confusion regarding the application of forces and the resulting motion, particularly in distinguishing between the driving force and the restoring force of the spring. Participants are also grappling with the implications of their calculations and the assumptions made in the problem setup.

yecko
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Homework Statement


A mass of 2.0 kg hangs from a spring with a force constant of 50 N/m. An oscillating force F = (4.8 N) cos[(3.0 rad/s)t] is applied to the mass. What is the amplitude of the resulting oscillations? Neglect damping. Answer: 0.15 m

Homework Equations


F=kx , the mass only change the equilibrium position, amplitude is when max F thus cos is equal to 1

The Attempt at a Solution


F=4.8N, k=50N/m
x=F/k
=4.8/50
=0.096m.

What's wrong with my calculation? Thanks
 
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yecko said:

Homework Statement


A mass of 2.0 kg hangs from a spring with a force constant of 50 N/m. An oscillating force F = (4.8 N) cos[(3.0 rad/s)t] is applied to the mass. What is the amplitude of the resulting oscillations? Neglect damping. Answer: 0.15 m

Homework Equations


F=kx , the mass only change the equilibrium position, amplitude is when max F thus cos is equal to 1

The Attempt at a Solution


F=4.8N, k=50N/m
x=F/k
=4.8/50
=0.096m.

What's wrong with my calculation? Thanks
The mass experiences three forces, gravity, mg, spring force, (-kx) and the driving force F. Gravity changes the equilibrium position only, but F is not equal to kx.
 
yecko said:
the mass only change the equilibrium position
Yes, I understand and i have mentioned it in the first post
ehild said:
F is not equal to kx.
So what to calculate?
Thanks
 
Last edited:
yecko said:
Yes, I have mentioned it

So what to calculate?
Thanks
The amplitude of vibration.
 
ehild said:
The amplitude of vibration.
I know, frankly its the question~
i mean how i can calculate in order to get the amplitude of vibration (the answer!)
what are the formulae to be used and the steps for getting the answer, if my steps are wrong (whats wrong?? why F is not equal to k*dx ?)
thanks

(ps. what I mentioned in the first posts are things all i understand... please tell me what i have misunderstood...)
 
Really wish I could help you here, but no matter how I look at this question I get the same answer you do, and the only answer I can find online is formatting that makes no sense;

- A = ( F m ) ( 2 2 ) 2 ( b 2 m ) 2 = ( 4.8 N 2 kg ) [ ( 3 s 1 ) 2 ( 50 N m 2 kg ) ] 2 + = 0.15 m -

I'd keep trying but its like 2 am, hope somebody else helps you find your answer.
 
well, indeed the format goes wrong by directly copy and paste would make it unreadable

it seems we are on the another half of the planet:) it is still early here~
however i am going to have exam tomorrow, hope all problems in my exercises can be settled by today:)
thank you very much
 
Can you write Newton's second law for this object? Remember, you have to take two forces into account.
The object does SHM -simple harmonic motion. What is its displacement x in terms of time?
You certainly learned about driven oscillation. The object will oscillate with the frequency of the diving force after some time. What is the angular frequency of the external force?
See the x(t) function of the forced oscillator in http://www.physics.louisville.edu/cldavis/phys298/notes/resonance.html, with the damping b neglected.
 
ehild said:
Can you write Newton's second law for this object?
F=ma
ehild said:
What is its displacement x in terms of time?
x=A cos(ωt+Θ)
ehild said:
What is the angular frequency of the external force?
yecko said:
F = (4.8 N) cos[(3.0 rad/s)t]
ω=3
 
  • #10
yecko said:
F=ma

F = ma is valid if F means the net force applied to the object. What is that net force in this case?
yecko said:
x=A cos(ωt+Θ)ω=3

The driving force is FD=4.8 cos(3t) . The elastic force is Fe=-kx. What is the net force?
If what you said FD=kx, the net force is zero, and the acceleration is zero. The object can not oscillate.
 
  • #11
ehild said:
What is the net force?
Fd+Fe?
ehild said:
FD=kx, the net force is zero
Why? I mean when the spring length is amplitude, force is max, accerleration is max.
 
  • #12
yecko said:
Fd+Fe?
Yes, and what is in terms of time and x?
yecko said:
Why? I mean when the spring length is amplitude, force is max, accerleration is max.
You speak about the spring force. But there are two forces, applied on the object, and the acceleration is (Fd+Fe)/m.
 
  • #13
ehild said:
Yes, and what is in terms of time and x?
Fd=4.8 cos(3t) , Fe=-kx
F(net)=4.8 cos(3t) +-kx
ehild said:
Gravity changes the equilibrium position only
ehild said:
acceleration is (Fd+Fe)/m
are they contradict?
 
  • #14
yecko said:
Fd=4.8 cos(3t) , Fe=-kx
F(net)=4.8 cos(3t) -kx
Yes. So ma=4.8 cos(3t) -kx. And the object performs SHM, with the frequency of the driving force, so x=Bcos(3t+θ). What is the acceleration then?
 
  • #15
nothing
 
  • #16
nothing
 
  • #17
yecko said:
x=Acos(3t+θ) [A=amplitude?]
a=-9Acos(3t+θ)
yes. And ma=4.8cos(3t)-kx. What should be A and θ?
 
  • #18
A = 0.096(?) and θ=0
step:50Acos(3t+θ)=4.8cos(3t)
whats wrong with my calculation?
 
  • #19
yecko said:
A = 0.096(?) and θ=0
step:50Acos(3t+θ)=4.8cos(3t)
whats wrong with my calculation?
How did you get the equation 50Acos(3t+θ)=4.8cos(3t)? You think that the acceleration is zero? That is wrong.
 
  • #20
substitute a and x into ma=4.8cos(3t)-kx
 
  • #21
yecko said:
substitute a and x into ma=4.8cos(3t)-kx
Where is a?
 
  • #22
oh
i got it
i mixed up the plus and minus sign only
thank you very much for your help!
 

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