What is the frequency of the oscillation?

  • Thread starter physicsss
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  • #1
physicsss
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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?
 

Answers and Replies

  • #2
Pyrrhus
Homework Helper
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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]
 
  • #3
physicsss
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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:
  • #4
the frequency is:

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

and

[tex]
\omega = \sqrt{\frac{k}{m}}
[/tex]
 
Last edited by a moderator:
  • #5
Hmm.. Did they tell you what the spring constant is?
 
  • #6
physicsss
319
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no. :frown:
 
  • #7
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.
 
  • #8
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.
 
  • #9
physicsss
319
0
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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