Simple harmonic oscillations
Section: Oscillations | Syllabus: Cambridge A Level Physics (9702)
Describing Oscillations
An oscillation is a type of periodic motion that repeats regularly (e.g., up and down, back and forward). The equilibrium position is the position where an object returns when oscillation stops; the resultant force at this position is zero.
Key Terms:
- Displacement (x): Distance from the equilibrium position.
- Amplitude (x_0): Maximum displacement from equilibrium.
- Period (T): Time for one complete oscillation (cycle).
- Frequency (f): Number of oscillations per unit time (f = 1/T).
- Angular Frequency (ω): The rate of change of phase angle, measured in rad s^-1.
Defining Simple Harmonic Motion (SHM)
Simple harmonic motion occurs when the restoring force (and thus acceleration) is proportional to the displacement and acts in the opposite direction.
a = -ω^2 xThe negative sign indicates that acceleration a and displacement x are always in opposite directions (acceleration is always directed towards the equilibrium position).
Figure 17.1: Simple Harmonic Motion GraphsGraphs showing the relationship between displacement, velocity, and acceleration over time. Displacement as a sine curve, velocity as a cosine curve, and acceleration as a negative sine curve.
Equations of Motion
The displacement x depends on the starting position of the oscillation:
- Starting at equilibrium: x = x_0sin(ω t)
- Starting at maximum displacement: x = x_0cos(ω t)
Velocity (v) and Acceleration (a) are derived from these:
v = v_0cos(ω t) (where v_0 = ω x_0 is the max velocity)
v = ± ω √(x_0^2 - x^2)
a = -ω^2 x = -ω^2 x_0sin(ω t)
Worked Examples
Worked Example: Ball in a Curved BowlQuestion: A ball oscillates in a curved bowl starting at the center (displacement zero). Amplitude is 5.0 cm and period is 860 ms. Calculate the displacement at t = 250 ms.
Solution1. ω = 2π / T = 2π / 0.86 = 7.31 rad s^-1.
2. Since it starts at equilibrium, x = x_0sin(ω t).
3. x = 0.050 × sin(7.31 × 0.25) = 0.048 m (or 4.8 cm).
Note: Ensure your calculator is in RADIANS mode.
Worked Example: Floating Test TubeQuestion: A test tube is lifted up by 2 cm and released. It oscillates with period 0.80 s. Sketch the displacement-time graph for two cycles.
Solution SketchSince the motion starts at max displacement (x = +2 at t = 0), the graph is a cosine curve starting at +2. Period is 0.8 s, so two cycles end at 1.6 s.
Figure 17.2: Test Tube Oscillation GraphA cosine graph for x vs t. Y-axis: +2 to -2 cm. X-axis: Mark 0.4, 0.8, 1.2, 1.6 s.
The Vertical Mass-Spring System
A mass m on a spring with constant k obeys F = -kx. By Newton's Second Law (F=ma):
ma = -kx a = -(k/m)x
Since a = -ω^2 x, then ω^2 = (k/m).
T = 2π√(m/k)Note: The period is independent of gravity (g) and the amplitude.
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