Harmonic Oscillator - Modeling and Observations

In summary, the conversation discussed a steel spring that extends by 10cm in equilibrium when a weight of 100g is attached to its lower end. Using the equation for the harmonic oscillator, the period of the up and down oscillation was determined to be 0.63 seconds. The assumptions made by employing these equations include linearity, constant rate of restoring force, harmonic behavior, and no damping or friction. The equations will be valid for small amplitudes. It was also discussed that suspending a weight of 1kg would result in a longer period of oscillation, calculated to be 0.77 seconds.
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Homework Statement


Consider a steel spring with the property that it extends by 10cm (0.1m) in equilibrium when you attach the upper end of the spring to a fixed support and hang a weight of 100g (0.1kg) at the springs lower end.
1) Use the equation for the harmonic oscillator to determine the period of the up and down oscillation you observe when you give the weight a displacement from the equilibrium position by lifting it up and slightly letting it go.
2) Write down different modelling assumptions that you have made by employing these equations.
3) Over what range of amplitude will these equations be valid?
4) Using the same spring, what would you expect to observe when you suspend instead a weight of 1kg?


Homework Equations





The Attempt at a Solution


1) F = kx = mg so k = mg/x = 0.1*9.81/0.1 = 9.81 N/m.
Period T = 2[itex]\pi[/itex]sqrt(m/k) = 2[itex]\pi[/itex]sqrt(0.1/9.81) = 0.63 seconds (2dp).
 
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  • #2
2) Assumptions: The spring is linear and has a constant rate of restoring force; the spring behaves in a harmonic manner; there is no damping or friction; the applied force is small enough to be considered as a perturbation.3) The equations will be valid for small amplitudes (less than 0.1m).4) We would expect to observe a longer period of oscillation, as the mass is increased from 0.1kg to 1kg. The period can be calculated using the same equation as before, i.e. T = 2\pisqrt(m/k) = 2\pisqrt(1/9.81) = 0.77 seconds (2dp).
 

Related to Harmonic Oscillator - Modeling and Observations

1. What is a harmonic oscillator?

A harmonic oscillator is a system that oscillates or vibrates at a regular frequency. It is typically described as a mass attached to a spring, where the restoring force of the spring causes the mass to oscillate back and forth.

2. What is the equation for a harmonic oscillator?

The equation for a harmonic oscillator is F = -kx, where F is the restoring force, k is the spring constant, and x is the displacement from equilibrium.

3. What is the period of a harmonic oscillator?

The period of a harmonic oscillator is the time it takes for one complete oscillation, or cycle. It is calculated as T = 2π√(m/k), where m is the mass and k is the spring constant.

4. How does amplitude affect a harmonic oscillator?

The amplitude of a harmonic oscillator is the maximum displacement from equilibrium. As the amplitude increases, the restoring force also increases, leading to a higher frequency and shorter period of oscillation.

5. What is the relationship between a harmonic oscillator and simple harmonic motion?

A harmonic oscillator is an example of simple harmonic motion, which is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium. All harmonic oscillators exhibit simple harmonic motion, but not all systems with simple harmonic motion are harmonic oscillators.

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