What is the frequency of the oscillation?

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Homework Help Overview

The discussion revolves around a mass-spring system where a small object oscillates after being released from an initial position. The problem includes determining the frequency of oscillation, the speed of the object at a certain position, and the effects of adding mass to the system on its frequency and equilibrium position.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss applying Newton's 2nd Law to derive the equation of motion and explore the relationship between force, mass, and acceleration. There are attempts to relate the spring constant to gravitational force and oscillation frequency. Questions arise regarding the spring constant and its role in the calculations.

Discussion Status

The conversation is ongoing, with various approaches being explored. Some participants have provided equations and reasoning, while others question the assumptions made, particularly regarding the spring constant and its implications for the frequency calculation.

Contextual Notes

There is a noted lack of information regarding the spring constant, which is critical for solving the frequency and speed questions. Participants are also considering the implications of adding mass to the system and how it affects the oscillation characteristics.

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A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially
held at rest in a position yi such that the spring is at its rest length. The object is then released from yi and
oscillates up and down, with its lowest position being 10 cm below yi.
(a) What is the frequency of the oscillation?
(b) What is the speed of the object when it is 8.0 cm below the
initial position?
(c) An object of mass 300 g is attached to the first object, after which the system oscillates
with half the original frequency. What is the mass of the first object?
(d) Relative to yi, where is the new
equilibrium (rest) position with both objects attached to the spring?
 
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Apply Newton's 2nd Law to get the equation of motion for this system.

[tex]\sum_{i=1}^{n} \vec{F}_{i} = m \vec{a}[/tex]

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

[tex]g - \frac{k}{m}x = \ddot{x}[/tex]

this gives a second order differential equation with the following solution:

[tex]x = A \sin (\omega t - \phi) + B[/tex]
 
For finding the frequency, I did mg=kx, and k=mg/x. since f=1/2pi*sqrt(k/m) I substitute mg/x with k. But it was wrong.
 
Last edited:
the frequency is:

[tex] f = \frac{\omega}{2\pi}[/tex]

and

[tex] \omega = \sqrt{\frac{k}{m}}[/tex]
 
Last edited by a moderator:
Hmm.. Did they tell you what the spring constant is?
 
no. :frown:
 
Actually, if it has the spring got a gravitational acceleration then you can say that.

[tex] g = \omega ^2x[/tex]

But i am still not sure about this.
 
If you have omega then:

for b) you can let x(t) = 8 and find t.
then take the derivative of x(t) and find the velocity.

I think that should work.
 
how did you come up with the equation g=w^2*x?
 
  • #10
well - if u take the second derivative of x(t)
 

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