# Homework Help: Time period of oscillation and displacement from O

1. Apr 14, 2012

1. The problem statement, all variables and given/known data
In the attachment!

3. The attempt at a solution
My big hangup in this situation is why T, the period, of the oscillation doesn't change?
If you increase the displacement by 2A does this mean that since the force acting on the spring also increases proportionally there's just no change in frequency and the time period of oscillation? I understand however that the energy in the system increases is quadrupled. graphing Force with respect to distance, it's 4 times the area underneath the graph if you double A, the displacement, according to hooke's law.

So! Why does the Time period of oscillation not change at all? This is what confuses me! Thankyou for any response :)

#### Attached Files:

• ###### Hooke's law and Time periods.jpg
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2. Apr 14, 2012

### tiny-tim

same as the reason why the period of a pendulum doesn't change (that's why a pendulum clock works steadily even though it's running down) …

it's a simple harmonic oscillator: the equation is x'' = -kx, and the solutions are harmonic functions with frequency √k, times constants

3. Apr 14, 2012

### jambaugh

This is a feature of the simple harmonic oscillator, the period is independent of the amplitude. In terms of the mechanics, the greater maximum displacement means a greater restoring force so a faster trip through the cycle.

Imagine different spring force relationships, F = f(x). If the restoring force gets larger much faster as x increases then for larger displacements you get a faster cycle. (Imagine a rubber ball swinging on a pendulum and bouncing between two walls until it losses enough energy and simply swings at a slower rate.

Likewise if the restoring force grows more slowly with x you'll have what you intuition might imagine, larger amplitudes have longer periods. Example there, orbiting planets.

Somewhere in between there should be a spring force relationship where the period for all amplitudes match. That turns out to be the Hook's law case.

4. Apr 14, 2012

### jambaugh

Actually a true pendulum does change. We take the SHO approximation for small oscillations $sin(\theta) \approx \theta$.

The "rule of thumb" is when the pendulum swings more than 10° the SHO approximation breaks down and you shouldn't assume it has constant period. In the extreme case, if you start the pendulum all the way at the top it has infinite period. If it's not quite all the way up, you get a long wait while it slowly moves further off center and then the big swing.

5. Apr 15, 2012