describe the interchange between kinetic and potential energy during simple harmonic motion

Simple Harmonic Motion – Energy Interchange (Cambridge AS & A Level Physics 9702)

Learning Objectives

• State the definition of simple harmonic motion (SHM) and the restoring‑force law.
• Write the displacement, velocity and acceleration equations and derive the angular frequency \(\omega\).
• Obtain the expressions for kinetic energy \((E_k)\), elastic potential energy \((E_p)\) and total mechanical energy \((E)\).
• Interpret energy‑versus‑position and energy‑versus‑time graphs and relate them to the motion.
• Analyse the standard \(x\), \(v\) and \(a\) graphs for SHM (AO2).
• Connect the energy discussion to related topics (waves, damping) and to the required mass‑spring experiment (AO3).

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1. Definition and Restoring‑Force Law

Simple harmonic motion is the oscillation of a system in which the restoring force is directly proportional to the displacement from the equilibrium position and is directed towards that position:

\[ F = -kx \qquad\text{(Hooke’s law)} \] where \(k\) is the force constant (N m\(^{-1}\)) and \(x\) is the instantaneous displacement (m).

2. Equations of Motion

Integrating the force law with Newton’s second law (\(F = ma\)) gives:

\[ m\ddot{x} = -kx \;\Longrightarrow\; \ddot{x} + \omega^{2}x = 0, \qquad\text{with}\qquad \boxed{\;\omega = \sqrt{\frac{k}{m}}\;} \] \[ \text{General solution:}\qquad x(t)=A\cos(\omega t+\phi) \]

• \(A\) – amplitude (m)
• \(\omega\) – angular frequency (rad s\(^{-1}\))
• \(\phi\) – phase constant (rad)

First and second derivatives give the velocity and acceleration:

\[ v(t)=\dot{x}= -A\omega\sin(\omega t+\phi),\qquad a(t)=\ddot{x}= -\omega^{2}x(t). \]

3. Energy in SHM

Instantaneous kinetic and elastic potential energies are

\[ E_k = \frac12 mv^{2}, \qquad E_p = \frac12 kx^{2}. \]

Substituting \(k=m\omega^{2}\) and the expressions for \(x(t)\) and \(v(t)\) gives

\[ E_k = \frac12 mA^{2}\omega^{2}\sin^{2}(\omega t+\phi),\qquad E_p = \frac12 mA^{2}\omega^{2}\cos^{2}(\omega t+\phi). \]

Adding them yields the total mechanical energy, which is constant:

\[ \boxed{E = E_k+E_p = \frac12 m\omega^{2}A^{2}= \frac12 kA^{2}}. \]

Key quantitative facts (AO1)

• Maximum kinetic energy (at equilibrium, \(x=0\)): \[ E_{k,\max}= \frac12 kA^{2},\;E_p=0. \]
• Maximum potential energy (at the extremes, \(x=\pm A\)): \[ E_{p,\max}= \frac12 kA^{2},\;E_k=0. \]
• Equal split of energy (when \(|x|=A/\sqrt2\)): \[ E_k = E_p = \frac14 kA^{2}. \]
• The phase constant \(\phi\) does not affect the energy values because \(\sin^{2}\theta+\cos^{2}\theta=1\).

4. Energy‑versus‑Time and Energy‑versus‑Position Graphs (AO2)

Interpretation:

• Both energies oscillate sinusoidally with the same period \(T=2\pi/\omega\) but are out of phase by \(\pi/2\).
• At any position \(x\) the kinetic energy is the “missing” part of the constant total energy: \(E_k = E - E_p\).
• The parabolic \(E_k\)–\(x\) curve shows that kinetic energy is largest at the centre and falls to zero at the extremes.

5. Standard \(x\), \(v\) and \(a\) Graphs (AO2)

Key points to note when analysing these graphs:

• Maximum \(|x|\) ↔ zero velocity ↔ maximum magnitude of acceleration.
• Zero displacement ↔ maximum speed ↔ zero acceleration.
• The area under the \(v\)–\(t\) curve between two successive zero‑velocity points equals the amplitude \(A\).

6. Qualitative Energy Interchange (AO2)

1. At the extreme positions (\(x=\pm A\)) the mass momentarily stops; all the mechanical energy is stored as elastic potential energy.
2. As the mass moves toward equilibrium the spring releases this stored energy, converting \(E_p\) into \(E_k\); the speed therefore increases.
3. When the mass passes the equilibrium point (\(x=0\)) the spring is neither stretched nor compressed, so \(E_p=0\) and the entire energy is kinetic.
4. Inertia carries the mass beyond equilibrium, stretching/compressing the spring in the opposite sense; kinetic energy is again being reconverted into potential energy.
5. The cycle repeats, producing the characteristic sinusoidal motion.

7. Worked Example (AO3 – Application)

Problem: A 0.50 kg mass is attached to a spring with \(k=200\;\text{N m}^{-1}\). It is pulled 0.10 m from equilibrium and released from rest.

1. Amplitude: \(A = 0.10\;\text{m}\).
2. Angular frequency: \[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.50}} = 20\;\text{rad s}^{-1}.
3. Period: \(T = \dfrac{2\pi}{\omega}= \dfrac{2\pi}{20}=0.314\;\text{s}\).
4. Total mechanical energy: \[ E = \tfrac12kA^{2}= \tfrac12(200)(0.10)^{2}=1.0\;\text{J}. \]
5. Maximum speed (at equilibrium): \[ v_{\max}= \sqrt{\frac{2E}{m}} = \sqrt{\frac{2(1.0)}{0.50}} = 2.0\;\text{m s}^{-1}. \]
6. Energy at \(x=\pm A/\sqrt2\): \[ E_k = E_p = \frac14kA^{2}=0.5\;\text{J},\qquad v = \frac{v_{\max}}{\sqrt2}=1.41\;\text{m s}^{-1}. \]

This quantitative analysis confirms the energy interchange described in the tables and graphs.

8. Summary Table – Energy at Characteristic Positions

Position \(x\)	Velocity \(v\)	Kinetic Energy \(E_k\)	Potential Energy \(E_p\)	Energy State
\(x=0\) (equilibrium)	\(\pm A\omega\) (maximum)	\(\displaystyle\frac12kA^{2}\) (max)	0 (min)	All energy kinetic
\(x=\pm\frac{A}{\sqrt2}\)	\(\pm\frac{A\omega}{\sqrt2}\)	\(\displaystyle\frac14kA^{2}\)	\(\displaystyle\frac14kA^{2}\)	Energy equally split
\(x=\pm A\) (extrema)	0 (min)	0 (min)	\(\displaystyle\frac12kA^{2}\) (max)	All energy potential

9. Links to Other Topics (AO2)

• Waves: The sinusoidal displacement of a point in SHM is the basis for the transverse and longitudinal wave equations. The angular frequency \(\omega\) appears in the wave speed relation \(v = \lambda f = \omega/k\).
• Damping: Real systems lose energy to friction or air resistance. Introducing a damping term \(-b\dot{x}\) modifies the energy balance: \(E\) decreases exponentially, and the amplitude decays.
• Resonance: When a periodic driving force has frequency \(\omega\) equal to the natural frequency \(\sqrt{k/m}\), the system absorbs energy efficiently, leading to large amplitudes.

10. Experimental Investigation (AO3)

The Cambridge syllabus requires a practical “mass‑spring” investigation. The key steps are:

1. Measure the period \(T\) for several different masses \(m\) (keeping the spring unchanged).
2. Plot \(T^{2}\) against \(m\); the slope should be \(\displaystyle\frac{4\pi^{2}}{k}\), allowing determination of the spring constant \(k\).
3. From the measured amplitude \(A\) and calculated \(k\), compute the total mechanical energy \(E=\tfrac12kA^{2}\) and compare it with the kinetic energy obtained from the maximum speed (using a motion sensor or photogate).
4. Discuss sources of error (air resistance, non‑ideal spring, friction at the support) and how they lead to a gradual loss of mechanical energy (damping).

11. Suggested Diagram for the Notes

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These notes cover all required sub‑topics for the Cambridge AS & A Level Physics (9702) specification, provide the necessary derivations, include the required graphs, and link the energy discussion to related concepts and the mandatory experimental investigation.