Simple Harmonic Motion (S.H.M)

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Simple Harmonic Motion (S.H.M) & Resonance

Objectives:

1. Establish the relationship between period and frequency.
2. Analyse the energy changes occurring during S.H.M.
3. Identify different types of forced vibration.
4. Enumerate applications of resonance.

Key Concepts and Formulas

Definition of S.H.M: Motion in which the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.

Angular Frequency (ω): ω = 2πf, where f is the frequency (in Hz).

Period (T) and Frequency (f): T = 1/f; T is the time for one complete cycle.

Amplitude (A): The maximum displacement from the equilibrium position.

Displacement Equation: x = A cos(ωt + φ), where φ is the phase constant.

Velocity in S.H.M: v = -ωA sin(ωt + φ).

Acceleration in S.H.M: a = -ω²A cos(ωt + φ).

Energy in S.H.M: Total energy E = ½kA² = ½mω²A², where k is the spring constant and m is the mass.

Force Vibration & Resonance: Resonance occurs when the frequency of an external force matches the natural frequency of the system, resulting in maximum amplitude.

Worked Examples

The following 6 worked examples (displayed in green boxes) illustrate the concepts and formulas of S.H.M and resonance.

Example 1:
A mass-spring system oscillates such that the mass moves through 6.28 radians in 2 s. Find its frequency and angular frequency.
Solution: f = 1/T = 1/(2 s) = 0.5 Hz; ω = 2πf = 2π×0.5 = π rad/s.

Example 2:
A pendulum of length 1 m oscillates in S.H.M. (approximation). If g = 9.8 m/s², find its period.
Solution: T = 2π√(L/g) = 2π√(1/9.8) ≈ 2.01 s; f = 1/T ≈ 0.50 Hz.

Example 3:
A 5 kg object attached to a spring oscillates with an amplitude of 0.2 m and angular frequency 4 rad/s. Calculate its maximum speed and acceleration.
Solution: v_max = ωA = 4×0.2 = 0.8 m/s; a_max = ω²A = 16×0.2 = 3.2 m/s².

Example 4:
In a mass-spring system, if the amplitude is 0.1 m and the spring constant is 200 N/m, find the total energy assuming m = 2 kg.
Solution: ω = √(k/m) = √(200/2) = √100 = 10 rad/s; E = ½mω²A² = ½×2×(10²)×(0.1²) = 1 J.

Example 5:
A system is forced by an external periodic force. When the forcing frequency equals the natural frequency, resonance occurs. Explain briefly its effect on amplitude.
Solution: At resonance, the amplitude reaches a maximum because the external force adds energy efficiently. (No numeric answer.)

Example 6:
A 1500 kg car on a curved road (radius 50 m) moves at 20 m/s. Calculate the centripetal (radial) acceleration and force.
Solution: a_c = v²/r = 400/50 = 8 m/s²; F_c = ma_c = 1500×8 = 12000 N.

JAMB CBT Quiz on Simple Harmonic Motion & Resonance

Total time: 900 seconds

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This lesson covers: Definition: S.H.M is a periodic motion where the restoring force is proportional to displacement. Angular velocity: ω = 2πf Period and Frequency: T = 1/f Amplitude: Maximum displacement (A) Velocity: v = -ωA sin(ωt + φ) Acceleration: a = -ω²A cos(ωt + φ) Energy: E = ½mω²A² = ½kA² Forced vibration and resonance: Maximum amplitude occurs when forcing frequency equals natural frequency. The worked examples illustrate numerical problems and real-life applications involving period, frequency, energy changes, and resonance in S.H.M.