Simple harmonic oscillator and pendulum

​​In a nutshell

The time period of a mass-spring system can be calculated if the mass and the spring constant are known. The time period of a pendulum can be calculated if the length of the string is known and is independent of mass or amplitude of the swing. 

Equations

Constants

Variable definitions

Mass on a spring

When a mass on a spring is pulled away from its resting position and released, the system will undergo simple harmonic motion.

The time period of the oscillations can be calculated if the masses and spring constant are known.

​$$T=2\pi \sqrt {\frac mk}$$​​

The higher the spring constant, the smaller the time period, however the heavier the mass used, the longer the time period. 

Note: The spring constant is a measure of stiffness. 

​​Example

A $$200 \, g$$ mass stack is attached to a spring whose spring constant is $$670 \, N\,m^{-1}$$. The mass stack is displaced from its resting position and released. The system undergoes simple harmonic motion. Calculate the time period of the oscillations, assume frictional forces are negligible.

Firstly, write down the known values: 

$$m=200\, g = 0.20 \, kg \newline \\[0.1in]k=670 \, N\,m^{-1}$$

​Next, write down the equations needed and rearrange if necessary:

$$T=2\pi \sqrt {\frac mk}$$

Then, substitute the values into the equation:

$$T=2\pi \sqrt {\frac {0.20}{670}} \newline \\[0.1in]T=0.1085569...$$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$$time \ period, \ T=0.11 \, s$$​​​​​

The time period of the oscillation is $$\underline {0.11 \, s}.$$​

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The pendulum

When a pendulum is allowed to swing, the motion can be described as simple harmonic motion. 

The time period of the pendulum can be calculated if the length of the string is known. 

​$$T=2\pi \sqrt {\frac lg}$$​​

The only variable which affects the pendulum is the length of the string. It is independent of mass and amplitude of the swing, which is a common misconception. 

Note: If the pendulum was swung on a different planet or the Moon, then the time period would change, due to different gravitational field strengths. However, it is  for Earth. 

​​Example

A bob is attached to a piece of string with a length of $$145 \, cm$$. The bob is displaced from its resting position and released. The system undergoes simple harmonic motion. Calculate the time period of the oscillations, assume frictional forces are negligible.

Firstly, write down the known values: 

$$l=145 \, cm = 1.45\,m$$

​Next, write down the equations needed and rearrange if necessary:

$$T=2\pi \sqrt {\frac lg}$$

Then, substitute the values into the equation:

$$T=2\pi \sqrt {\frac {1.45}{9.81}} \newline \\[0.1in]T=2.41562...$$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$$time \ period, \ T=2.42 \, s$$​​​​​

The time period of the oscillation is $$\underline {2.42 \, s}.$$​

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