Motion equation for harmonic oscillator

In summary, the mass is oscillating up and down due to the forces of gravity and a spring. The equilibrium position is when the forces are equal to each other.
  • #1
phys2
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0

Homework Statement



A mass m is attached to a spring of stiffness k. The spring is attached to the ceiling and the mass hangs freely from the spring under the force of gravity.

(a) Derive the equation of motion for this system.
(b) Find an expression for the equilibrium position of the mass ( in terms of the equilibrium position in the absence of gravity).
(c) Show that the equation of motion is equivalent to the mass plus spring system in the absence of gravity.

Homework Equations



F = -kx, Euler Lagrange equation and L = T - V

The Attempt at a Solution



So for (a), I attempted to use the Lagrangian method. Taking the only degree of freedom as up and down (y direction), my velocity vector is v = dy/dx times the unit vector y hat.

So T (kinetic energy) = 0.5m (ydot)^2 where y dot = dy/dx

So now I need V and this is where I am confused, don't I have two potentials here; gravity and the spring potential? Which one should I take? I am guessing that I cannot take both?
 
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  • #2
If I understand the setup correctly, there is just a mass hanging from a spring. It shouldn't be necessary to go into the Lagrangian. You know the forces. There are only two of them on the spring. Set their sum equal to ma, and you have the equation of motion. What's the condition for equilibrium? That the forces are equal to each other. Write the equation for that and you will find the equilibrium position.
 
  • #3
phys2 said:

The Attempt at a Solution



So for (a), I attempted to use the Lagrangian method. Taking the only degree of freedom as up and down (y direction), my velocity vector is v = dy/dx times the unit vector y hat.

So T (kinetic energy) = 0.5m (ydot)^2 where y dot = dy/dx
You should be differentiating with respect to time, right?

So now I need V and this is where I am confused, don't I have two potentials here; gravity and the spring potential? Which one should I take? I am guessing that I cannot take both?
You include both. Why wouldn't you?
 
  • #4
You should be differentiating with respect to time, right?

Oh yea, typo over there, it should be dy/dt.

You include both. Why wouldn't you?

Oh alright. I just sort of felt it might be wrong to have two potentials, no real reason.

So for V, I am just supposed to add both together? V = -1/2 kx^2 + mgx
 
  • #5
Yes, just add them together. Make sure you get the signs right, though. How are you defining your coordinates?
 
  • #6
I am taking downwards as my positive direction. It should really be V = -1/2 ky^2 + mgy since the oscillator is moving up and down the y-axis. Thanks for your help btw...
 
  • #7
So for part (a), I got my V = -0.5my^2 + mgy and so differentiating V with respect to y, I get dV/dy = -ky + mg

Now using the Euler Lagrange equation d/dt [dL/dydot) - dL/dy = 0, where dL/dydot = mydot and dL/dy = -ky + mg, d/dt [dL/dydot] = md2y/dt2

So now the equation of motion is md2y/dt2 - [-ky + mg] = 0 or md2y/dt2 + ky = mg. Would that be right? The thing is I have no idea whether my answers are right! Is there any way to do a check whether it actually makes sense or not?

Ok, now for (b), what I did was take the net forces (between gravity and the spring force) to be zero. So it would be -ky + mg = 0 and y = mg/k which is the equilibrium position of the mass? I don't get why they talk about the absence of gravity though? What exactly does that mean for the question?

Sorry for all the questions but I am really trying to get my head around harmonic oscillators...seems like the worst topic for me in physics.
 
  • #8
The correct potential is ##V = \frac{1}{2} ky^2 - mgy##. You made other sign errors, which ended up canceling out, but you should go back and fix everything up.

Your equation of motion is ##m\frac{d^2 y}{dt^2} = mg - ky##. Hopefully, you recognize that as Newton's 2nd law.

You might want to go back to your freshman physics book and review the harmonic oscillator.
 
Last edited:
  • #9
Careful, vela. You've got acceleration equal to force.
 
  • #10
Good catch. Fixed it. Thanks.
 
  • #11
Yea, I see what I have done wrong. Thanks again vela.
 

Related to Motion equation for harmonic oscillator

1. What is the motion equation for a harmonic oscillator?

The motion equation for a harmonic oscillator is x(t) = A*cos(ωt + φ), where x(t) is the position of the oscillator at time t, A is the amplitude, ω is the angular frequency, and φ is the phase angle.

2. How is the frequency of a harmonic oscillator related to its mass and spring constant?

The frequency of a harmonic oscillator is directly proportional to the square root of its spring constant and inversely proportional to the square root of its mass. This relationship is given by the equation f = (1/2π)*√(k/m), where f is the frequency, k is the spring constant, and m is the mass.

3. What is the significance of the phase angle in the motion equation for a harmonic oscillator?

The phase angle in the motion equation for a harmonic oscillator represents the initial displacement of the oscillator from its equilibrium position. It determines the starting point of the motion and helps describe the phase difference between two oscillators.

4. How can the motion equation for a harmonic oscillator be used to calculate the period of oscillation?

The period of oscillation of a harmonic oscillator can be calculated by taking the inverse of the frequency, which is given by the equation T = 2π/ω. This means that the period is directly proportional to the angular frequency and inversely proportional to the amplitude of the oscillator.

5. What happens to the motion of a harmonic oscillator when the amplitude is increased?

When the amplitude of a harmonic oscillator is increased, the maximum displacement of the oscillator from its equilibrium position also increases. This results in an increase in the energy and frequency of the oscillator, causing it to oscillate more rapidly.

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