Simple Harmonic Oscillator: Calculating Particle Position as a Function of Time

In summary: Clearly the real lesson of this problem is that you should never trust your friends :tongue2: (just kidding of course)Clearly the real lesson of this problem is that you should never trust your friends :tongue2: (just kidding of course)
  • #1
Je m'appelle
120
0
I need someone to please verify my work.

Homework Statement



A particle of mass [tex]m[/tex] is suspended from the ceiling by a spring of constant [tex]k[/tex] and initially relaxed length [tex]l_0[/tex]. The particle is then let go from rest with the spring initially relaxed. Taking the z-axis as vertically oriented downwards with it's origin in the ceiling, calculate the position 'z' of the particle as a function of time 't'.

Homework Equations



SHO equation:

[tex]m\ddot{x}(t) = -kx(t) [/tex]

The Attempt at a Solution



In this case, the acting forces on the mass are the weight directed downwards and the elastic force of the spring directed upwards, so

[tex]m\ddot{x}(t) = mg - kx(t) [/tex]

[tex]\ddot{x}(t) = g - \omega^2 x(t) [/tex]

[tex]\ddot{x}(t) + \omega^2 x(t) = g [/tex]

It's important to note that I'm assuming 'x' as the displacement of the spring and NOT the displacement of the mass.

[tex]\ddot{x}(t) + \omega^2 x(t) = g [/tex]

[tex]x(t) = x_p(t) + x_h(t) [/tex]

The homogeneous solution will be

[tex]x_h(t) = acos(\omega t) + bsin(\omega t) [/tex]

And the particular solution will be

[tex]x_p(t) = \frac{g}{\omega^2}[/tex]

By using the initial conditions [tex]x(0) = 0, \dot{x}(0) = 0[/tex] we get

[tex]x(0) = 0 = a + \frac{g}{\omega^2} [/tex]

[tex]a = - \frac{g}{\omega^2} [/tex]

[tex]\dot{x}(0) = 0 = b\omega,\ b = 0[/tex]

So by rearranging we get to

[tex]x(t) = -\frac{g}{\omega^2}cos(\omega t) + \frac{g}{\omega^2} [/tex]

But notice that this is the displacement of the spring and not of the mass, in order to find the displacement of the mass we would have to add the initial length of the relaxed spring, that is, [tex]l_0[/tex].

So the final answer for the displacement of the mass in the z-axis will be

[tex]z(t) = l_0 - \frac{mg}{k} (cos(\sqrt{\frac{k}{m}} t) - 1) [/tex]

Could someone please verify this, as there is a different answer in a solution manual I found, and I'm not so sure which one is correct.
 
Physics news on Phys.org
  • #2
Hm, it seems reasonable to me. What's the other answer you found?
 
  • #3
diazona said:
Hm, it seems reasonable to me. What's the other answer you found?

Well, I think it is correct after all.

I had two solutions, mine and a friend's which he claimed to have obtained from a solutions manual, but I think that wasn't the case.

I'm sure now that the answer I've provided here is the correct one, as I've analyzed the case when [tex]t = 0[/tex], that is when the system is at rest, the displacement of the mass is simply the length of the relaxed spring which is reasonable.

Thanks anyway diazona
 
  • #4
Ah. Clearly the real lesson of this problem is that you should never trust your friends :tongue2: (just kidding of course)
 
  • #5
Je m'appelle said:
Well, I think it is correct after all.

I had two solutions, mine and a friend's which he claimed to have obtained from a solutions manual, but I think that wasn't the case.

I'm sure now that the answer I've provided here is the correct one, as I've analyzed the case when [tex]t = 0[/tex], that is when the system is at rest, the displacement of the mass is simply the length of the relaxed spring which is reasonable.

Thanks anyway diazona

Are you sure it is not -l0,which direction are you taking it positive?
 
Last edited:

1. What is a Simple Harmonic Oscillator?

A Simple Harmonic Oscillator is a system that exhibits repetitive, back-and-forth motion around a central equilibrium point. It is characterized by a restoring force that is directly proportional to the displacement of the object from its equilibrium position, and the motion is described by a sinusoidal function.

2. What are some real-life examples of Simple Harmonic Oscillators?

Examples of Simple Harmonic Oscillators include a swinging pendulum, a mass attached to a spring, and the motion of a guitar string. Other examples can be found in nature, such as the motion of tides, the vibrations of atoms in a solid, and the motion of planets in orbit around a star.

3. How is the period of a Simple Harmonic Oscillator determined?

The period of a Simple Harmonic Oscillator is determined by the mass of the object, the stiffness of the restoring force, and the amplitude of the motion. It can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.

4. What is the relationship between Simple Harmonic Oscillators and energy?

In a Simple Harmonic Oscillator, energy is constantly being exchanged between potential energy (stored in the spring or other restoring force) and kinetic energy (associated with the motion of the object). The total energy remains constant, but it is constantly being converted between these two forms.

5. How is a Simple Harmonic Oscillator different from a damped or driven oscillator?

A damped oscillator experiences a decrease in amplitude over time due to the dissipation of energy, while a driven oscillator is subject to an external driving force that affects its motion. In contrast, a Simple Harmonic Oscillator is an idealized system with no external forces acting on it and no energy loss, resulting in a constant amplitude of motion.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
166
  • Introductory Physics Homework Help
Replies
16
Views
335
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
405
  • Introductory Physics Homework Help
Replies
4
Views
685
  • Introductory Physics Homework Help
Replies
10
Views
831
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
280
Replies
31
Views
593
Back
Top