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Wave Behaviour - Simple Harmonic Motion

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Simple Harmonic Motion (SHM) is defined as periodic motion where the acceleration aa of the system is directly proportional to its displacement xx from a fixed equilibrium position and is always directed towards that position: a∝−xa \propto -x.

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The defining equation for SHM is a=−ω2xa = -\omega^2 x, where ω\omega is the angular frequency measured in rad s−1\text{rad s}^{-1}.

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SHM is isochronous, meaning the period TT of the oscillation is independent of the amplitude x0x_0.

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Displacement, velocity, and acceleration vary sinusoidally. Velocity is the gradient of the displacement-time graph, and acceleration is the gradient of the velocity-time graph.

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Energy in SHM: In the absence of damping, total energy EtotalE_{total} is conserved. It oscillates between kinetic energy EkE_k (maximum at equilibrium) and potential energy EpE_p (maximum at maximum displacement).

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Phase difference ϕ\phi measures the fraction of a cycle by which one oscillation moves relative to another. One full cycle corresponds to 2π2\pi radians.

📐Formulae

a=−ω2xa = -\omega^2 x

ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f

x=x0sin⁡(ωt) or x=x0cos⁡(ωt)x = x_0 \sin(\omega t) \text{ or } x = x_0 \cos(\omega t)

v=v0cos⁡(ωt)=±ωx02−x2v = v_0 \cos(\omega t) = \pm \omega \sqrt{x_0^2 - x^2}

vmax=ωx0v_{max} = \omega x_0

amax=ω2x0a_{max} = \omega^2 x_0

ET=12mω2x02E_T = \frac{1}{2} m \omega^2 x_0^2

T=2πlg (Simple Pendulum)T = 2\pi \sqrt{\frac{l}{g}} \text{ (Simple Pendulum)}

T=2πmk (Mass-Spring System)T = 2\pi \sqrt{\frac{m}{k}} \text{ (Mass-Spring System)}

💡Examples

Problem 1:

A mass of 0.20 kg0.20 \text{ kg} is attached to a spring and undergoes SHM with an amplitude of 0.05 m0.05 \text{ m} and a period of 0.40 s0.40 \text{ s}. Calculate the maximum restoring force acting on the mass.

Solution:

  1. Find angular frequency: ω=2πT=2π0.40=5π rad s−1\omega = \frac{2\pi}{T} = \frac{2\pi}{0.40} = 5\pi \text{ rad s}^{-1}.
  2. Calculate maximum acceleration: amax=ω2x0=(5π)2×0.05≈12.34 m s−2a_{max} = \omega^2 x_0 = (5\pi)^2 \times 0.05 \approx 12.34 \text{ m s}^{-2}.
  3. Calculate maximum force using F=maF = ma: Fmax=0.20×12.34=2.47 NF_{max} = 0.20 \times 12.34 = 2.47 \text{ N}.

Explanation:

The maximum restoring force occurs at the maximum displacement (amplitude), where acceleration is at its peak.

Problem 2:

A particle executes SHM with an amplitude x0x_0. At what displacement xx is the kinetic energy of the particle equal to its potential energy?

Solution:

Ek=EpE_k = E_p Since ET=Ek+EpE_T = E_k + E_p, then ET=2EpE_T = 2 E_p. 12mω2x02=2(12mω2x2)\frac{1}{2} m \omega^2 x_0^2 = 2 (\frac{1}{2} m \omega^2 x^2) x02=2x2x_0^2 = 2x^2 x=x02≈0.707x0x = \frac{x_0}{\sqrt{2}} \approx 0.707 x_0

Explanation:

Energy is shared equally between kinetic and potential forms when the displacement is approximately 71%71\% of the amplitude.

Simple Harmonic Motion - Revision Notes & Key Formulas | IB Grade 11 Physics