Simple Harmonic Motion

Specification

info

Standard level and higher level (3 hours): Students should understand:

• Conditions that lead to simple harmonic motion
• The defining equation of simple harmonic motion: $a = -\omega^2 x$
• A particle undergoing SHM can be described using time period $T$, frequency $f$, angular frequency $\omega$, amplitude, equilibrium position, and displacement
• The time period in terms of frequency and angular frequency: $T = \frac{1}{f} = \frac{2\pi}{\omega}$
• The time period of a mass-spring system: $T = 2\pi \sqrt{\frac{m}{k}}$
• The time period of a simple pendulum: $T = 2\pi \sqrt{\frac{L}{g}}$
• A qualitative approach to energy changes during one cycle of an oscillation

Additional higher level (4 hours): Students should understand:

• That a particle undergoing SHM can be described using phase angle
• That problems can be solved using the equations for SHM:

$$x = x_0 \sin(\omega t + \Phi),\quad v = \omega x_0 \cos(\omega t + \Phi),\quad v = \pm\omega\sqrt{x_0^2 - x^2},\quad E_T = \tfrac{1}{2}m\omega^2 x_0^2,\quad E_P = \tfrac{1}{2}m\omega^2 x^2$$

Throughout this file, amplitude is $A$ and initial phase is $\phi_0$, corresponding to the data booklet notation $x_0$ and $\Phi$.

Fundamental Principles

Simple Harmonic Motion (SHM) is a periodic oscillation about a stable equilibrium position, characterized by a restoring force $F$ directly proportional to the displacement $x$ from equilibrium and directed oppositely to the displacement. This yields Newton's second law:

\begin`\{aligned}` F \propto -x \\ F_{\mathrm{net}} = -kx = m \frac{d^2x}{dt^2}, \end`\{aligned}`

where $k > 0$ is the stiffness constant (e.g., spring constant). Rearranged as the equation of motion:

$$\frac{d^2x}{dt^2} + \omega^2 x = 0, \quad \omega = \sqrt{\frac{k}{m}}. \tag{1}$$

Here, $\omega$ is the angular frequency (rad s$^{-1}$), governing the system's temporal evolution.

Key Characteristics:

• Equilibrium Position: Point where net force vanishes ($F_{\mathrm{net}} = 0$).
• Amplitude ($A$): Maximum displacement from equilibrium ($|x|_{\mathrm{max}} = A$).
• Isochrony: Period $T$ is amplitude-independent for ideal SHM.

Conditions for Ideal SHM:

1. Restoring force obeys Hooke's law: $F = -kx$.
2. Zero dissipative forces (undamped motion).
3. Constant total mechanical energy.

Kinematic Relations

The general solution to Equation (1) is:

$$x(t) = A \cos(\omega t + \phi_0), \tag{2}$$

where $\phi_0$ is the initial phase angle. Velocity $v$ and acceleration $a$ follow by differentiation:

v(t) = \frac`\{dx}``\{dt}` = -\omega A \sin(\omega t + \phi_0), \tag{3} $$a(t) = \frac{d^2x}{dt^2} = -\omega^2 A \cos(\omega t + \phi_0) = -\omega^2 x. \tag{4}$$

Phase Relationships:

• Displacement-Velocity: $v = \pm \omega \sqrt{A^2 - x^2}$ (from energy conservation).
• Displacement-Acceleration: $a = -\omega^2 x$ (definitive property of SHM).
• Extrema:
  • $|v|_{\mathrm{max}} = \omega A$ at $x = 0$ (equilibrium).
  • $|a|_{\mathrm{max}} = \omega^2 A$ at $x = \pm A$ (max displacement).

Graphical Interpretation:

• $x(t)$, $v(t)$, and $a(t)$ are phase-shifted sinusoids.
• $a(t)$ is inverted relative to $x(t)$ due to $a \propto -x$.

Energy Conservation

Total mechanical energy $E_{\mathrm{total}}$ is conserved:

$$E_{\mathrm{total}} = K + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2. \tag{5}$$

Substituting Equations (2)--(4) yields: Kinetic Energy ($K$):

$$K = \frac{1}{2}m \omega^2 A^2 \sin^2(\omega t + \phi_0) = \frac{1}{2}m\omega^2 (A^2 - x^2). \tag{6}$$

Potential Energy ($U$):

$$U = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2. \tag{7}$$

Total Energy:

$$E_{\mathrm{total}} = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2. \tag{8}$$

Energy Oscillations:

• $K_{\mathrm{max}} = E_{\mathrm{total}}$ at $x = 0$.
• $U_{\mathrm{max}} = E_{\mathrm{total}}$ at $x = \pm A$.

Example Systems

Simple Pendulum

Description: Point mass $m$ suspended on a massless string of length $L$ in gravitational field $g$. Equation of Motion: For small $\theta$ ($\sin\theta \approx \theta$):

$$\frac{d^2\theta}{dt^2} + \frac{g}{L}\theta = 0. \tag{9}$$

This matches Equation (1) with $\omega = \sqrt{g/L}$. Period:

$$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L}{g}}. \tag{10}$$

Properties:

• $T \propto \sqrt{L}$; $T \propto 1/\sqrt{g}$; independent of $m$ and $A$ (for $\theta \ll 1$ rad).

Mass-Spring System

Masses and Springs

Explore how mass, spring stiffness, and damping affect oscillations. Hang different masses from springs and observe how the period and amplitude change in real time.

Description: Mass $m$ attached to a spring of stiffness $k$. Equation of Motion: From Hooke's law:

$$m\frac{d^2x}{dt^2} = -kx \implies \frac{d^2x}{dt^2} + \frac{k}{m}x = 0. \tag{11}$$

Period:

$$T = 2\pi \sqrt{\frac{m}{k}}. \tag{12}$$

Properties:

• $T \propto \sqrt{m}$; $T \propto 1/\sqrt{k}$; independent of $A$.

Angular Frequency and Phase

Angular Frequency ($\omega$):

$$\omega = 2\pi f = \frac{2\pi}{T}, \tag{13}$$

where $f$ is linear frequency (Hz). Converts temporal periodicity to angular speed.

Phase Angle ($\phi$): Generalizes Equation (2):

$$x(t) = A \cos(\omega t + \phi_0).$$

• Phase Difference ($\Delta\phi$): Temporal shift between two SHMs: $$\Delta\phi = \omega \Delta t = \frac{2\pi \Delta t}{T}. \tag{14}$$
• Measured in radians (1 rad $\approx$ 57.3°).

Summary of Key Equations

Quantity	Expression
Displacement	$x = A \cos(\omega t + \phi_0)$
Velocity	$v = -\omega A \sin(\omega t + \phi_0)$
Acceleration	$a = -\omega^2 x$
Angular Frequency	$\omega = \sqrt{k/m}$ (spring), $\omega = \sqrt{g/L}$ (pendulum)
Period	$T = 2\pi / \omega$
Kinetic Energy	$K = \frac{1}{2}m\omega^2(A^2 - x^2)$
Potential Energy	$U = \frac{1}{2}m\omega^2 x^2$
Total Energy	$E_{\mathrm{total}} = \frac{1}{2}m\omega^2 A^2$

Derivation of the SHM Solution

Verification by Direct Substitution

Claim: $x(t) = A\cos(\omega t + \phi_0)$ satisfies $\frac{d^2x}{dt^2} + \omega^2 x = 0$ for any constants $A$, $\omega$, and $\phi_0$.

Proof. First derivative:

\frac`\{dx}``\{dt}` = -\omega A \sin(\omega t + \phi_0)

Second derivative:

$$\frac{d^2x}{dt^2} = -\omega^2 A \cos(\omega t + \phi_0) = -\omega^2 x$$

Substituting into the equation of motion:

$$\frac{d^2x}{dt^2} + \omega^2 x = -\omega^2 x + \omega^2 x = 0 \quad \blacksquare$$

Sine Form Equivalence

Claim: $x(t) = A\sin(\omega t + \phi_0)$ is equally valid as a general solution.

Using the identity $\sin\theta = \cos\!\left(\theta - \frac{\pi}{2}\right)$:

$$A\sin(\omega t + \phi_0) = A\cos\!\left(\omega t + \phi_0 - \frac{\pi}{2}\right)$$

This is the cosine form with a shifted phase. Since $\phi_0$ is already an arbitrary constant, absorbing a constant offset of $-\pi/2$ does not reduce generality. Both forms span the full two-parameter solution space $(A, \phi_0)$.

Direct substitution confirms:

$$\frac{d^2}{dt^2}\!\left[A\sin(\omega t + \phi_0)\right] = -\omega^2 A\sin(\omega t + \phi_0) = -\omega^2 x \quad \blacksquare$$

Choosing Sine vs Cosine

The two forms are physically equivalent; the choice is a matter of convenience based on initial conditions.

Initial condition	Preferred form	Rationale
Released from $x = A$ with $v = 0$	$x = A\cos(\omega t)$	$\cos(0) = 1$, $\sin(0) = 0$
Released from $x = 0$ with $v = v_{\max}$	$x = A\sin(\omega t)$	$\sin(0) = 0$, $\cos(0) = 1$
General initial state	Either with appropriate $\phi_0$	Phase angle absorbs the offset

The IB data booklet uses the sine convention ($x = x_0 \sin(\omega t + \Phi)$). Both conventions are correct. Pick one and remain consistent within a single problem. Switching conventions mid-problem introduces a systematic phase error of $\pm\pi/2$.

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Worked Examples — SL Level

Example 1: Period and Frequency of a Mass-Spring System

A spring of stiffness $k = 200\mathrm{ N/m}$ has a $0.50\mathrm{ kg}$ mass attached. Find the period $T$ and frequency $f$.

$$T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.50}{200}} = 2\pi(0.0500) = 0.314\mathrm{ s}$$ $$f = \frac{1}{T} = \frac{1}{0.314} = 3.18\mathrm{ Hz}$$

Example 2: Displacement and Velocity at a Given Time

A mass-spring system has amplitude $A = 0.10\mathrm{ m}$ and period $T = 0.50\mathrm{ s}$. The mass is released from maximum displacement at $t = 0$. Find $x$ and $v$ at $t = 0.125\mathrm{ s}$.

Angular frequency:

$$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.50} = 4\pi\mathrm{ rad/s}$$

With $x(0) = A$, use the cosine form: $x = A\cos(\omega t)$.

$$x(0.125) = 0.10\cos(4\pi \times 0.125) = 0.10\cos\!\left(\frac{\pi}{2}\right) = 0\mathrm{ m}$$ $$v(0.125) = -\omega A \sin(\omega t) = -4\pi(0.10)\sin\!\left(\frac{\pi}{2}\right) = -1.26\mathrm{ m/s}$$

The mass passes through equilibrium at $t = T/4 = 0.125\mathrm{ s}$, moving in the negative direction with speed $|v| = \omega A = 1.26\mathrm{ m/s}$.

Example 3: Amplitude from Energy

A $0.30\mathrm{ kg}$ mass oscillates on a spring with $k = 120\mathrm{ N/m}$. At equilibrium, the speed is $1.6\mathrm{ m/s}$. Determine the amplitude.

At $x = 0$, all energy is kinetic:

$$E_{\mathrm{total}} = \frac{1}{2}mv_{\max}^2 = \frac{1}{2}(0.30)(1.6)^2 = 0.384\mathrm{ J}$$

At $x = A$, all energy is potential:

$$E_{\mathrm{total}} = \frac{1}{2}kA^2 \implies A = \sqrt{\frac{2E_{\mathrm{total}}}{k}} = \sqrt{\frac{2(0.384)}{120}} = \sqrt{0.00640} = 0.0800\mathrm{ m}$$

Example 4: Simple Pendulum Period

A simple pendulum on Earth has a period of $2.00\mathrm{ s}$. Determine its length. Use $g = 9.81\mathrm{ m/s}^2$.

$$L = \frac{gT^2}{4\pi^2} = \frac{(9.81)(2.00)^2}{4\pi^2} = \frac{39.2}{39.5} = 0.993\mathrm{ m}$$

This is close to $1.00\mathrm{ m}$, which is why a "seconds pendulum" ($T = 2\mathrm{ s}$) is approximately $1\mathrm{ m}$ long on Earth.

Example 5: Phase Difference Between Two Oscillators

Two identical mass-spring systems oscillate with the same amplitude and frequency. System A is released from maximum displacement at $t = 0$. System B is released from equilibrium, moving in the positive direction, at $t = 0$. Find the phase difference $\Delta\phi$.

System A (cosine form): $x_A = A\cos(\omega t)$

System B (starts at $x = 0$ with positive velocity, sine form): $x_B = A\sin(\omega t) = A\cos\!\left(\omega t - \frac{\pi}{2}\right)$

Comparing phases: $\phi_A = 0$, $\phi_B = -\pi/2$.

|\Delta\phi| = \left|`0 - \left(-\frac{\pi}{2}\right)\right`| = \frac{\pi}{2}\mathrm{ rad} = 90^{\circ}

System A leads System B by $90^\circ$.

Example 6: Energy Partition at a Given Displacement

A mass-spring system has $A = 0.15\mathrm{ m}$, $k = 80.0\mathrm{ N/m}$, and $m = 0.20\mathrm{ kg}$. Find the kinetic energy, potential energy, and speed at $x = 0.090\mathrm{ m}$.

$$E_{\mathrm{total}} = \frac{1}{2}kA^2 = \frac{1}{2}(80.0)(0.15)^2 = 0.900\mathrm{ J}$$ $$U = \frac{1}{2}kx^2 = \frac{1}{2}(80.0)(0.090)^2 = 0.324\mathrm{ J}$$ $$K = E_{\mathrm{total}} - U = 0.900 - 0.324 = 0.576\mathrm{ J}$$ $$v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(0.576)}{0.20}} = \sqrt{5.76} = 2.40\mathrm{ m/s}$$

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HL Extension: Phase and Advanced Problems

Determining Phase Angle from Initial Conditions

Given initial displacement $x(0) = x_0$ and initial velocity $v(0) = v_0$, we determine both the amplitude $A$ and the phase angle $\phi_0$.

Using the cosine form $x = A\cos(\omega t + \phi_0)$:

$$x(0) = A\cos\phi_0 = x_0 \tag{i}$$ v(0) = -\omega A\sin\phi_0 = v_0 \tag`\{ii}`

Squaring and adding (i) and (ii):

$$A^2\cos^2\phi_0 + A^2\sin^2\phi_0 = x_0^2 + \frac{v_0^2}{\omega^2}$$ $$A = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}} \tag{15}$$

Dividing (ii) by (i):

$$\tan\phi_0 = -\frac{v_0}{\omega x_0} \tag{16}$$

warning

Equation (16) alone is ambiguous: $\arctan$ returns values in $(-\pi/2,\, \pi/2)$, but the actual phase could be in any quadrant. Always verify the signs of $\cos\phi_0 = x_0/A$ and $\sin\phi_0 = -v_0/(\omega A)$ to resolve the correct quadrant.

HL Example 1: Phase Angle Determination

A $0.40\mathrm{ kg}$ mass on a spring with $k = 160\mathrm{ N/m}$ has initial conditions $x(0) = 0.060\mathrm{ m}$ and $v(0) = 1.20\mathrm{ m/s}$ (positive). Find the amplitude and phase angle.

$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{160}{0.40}} = 20.0\mathrm{ rad/s}$$ $$A = \sqrt{(0.060)^2 + \frac{(1.20)^2}{(20.0)^2}} = \sqrt{0.00360 + 0.00360} = \sqrt{0.00720} = 0.0849\mathrm{ m}$$ $$\tan\phi_0 = -\frac{v_0}{\omega x_0} = -\frac{1.20}{(20.0)(0.060)} = -1.00$$

So $\phi_0 = -\pi/4$ or $\phi_0 = 3\pi/4$. Check quadrant: $\cos\phi_0 = x_0/A = 0.707 > 0$ (first or fourth quadrant). $\sin\phi_0 = -v_0/(\omega A) = -0.707 < 0$ (third or fourth quadrant). Both conditions point to the fourth quadrant: $\phi_0 = -\pi/4 = -0.785\mathrm{ rad}$.

HL Example 2: Velocity at Given Displacement

A pendulum oscillates with amplitude $A = 0.120\mathrm{ m}$ and angular frequency $\omega = 3.50\mathrm{ rad/s}$. Find the speed at $x = 0.050\mathrm{ m}$.

$$v = \pm\omega\sqrt{A^2 - x^2} = \pm 3.50\sqrt{(0.120)^2 - (0.050)^2} = \pm 3.50\sqrt{0.0119} = \pm 0.382\mathrm{ m/s}$$

The $\pm$ reflects that the mass passes through $x = 0.050\mathrm{ m}$ twice per cycle: once toward equilibrium and once away.

HL Example 3: Energy at Arbitrary Displacement

A $0.50\mathrm{ kg}$ mass on a spring ($k = 200\mathrm{ N/m}$) oscillates with amplitude $0.10\mathrm{ m}$. Find the kinetic energy, potential energy, and speed at $x = 0.060\mathrm{ m}$.

$$E_T = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}(200)(0.10)^2 = 1.00\mathrm{ J}$$ $$E_P = \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}(200)(0.060)^2 = 0.360\mathrm{ J}$$ $$K = E_T - E_P = 1.00 - 0.360 = 0.640\mathrm{ J}$$ $$v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(0.640)}{0.50}} = 1.60\mathrm{ m/s}$$

Verification: $\omega = \sqrt{k/m} = 20.0\mathrm{ rad/s}$, $v = 20.0\sqrt{(0.10)^2 - (0.060)^2} = 1.60\mathrm{ m/s}$ $\checkmark$

HL Example 4: Period, Amplitude, and Total Energy

A $0.50\mathrm{ kg}$ object undergoes SHM with total energy $2.0\mathrm{ J}$ and amplitude $0.20\mathrm{ m}$. Determine $\omega$, $k$, and the speed at $x = 0.10\mathrm{ m}$.

$$E_T = \frac{1}{2}m\omega^2 A^2 \implies \omega^2 = \frac{2E_T}{mA^2} = \frac{2(2.0)}{(0.50)(0.20)^2} = 200$$ $$\omega = \sqrt{200} = 14.1\mathrm{ rad/s}, \quad k = m\omega^2 = (0.50)(200) = 100\mathrm{ N/m}$$ $$v = \omega\sqrt{A^2 - x^2} = 14.1\sqrt{(0.20)^2 - (0.10)^2} = 14.1(0.173) = 2.44\mathrm{ m/s}$$

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Damping — Qualitative Treatment

Real oscillators lose energy to their surroundings. The rate of energy loss determines the damping regime. IB students may be asked to sketch displacement-time graphs for each regime.

Light Damping (Underdamping)

The amplitude decays exponentially: $A(t) = A_0 e^{-bt}$, where $b$ is the damping coefficient.

• The system oscillates with gradually decreasing amplitude.
• The period is slightly longer than the undamped period: $T_{\mathrm{damped}} \gt T_0$.
• Energy is dissipated each cycle, primarily as heat via friction or air resistance.
• The quality factor $Q = 2\pi \times \frac{\mathrm{energy stored}}{\mathrm{energy lost per cycle}}$ quantifies how underdamped the system is. High $Q$ means low energy loss per cycle.

Critical Damping

The damping force is just sufficient to prevent oscillation. The system returns to equilibrium in the shortest possible time without overshooting.

• Displacement decays as $x(t) = (c_1 + c_2 t)e^{-\alpha t}$ for constants $c_1$, $c_2$, $\alpha$.
• No oscillation occurs.
• Applications: car shock absorbers, door closers, instrument mechanisms.

Heavy Damping (Overdamping)

The damping force exceeds the critical value. The system returns to equilibrium more slowly than critical damping.

• No oscillation.
• The return to equilibrium is sluggish.
• Example: a pendulum immersed in viscous oil.

Comparison of Damping Regimes

Regime	Oscillation	Return speed	Period vs undamped
Light	Yes (decaying $A$)	Moderate	Slightly longer
Critical	No	Fastest	N/A
Heavy	No	Slowest	N/A

Energy Dissipation

In all damped systems, mechanical energy converts to thermal energy through resistive forces. The rate of energy loss is proportional to $v^2$. For light damping, $E(t) \approx E_0 e^{-2bt}$; since $E \propto A^2$, the amplitude decays as $A(t) \propto e^{-bt}$.

Phase Space (Qualitative)

An undamped SHM traces a closed ellipse in the phase plane ($x$ vs $v$). Under light damping the ellipse spirals inward. Under critical or heavy damping the trajectory converges directly without loops.

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Common Pitfalls

1. Confusing $\omega$ and $f$

$\omega = 2\pi f$. The angular frequency $\omega$ (rad/s) is $2\pi$ times the linear frequency $f$ (Hz). When a question asks for "frequency," it means $f$, not $\omega$. When computing $v_{\max} = \omega A$, you need $\omega$.

2. Pendulum Period is Independent of Mass

$T = 2\pi\sqrt{L/g}$ contains no $m$. A heavier bob does not swing faster or slower (for small angles). Contrast this with the mass-spring system, where $T \propto \sqrt{m}$. The physical reason is that gravitational force ($mg$) and inertia ($m$) both scale with mass, cancelling out.

3. Sign Errors in the Restoring Force

The defining characteristic of SHM is $F = -kx$ or $a = -\omega^2 x$. The negative sign guarantees the force points toward equilibrium. Dropping it gives exponential growth, not oscillation. Always verify the restoring force direction opposes displacement.

4. Mixing Up Phase Angle Conventions

The cosine form $x = A\cos(\omega t + \phi_0)$ and sine form $x = A\sin(\omega t + \Phi)$ yield different phase constants for the same physical state. If $x(0) = 0$ using the cosine form, then $\phi_0 = \pm\pi/2$, not $0$.

5. Small-Angle Approximation Validity

$T = 2\pi\sqrt{L/g}$ relies on $\sin\theta \approx \theta$, accurate to within 1% for $\theta \lt 10^\circ$ and 0.1% for $\theta \lt 5^\circ$. Beyond approximately $15^\circ$, the true period exceeds the prediction and the motion is no longer strictly SHM.

6. Misidentifying Equilibrium Position

The equilibrium position is where the net force is zero, not where the spring is at its natural length. For a vertical spring-mass system, equilibrium is where the spring is stretched by $mg/k$ below its natural length. SHM occurs about this equilibrium, not the natural length.

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Problem Set

Problem 1 (SL)

A $0.25\mathrm{ kg}$ mass is attached to a spring with $k = 400\mathrm{ N/m}$. Calculate: (a) the period of oscillation, (b) the frequency, (c) the angular frequency.

Solution

(a)

$$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.25}{400}} = 2\pi(0.0250) = 0.157\mathrm{ s}$$

(b)

$$T_{\mathrm{Earth}} = 2\pi\sqrt{\frac{1.50}{9.81}} = 2.46\mathrm{ s}$$

On Mars:

$$T_{\mathrm{Mars}} = 2\pi\sqrt{\frac{1.50}{3.71}} = 4.00\mathrm{ s}$$

The period is longer on Mars due to the weaker gravitational field.

Problem 3 (SL)

A mass-spring system has amplitude $0.080\mathrm{ m}$ and spring constant $k = 500\mathrm{ N/m}$ with mass $0.50\mathrm{ kg}$. Find the kinetic energy and potential energy when the displacement is $0.040\mathrm{ m}$.

Solution

$$E_T = \frac{1}{2}kA^2 = \frac{1}{2}(500)(0.080)^2 = 1.60\mathrm{ J}$$$$U = \frac{1}{2}kx^2 = \frac{1}{2}(500)(0.040)^2 = 0.400\mathrm{ J}$$$$K = E_T - U = 1.60 - 0.400 = 1.20\mathrm{ J}$$

At half the amplitude, the split is 75% kinetic, 25% potential (energy scales as $x^2$).

Problem 4 (SL)

Two identical pendulums are released simultaneously. Pendulum P is released from an angle of $5^\circ$ and pendulum Q from an angle of $10^\circ$. Both are within the small-angle regime. Compare their periods.

Solution

For ideal SHM, the period is independent of amplitude:

$$T = 2\pi\sqrt{\frac{L}{g}}$$

Both pendulums have the same length and gravitational field, so $T_P = T_Q$. This is the property of isochrony. In practice, Q's period is very slightly longer because the small-angle approximation is less accurate at $10^\circ$, but this difference is negligible at the IB level.

Problem 5 (SL)

A $2.0\mathrm{ kg}$ object on a spring oscillates with amplitude $0.30\mathrm{ m}$. At $x = 0.20\mathrm{ m}$, the speed is $2.0\mathrm{ m/s}$. Find the total energy and the spring constant.

Solution

From $v^2 = \omega^2(A^2 - x^2) = \frac{k}{m}(A^2 - x^2)$:

$$(2.0)^2 = \frac{k}{2.0}\!\left((0.30)^2 - (0.20)^2\right) = \frac{k}{2.0}(0.050)$$$$4.0 = 0.025k \implies k = 160\mathrm{ N/m}$$

Total energy:

$$E_T = \frac{1}{2}kA^2 = \frac{1}{2}(160)(0.30)^2 = 7.2\mathrm{ J}$$

Problem 6 (HL)

A particle undergoes SHM with $\omega = 8.0\mathrm{ rad/s}$. At $t = 0$, $x = 0.030\mathrm{ m}$ and $v = -0.20\mathrm{ m/s}$. Determine the amplitude, the phase angle (cosine form), and the displacement at $t = 0.50\mathrm{ s}$.

Solution

Amplitude:

$$A = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}} = \sqrt{(0.030)^2 + \frac{(0.20)^2}{(8.0)^2}} = \sqrt{0.000900 + 0.000625} = \sqrt{0.001525} = 0.0391\mathrm{ m}$$

Phase angle:

$$\tan\phi_0 = -\frac{v_0}{\omega x_0} = -\frac{-0.20}{(8.0)(0.030)} = 0.833$$

$\phi_0 = \arctan(0.833) = 0.694\mathrm{ rad}$. Quadrant check: $\cos\phi_0 = x_0/A = 0.767 > 0$ and $\sin\phi_0 = -v_0/(\omega A) = 0.639 > 0$. Both positive, first quadrant confirmed.

Displacement at $t = 0.50\mathrm{ s}$:

$$x = 0.0391\cos(8.0 \times 0.50 + 0.694) = 0.0391\cos(4.694) = 0.0391(-0.0296) = -0.00116\mathrm{ m}$$

Problem 7 (HL)

A $0.60\mathrm{ kg}$ mass on a spring has total energy $0.48\mathrm{ J}$ and amplitude $0.040\mathrm{ m}$. (a) Find the spring constant. (b) Find the maximum speed. (c) Find the speed when $x = 0.020\mathrm{ m}$. (d) At what displacement is $K = U$?

Solution

(a)

$$k = \frac{2E_T}{A^2} = \frac{2(0.48)}{(0.040)^2} = 600\mathrm{ N/m}$$

(b)

$$v_{\max} = \omega A = A\sqrt{\frac{k}{m}} = 0.040\sqrt{\frac{600}{0.60}} = 0.040(31.6) = 1.26\mathrm{ m/s}$$

(c)

$$v = \omega\sqrt{A^2 - x^2} = 31.6\sqrt{(0.040)^2 - (0.020)^2} = 31.6\sqrt{0.00120} = 1.10\mathrm{ m/s}$$

(d) $K = U \implies E_T = 2U$:

$$\frac{1}{2}kA^2 = 2 \times \frac{1}{2}kx^2 \implies x = \frac{A}{\sqrt{2}} = \frac{0.040}{\sqrt{2}} = 0.0283\mathrm{ m}$$

This holds for any SHM regardless of system parameters.

Problem 8 (HL)

A simple pendulum of length $2.00\mathrm{ m}$ is released from a small angle on Earth. At the lowest point, the bob has speed $3.13\mathrm{ m/s}$. Determine: (a) the amplitude (arc length), (b) the maximum acceleration, (c) the speed when the bob is $0.50\mathrm{ m}$ below the release point.

Solution

$$A = \frac{v_{\max}}{\omega} = \frac{3.13}{2.21} = 1.42\mathrm{ m}$$

This corresponds to an angular amplitude of $0.71\mathrm{ rad} \approx 41^\circ$, exceeding the small-angle regime. The SHM model is approximate here.

(b)

$$a_{\max} = \omega^2 A = (4.905)(1.42) = 6.97\mathrm{ m/s}^2$$

(c) The release point is at $x = A = 1.42\mathrm{ m}$. A point $0.50\mathrm{ m}$ below corresponds to $x = A - 0.50 = 0.92\mathrm{ m}$.

$$v = \omega\sqrt{A^2 - x^2} = 2.21\sqrt{(1.42)^2 - (0.92)^2} = 2.21\sqrt{1.170} = 2.21(1.082) = 2.39\mathrm{ m/s}$$

For the A-Level treatment of this topic, see Oscillations.

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