What is the frequency of the oscillation?

In summary, the conversation discusses a system consisting of a massless spring hanging from the ceiling with a small object attached to its lower end, which is initially at rest at a position yi. The object is then released and oscillates up and down, with its lowest position being 10 cm below yi. The frequency of the oscillation is determined using the equation f = \omega / 2\pi, where \omega = \sqrt{k/m}. The speed of the object when it is 8.0 cm below the initial position is found by letting x(t) = 8 and finding t, then taking the derivative of x(t) to find the velocity. It is also mentioned that a second
  • #1
physicsss
319
0
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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  • #2
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
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.
 
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  • #4
the frequency is:

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

and

[tex]
\omega = \sqrt{\frac{k}{m}}
[/tex]
 
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  • #5
Hmm.. Did they tell you what the spring constant is?
 
  • #6
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
how did you come up with the equation g=w^2*x?
 
  • #10
well - if u take the second derivative of x(t)
 

FAQ: What is the frequency of the oscillation?

What is the definition of frequency in terms of oscillation?

The frequency of oscillation is the number of complete cycles or vibrations that occur in a given amount of time. It is measured in Hertz (Hz).

How is frequency related to the amplitude of an oscillation?

Frequency and amplitude are inversely related. This means that as frequency increases, the amplitude decreases, and vice versa.

Can frequency of oscillation be changed?

Yes, the frequency of oscillation can be changed by altering the energy or force acting on the oscillating object. For example, increasing the force applied to a spring will increase its frequency of oscillation.

What factors affect the frequency of oscillation?

The frequency of oscillation is affected by the mass of the object, the stiffness of the oscillating system, and any external forces acting on the system.

How is frequency of oscillation measured?

Frequency of oscillation can be measured using various devices such as an oscilloscope or frequency counter. It can also be calculated by dividing the number of cycles by the time taken to complete those cycles.

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