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Wave Behaviour - Simple Harmonic Motion

Grade 12IBPhysics

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 is proportional to the displacement xx from the equilibrium position and is directed towards that position (a∝−xa \propto -x).

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The angular frequency ω\omega is related to the period TT and frequency ff by the relations ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f.

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

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Displacement, velocity, and acceleration follow sinusoidal patterns. Velocity vv leads displacement xx by a phase difference of π2\frac{\pi}{2} radians, and acceleration aa leads displacement xx by π\pi radians.

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The total energy ETE_T in SHM is the sum of kinetic energy EkE_k and potential energy EpE_p, and remains constant if no dissipative forces (damping) are present: ET=Ek+EpE_T = E_k + E_p.

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Maximum velocity vmaxv_{max} occurs at the equilibrium position (x=0x = 0), while maximum acceleration amaxa_{max} occurs at maximum displacement (x=±x0x = \pm x_0).

📐Formulae

a=−ω2xa = -\omega^2 x

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

v=v0cos⁡(ωt)=ωx0cos⁡(ωt)v = v_0 \cos(\omega t) = \omega x_0 \cos(\omega t)

v=±ωx02−x2v = \pm \omega \sqrt{x_0^2 - x^2}

T=2πωT = \frac{2\pi}{\omega}

Ek=12mω2(x02−x2)E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2)

Ep=12mω2x2E_p = \frac{1}{2} m \omega^2 x^2

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

💡Examples

Problem 1:

A mass of 0.20 kg0.20 \text{ kg} is attached to a spring and undergoes SHM with an amplitude of 0.15 m0.15 \text{ m} and a period of 2.0 s2.0 \text{ s}. Calculate the maximum restoration force acting on the mass.

Solution:

First, calculate the angular frequency: ω=2πT=2π2.0=π rad s−1\omega = \frac{2\pi}{T} = \frac{2\pi}{2.0} = \pi \text{ rad s}^{-1}. The maximum acceleration is amax=ω2x0=(π)2×0.15≈1.48 m s−2a_{max} = \omega^2 x_0 = (\pi)^2 \times 0.15 \approx 1.48 \text{ m s}^{-2}. Using Newton's Second Law, Fmax=mamax=0.20×1.48=0.296 NF_{max} = m a_{max} = 0.20 \times 1.48 = 0.296 \text{ N}.

Explanation:

The maximum force in SHM occurs at maximum displacement, where acceleration is also at its peak. We relate the period to angular frequency and then apply the defining SHM acceleration formula.

Problem 2:

An object oscillates with SHM. When its displacement is 0.04 m0.04 \text{ m}, its velocity is 0.12 m s−10.12 \text{ m s}^{-1}. If the maximum velocity is 0.15 m s−10.15 \text{ m s}^{-1}, find the amplitude x0x_0 of the motion.

Solution:

Using the relation v2=ω2(x02−x2)v^2 = \omega^2 (x_0^2 - x^2) and knowing vmax=ωx0  ⟹  ω=vmaxx0v_{max} = \omega x_0 \implies \omega = \frac{v_{max}}{x_0}. Substitute ω\omega: v2=(vmaxx0)2(x02−x2)v^2 = (\frac{v_{max}}{x_0})^2 (x_0^2 - x^2). Rearranging: (vvmax)2=1−(xx0)2(\frac{v}{v_{max}})^2 = 1 - (\frac{x}{x_0})^2. Plugging in values: (0.120.15)2=1−(0.04x0)2  ⟹  0.64=1−0.0016x02  ⟹  0.0016x02=0.36  ⟹  x0=0.00160.36≈0.0667 m(\frac{0.12}{0.15})^2 = 1 - (\frac{0.04}{x_0})^2 \implies 0.64 = 1 - \frac{0.0016}{x_0^2} \implies \frac{0.0016}{x_0^2} = 0.36 \implies x_0 = \sqrt{\frac{0.0016}{0.36}} \approx 0.0667 \text{ m}.

Explanation:

This problem uses the velocity-displacement relationship. By substituting the expression for ω\omega derived from the maximum velocity, we can solve for the unknown amplitude.

Simple Harmonic Motion - Revision Notes & Key Formulas | IB Grade 12 Physics