Finding Amplitude of the Oscillation

In summary: So, yes, your answer is correct.In summary, the problem involves a man applying a constant force to a spring block system, causing it to perform simple harmonic motion with a new equilibrium point. The equilibrium position is where the applied force is equal to the force from the spring, and the time period can be calculated using the given equations. The problem asks for the initial position of the block with respect to the new equilibrium point, and the resulting SHM can be described by the equation X = Asin(ωt + Φ), where the amplitude is equal to the displacement at t = 0.
  • #1
Hydrous Caperilla
62
7

Homework Statement

A
A man applies a Force F on a spring block system shown.towards right when the block is at rest and spring is relaxed .If F is constant then
[/B]

Homework Equations

: F=-kx[/B]

The Attempt at a Solution


The equilibrium position will be at the position where the disturbing and restoring fores are equal

F=kx
x=F/k

Also the time period will be 2pi√m/k.How should I proceed after this
 

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  • #2
What is the problem asking you to find?
 
  • #3
NFuller said:
What is the problem asking you to find?
The Amplitude of the resulting simple harmonic motion
 
  • #4
Hydrous Caperilla said:

Homework Statement

A
A man applies a Force F on a spring block system shown.towards right when the block is at rest and spring is relaxed .If F is constant then
[/B]

Homework Equations

: F=-kx[/B]

The Attempt at a Solution


The equilibrium position will be at the position where the disturbing and restoring fores are equal

F=kx
x=F/k

Also the time period will be 2pi√m/k.How should I proceed after this
What is the initial position of the block with rspect to the new equilibrium point?
 
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  • #5
ehild said:
What is the initial position of the block with rspect to the new equilibrium point?
ehild said:
What is the initial position of the block with rspect to the new equilibrium point?
The intial position of the block is at the point where the spring is relaxed
 
  • #6
If the force applied is constant, as stated, then the equilibrium position is where the applied force ##F_{a}## is equal to the force from the spring ##F_{s}##.
##F_{a}=F_{s}=kx##
You just need to solve for ##x##.
 
  • #7
Hydrous Caperilla said:
The intial position of the block is at the point where the spring is relaxed
Yes, and it will perform SHM about the new equilibrium point. At t=0, what is its displacement with respect to that point?
 
  • #8
NFuller said:
If the force applied is constant, as stated, then the equilibrium position is where the applied force ##F_{a}## is equal to the force from the spring ##F_{s}##.
##F_{a}=F_{s}=kx##
You just need to solve for ##x##.
I got the new equilibrium point from that...I am confused about how to proceed after that
 
  • #9
ehild said:
Yes, and it will perform SHM about the new equilibrium point. At t=0, what is its displacement with respect to that point?
F/k in my opinion
 
  • #10
Hydrous Caperilla said:
F/k in my opinion
Yes. Write the time dependence of the displacement of the box for t ≥ 0.
 
  • #11
Hydrous Caperilla said:
I got the new equilibrium point from that...I am confused about how to proceed after that
If the problem is asking for the SHM about this new equilibrium point, then you need to know something about the velocity of the block at that point. It seems like your post is missing half of the problem.
 
  • #12
NFuller said:
If the problem is asking for the SHM about this new equilibrium point, then you need to know something about the velocity of the block at that point. It seems like your post is missing half of the problem.
The block is at rest and the spring is relaxed initially. So you know the position and velocity at t =0. No need for the velocity at the new equilibrium.
 
  • #13
ehild said:
Yes. Write the time dependence of the displacement of the box for t ≥ 0.
The equation of displacement will be :X=Asin( ωt+ Φ). At t=0, F/k=Asin( Φ).
 
  • #14
ehild said:
The block is at rest and the spring is relaxed initially. So you know the position and velocity at t =0. No need for the velocity at the new equilibrium.
You said the block is at rest at this point...won't this mean this is the amplitude point because at amplitude velocity is zero?
 
  • #15
Hydrous Caperilla said:
You said the block is at rest at this point...won't this mean this is the amplitude point because at amplitude velocity is zero?
Yes, the displacement of the block from its new equilibrium is biggest at t=0, when it is in rest. And the biggest displacement from equilibrium is the amplitude.
 
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Related to Finding Amplitude of the Oscillation

1. What is amplitude of oscillation?

The amplitude of oscillation is the maximum displacement of an oscillating object from its equilibrium position. It is a measure of the magnitude of the oscillation and is usually represented by the symbol "A".

2. How is amplitude of oscillation calculated?

The amplitude of oscillation can be calculated using the formula A = (maximum displacement - equilibrium position). This means that the amplitude is equal to the difference between the highest point and the lowest point of the oscillation.

3. What factors affect the amplitude of oscillation?

The amplitude of oscillation can be affected by several factors, including the energy input into the system, the frequency of oscillation, and the damping force. In general, a higher energy input or a lower damping force will result in a larger amplitude of oscillation.

4. How is amplitude of oscillation related to period?

The amplitude of oscillation is not directly related to the period, which is the time it takes for one complete oscillation. However, the amplitude can affect the period by changing the time it takes for the oscillation to reach its maximum or minimum point.

5. What are practical applications of finding amplitude of oscillation?

Finding the amplitude of oscillation is important in many practical applications, including in engineering, physics, and music. It can help determine the stability of structures, the accuracy of timekeeping devices, and the intensity of sound waves.

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