Oscillations

Oscillations

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Board Coverage AQA Paper 2 | Edexcel CP2 | OCR (A) Paper 2 | CIE P2

Masses and Springs

Explore the simulation above to develop intuition for this topic.

1. Simple Harmonic Motion — Definition

Definition. A body undergoes simple harmonic motion (SHM) if its acceleration is directly proportional to its displacement from a fixed point and is always directed towards that point:

$\boxed{a = -\omega^2 x}$

where $\omega$ is the angular frequency of the oscillation and $x$ is the displacement from the equilibrium position.

The negative sign is crucial: it ensures the acceleration is always directed towards the equilibrium position (a restoring force).

2. The SHM Equation and Its Solutions

Starting from $a = -\omega^2 x$, and using $a = \frac{d^2x}{dt^2}$:

$\frac{d^2x}{dt^2} = -\omega^2 x$

This is a second-order linear ODE with constant coefficients. We propose a solution of the form $x = A\cos(\omega t + \phi)$.

Verification by double differentiation:

$\frac{dx}{dt} = -A\omega\sin(\omega t + \phi)$

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

Therefore:

$\boxed{x(t) = A\cos(\omega t + \phi)}$

where:

• $A$ is the amplitude (maximum displacement)
• $\omega$ is the angular frequency (rad s$^{-1}$)
• $\phi$ is the phase constant (determined by initial conditions)

An equivalent form is $x = A\sin(\omega t + \phi')$, which differs only by a phase shift of $\pi/2$.

Derived Quantities

Velocity:

$v = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi)$

Maximum velocity: $v_{\max} = A\omega$ (at $x = 0$, the equilibrium position).

Acceleration:

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

Maximum acceleration: $a_{\max} = A\omega^2$ (at $x = \pm A$, the turning points).

Velocity-Displacement Relation

From $x = A\cos(\omega t + \phi)$ and $v = -A\omega\sin(\omega t + \phi)$:

$\cos(\omega t + \phi) = \frac{x}{A}, \quad \sin(\omega t + \phi) = \frac◆LB◆-v◆RB◆◆LB◆A\omega◆RB◆$

Since $\sin^2 + \cos^2 = 1$:

$\frac{x^2}{A^2} + \frac◆LB◆v^2◆RB◆◆LB◆A^2\omega^2◆RB◆ = 1$

$\boxed{v = \pm\omega\sqrt{A^2 - x^2}}$

3. Energy in SHM

Kinetic Energy

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2 - x^2)$

$\boxed{E_k = \frac{1}{2}m\omega^2(A^2 - x^2)}$

Derivation. Substituting $v = \omega\sqrt{A^2 - x^2}$ into $E_k = \frac{1}{2}mv^2$. $\square$

Potential Energy

The potential energy at displacement $x$ is the work done against the restoring force from equilibrium to $x$:

$E_p = -\int_0^x F\,dx' = -\int_0^x (-m\omega^2 x')\,dx' = m\omega^2\int_0^x x'\,dx'$

$\boxed{E_p = \frac{1}{2}m\omega^2 x^2}$

Total Energy

$E_{\mathrm{total}} = E_k + E_p = \frac{1}{2}m\omega^2(A^2 - x^2) + \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2$

$\boxed{E_{\mathrm{total}} = \frac{1}{2}m\omega^2 A^2 = \mathrm{constant}}$

Proof that total energy is constant. The total energy depends only on $m$, $\omega$, and $A$ — none of which change during the motion. Therefore $E_{\mathrm{total}}$ is conserved. $\square$

Intuition. Energy oscillates between kinetic and potential forms. At the equilibrium ($x = 0$), all energy is kinetic. At the turning points ($x = \pm A$), all energy is potential. The total is always the same — this is just conservation of energy applied to SHM.

4. Mass-Spring System

Derivation of $\omega = \sqrt{k/m}$.

By Hooke's law, the restoring force on a mass $m$ attached to a spring of constant $k$ is $F = -kx$.

By Newton's second law: $ma = -kx$, i.e., $a = -\frac{k}{m}x$.

Comparing with $a = -\omega^2 x$:

$\boxed{\omega = \sqrt◆LB◆\frac{k}{m}◆RB◆, \qquad T = 2\pi\sqrt◆LB◆\frac{m}{k}◆RB◆}$

Intuition. A stiffer spring (larger $k$) gives faster oscillations. A heavier mass (larger $m$) gives slower oscillations. The period is independent of amplitude — this is a defining feature of SHM.

5. Simple Pendulum

Derivation of $\omega = \sqrt{g/L}$ for small angles.

Consider a pendulum of length $L$ with a bob of mass $m$ displaced by angle $\theta$ from the vertical.

The restoring force along the arc is $F = -mg\sin\theta$.

The displacement along the arc is $x = L\theta$.

Newton's second law: $ma = -mg\sin\theta$, where $a = L\ddot{\theta}$:

$mL\ddot{\theta} = -mg\sin\theta \implies \ddot{\theta} = -\frac{g}{L}\sin\theta$

Small angle approximation: For $\theta \lesssim 10^\circ$, $\sin\theta \approx \theta$ (in radians). Then:

$\ddot{\theta} = -\frac{g}{L}\theta$

This is SHM in the variable $\theta$ with angular frequency $\omega = \sqrt{g/L}$:

$\boxed{T = 2\pi\sqrt◆LB◆\frac{L}{g}◆RB◆}$

warning

Common Pitfall The period of a simple pendulum is independent of mass and independent of amplitude (for small oscillations). These are classic exam questions.

6. Phase Relationships

In SHM, displacement, velocity, and acceleration are out of phase:

• Displacement: $x = A\cos(\omega t + \phi)$
• Velocity: $v = -A\omega\sin(\omega t + \phi) = A\omega\cos(\omega t + \phi + \pi/2)$ — leads displacement by $\pi/2$
• Acceleration: $a = -A\omega^2\cos(\omega t + \phi) = A\omega^2\cos(\omega t + \phi + \pi)$ — leads displacement by $\pi$

Key phase differences:

• Velocity leads displacement by $90^\circ$ ($\pi/2$)
• Acceleration leads displacement by $180^\circ$ ($\pi$) — acceleration is always antiphase to displacement

7. Damping

Definition. Damping is the dissipation of energy from an oscillating system, usually due to friction or viscous forces.

Types of Damping

Type	Description	Displacement-Time Graph
Light (underdamped)	Oscillates with gradually decreasing amplitude	Decaying sinusoid
Critical	Returns to equilibrium in the shortest time without oscillating	Exponential decay, no overshoot
Heavy (overdamped)	Returns to equilibrium very slowly without oscillating	Slow exponential decay

Critical damping is the boundary between oscillatory and non-oscillatory decay. It occurs when the damping coefficient $b$ satisfies $b = 2\sqrt{km}$ (for a mass-spring system). Car suspension systems are designed to be critically damped.

8. Resonance

Definition. Resonance occurs when a periodic driving force has a frequency equal to the natural frequency of the oscillating system. At resonance, the amplitude of oscillation is maximised.

Key features:

• The amplitude is maximum when the driving frequency $f_d$ equals the natural frequency $f_0$.
• Lightly damped systems have a sharp resonance peak; heavily damped systems have a broad peak.
• At resonance, the system absorbs maximum energy from the driving force.

Examples: A child on a swing (driving at natural frequency), the Tacoma Narrows Bridge (aeroelastic resonance), tuning a radio (LC circuit resonance), MRI machines (nuclear magnetic resonance).

tip

Exam Technique In exam questions about damping and resonance, remember: more damping $\to$ lower maximum amplitude $\to$ broader resonance curve. Less damping $\to$ higher peak $\to$ narrower curve.

Problem Set

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Problem 1

A mass-spring system has $m = 0.40$ kg and $k = 160$ N m$^{-1}$. The mass is displaced $0.05$ m from equilibrium and released. Find: (a) the period, (b) the maximum velocity, (c) the maximum acceleration.

Answer. (a) $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.40/160} = 2\pi\sqrt{0.0025} = 2\pi \times 0.05 = 0.314$ s.

(b) $v_{\max} = A\omega = 0.05 \times \sqrt{160/0.40} = 0.05 \times 20 = 1.0$ m s$^{-1}$.

(c) $a_{\max} = A\omega^2 = 0.05 \times 400 = 20$ m s$^{-2}$.

If you get this wrong, revise: Mass-Spring System

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Problem 2

A simple pendulum has a period of $2.00$ s on Earth. What would be its period on the Moon ($g_{\mathrm{Moon}} = 1.62$ m s$^{-2}$)?

Answer. $T_E = 2\pi\sqrt{L/g_E} \implies L = g_E(T_E/2\pi)^2 = 9.81(2.00/2\pi)^2 = 9.81 \times 0.1013 = 0.994$ m.

$T_M = 2\pi\sqrt{0.994/1.62} = 2\pi \times 0.7827 = 4.92$ s.

If you get this wrong, revise: Simple Pendulum

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Problem 3

An object in SHM has amplitude $0.10$ m and frequency $5.0$ Hz. Find its speed and acceleration when it is $0.06$ m from the equilibrium position.

Answer. $\omega = 2\pi \times 5.0 = 31.4$ rad s$^{-1}$.

$v = \omega\sqrt{A^2 - x^2} = 31.4\sqrt{0.01 - 0.0036} = 31.4\sqrt{0.0064} = 31.4 \times 0.08 = 2.51$ m s$^{-1}$.

$a = -\omega^2 x = -(31.4)^2 \times 0.06 = -986 \times 0.06 = -59.2$ m s$^{-2}$.

If you get this wrong, revise: Velocity-Displacement Relation

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Problem 4

A pendulum of length $1.50$ m has a bob of mass $0.50$ kg. Calculate: (a) its period, (b) its total energy when swinging with amplitude $0.10$ m, (c) the maximum velocity.

Answer. (a) $T = 2\pi\sqrt{1.50/9.81} = 2\pi \times 0.391 = 2.46$ s.

(b) For small angles, $\omega = \sqrt{g/L} = \sqrt{9.81/1.50} = 2.557$ rad s$^{-1}$. $E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}(0.50)(2.557)^2(0.10)^2 = 0.0163$ J.

(c) $v_{\max} = A\omega = 0.10 \times 2.557 = 0.256$ m s$^{-1}$.

If you get this wrong, revise: Energy in SHM

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Problem 5

Show that the period of a mass-spring system is independent of the amplitude of oscillation.

Answer. $T = 2\pi/\omega = 2\pi\sqrt{m/k}$. The period depends only on $m$ and $k$, neither of which depends on the amplitude $A$. The amplitude $A$ is determined by initial conditions, not the inherent properties of the oscillator. $\square$

If you get this wrong, revise: Mass-Spring System

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Problem 6

A 50 g mass on a spring oscillates with amplitude 4.0 cm and period 0.80 s. Find the total energy and the speed when the displacement is 2.0 cm.

Answer. $\omega = 2\pi/0.80 = 7.854$ rad s$^{-1}$.

$E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}(0.050)(7.854)^2(0.04)^2 = \frac{1}{2}(0.050)(61.65)(0.0016) = 2.46 \times 10^{-3}$ J.

$v = \omega\sqrt{A^2 - x^2} = 7.854\sqrt{0.0016 - 0.0004} = 7.854 \times 0.0346 = 0.272$ m s$^{-1}$.

If you get this wrong, revise: Energy in SHM

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Problem 7

Sketch graphs showing how displacement, velocity, and acceleration vary with time for one complete cycle of SHM. Show the phase relationships clearly.

Answer. All three are sinusoidal curves of the same period.

• Displacement: $x = A\cos(\omega t)$ — starts at $A$, goes to $-A$, returns.
• Velocity: $v = -A\omega\sin(\omega t)$ — starts at $0$, goes to $-A\omega$, returns. Leads $x$ by $90^\circ$.
• Acceleration: $a = -A\omega^2\cos(\omega t)$ — starts at $-A\omega^2$, goes to $+A\omega^2$, returns. Antiphase to $x$.

If you get this wrong, revise: Phase Relationships

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Problem 8

A pendulum has a period of $1.00$ s at a location where $g = 9.81$ m s$^{-2}$. If the pendulum is taken to a location where $g = 9.40$ m s$^{-2}$, what is the new period?

Answer. $T \propto g^{-1/2}$. $T_2/T_1 = \sqrt{g_1/g_2} = \sqrt{9.81/9.40} = \sqrt{1.0436} = 1.0216$. $T_2 = 1.0216$ s.

If you get this wrong, revise: Simple Pendulum

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Problem 9

In an SHM system, the kinetic energy equals the potential energy when the displacement is $x = x_0$. Show that $x_0 = A/\sqrt{2}$.

Answer. $E_k = E_p$ when $\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2 x^2$. $A^2 - x^2 = x^2 \implies 2x^2 = A^2 \implies x = A/\sqrt{2}$. $\square$

If you get this wrong, revise: Energy in SHM

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Problem 10

Explain why a heavily damped system does not exhibit resonance at its natural frequency.

Answer. A heavily damped system has a very broad, low resonance curve. Energy is dissipated so rapidly that even when driven at the natural frequency, the amplitude build-up is minimal. The damping force is comparable to the restoring force, preventing significant oscillation. The peak of the resonance curve is very flat and occurs at a frequency slightly below the undamped natural frequency.

If you get this wrong, revise: Damping and Resonance

9. Energy in SHM — Alternative Forms and Frequency Doubling

Spring Constant Form

For a mass-spring system, the total energy can be written using the spring constant $k$:

$\boxed{E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2}$

Proof. Since $\omega^2 = k/m$ for a mass-spring system: $\frac{1}{2}m\omega^2 A^2 = \frac{1}{2}m \cdot \frac{k}{m} \cdot A^2 = \frac{1}{2}kA^2$. $\square$

Intuition. The total energy is set by the amplitude and the stiffness. A stiffer spring or larger amplitude stores more energy. The mass determines how energy is shared between KE and PE but does not affect the total.

KE and PE Oscillate at Twice the Displacement Frequency

Substituting $x = A\cos(\omega t + \phi)$ into the energy expressions:

$E_k(t) = \frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t + \phi)$

$E_p(t) = \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2 \cos^2(\omega t + \phi)$

Applying the double-angle identities $\sin^2\theta = \tfrac{1}{2}(1 - \cos 2\theta)$ and $\cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta)$:

$E_k(t) = \frac{1}{4}m\omega^2 A^2\bigl(1 - \cos(2\omega t + 2\phi)\bigr)$

$E_p(t) = \frac{1}{4}m\omega^2 A^2\bigl(1 + \cos(2\omega t + 2\phi)\bigr)$

Both $E_k$ and $E_p$ vary sinusoidally with angular frequency $2\omega$ — twice the displacement frequency.

Key observations:

• $E_k$ is maximum when $x = 0$ (equilibrium position), where all energy is kinetic.
• $E_p$ is maximum when $x = \pm A$ (turning points), where all energy is potential.
• KE and PE are always in antiphase with each other.
• At every instant, $E_k + E_p = \frac{1}{2}m\omega^2 A^2$ (constant).

Consider an oscillator with $\omega = 4\pi$ rad s$^{-1}$ (displacement frequency $f = 2$ Hz, period $T = 0.5$ s). KE varies as $\cos(8\pi t)$, giving a frequency of $4$ Hz — exactly double. In one displacement cycle, KE reaches its maximum twice: once as the mass passes through equilibrium moving left, and once moving right.

tip

tip at twice the displacement frequency, exactly out of phase, oscillating between $0$ and $E_{\mathrm{total}}$. Their sum is a horizontal line at $E_{\mathrm{total}}$.

10. Damping — Time Constant and Displacement-Time Graphs

The Damping Force

A damped oscillator experiences a resistive force proportional to velocity:

$F_d = -bv$

where $b$ is the damping coefficient (units: kg s$^{-1}$). The negative sign indicates the force opposes the motion.

The equation of motion becomes:

$m\ddot{x} + b\dot{x} + kx = 0$

Light Damping — Exponential Decay of Amplitude

For light damping ($b \lt 2\sqrt{km}$), the solution is:

$\boxed{x(t) = A_0 \, e^{-bt/(2m)} \cos(\omega' t + \phi)}$

where $\omega' = \sqrt◆LB◆\omega_0^2 - b^2/(4m^2)◆RB◆$ is the damped angular frequency, slightly less than the natural frequency $\omega_0 = \sqrt{k/m}$.

The amplitude decays inside the exponential envelope:

$A(t) = A_0 \, e^{-t/\tau}$

where the time constant is:

$\boxed{\tau = \frac{2m}{b}}$

Intuition. The time constant is the time for the amplitude to fall to $1/e \approx 37\%$ of its initial value. After one time constant, $A = A_0/e$; after two, $A = A_0/e^2 \approx 14\%$; after three, $A = A_0/e^3 \approx 5\%$. A larger $b$ means faster energy dissipation (shorter $\tau$). A larger $m$ means more inertia (longer $\tau$).

Comparison of Damping Regimes

Damping	Condition	Behaviour
Light (underdamped)	$b \lt 2\sqrt{km}$	Oscillates with exponentially decaying amplitude
Critical	$b = 2\sqrt{km}$	Fastest return to equilibrium without oscillation
Heavy (overdamped)	$b \gt 2\sqrt{km}$	No oscillation; returns to equilibrium more slowly than critical

Displacement-time description. For light damping, the graph shows a sinusoid shrinking inside an exponential envelope $A_0 e^{-t/\tau}$. The system crosses equilibrium many times. For critical damping, the displacement decreases smoothly to zero in the shortest possible time without overshooting. For heavy damping, the displacement also decreases smoothly to zero but takes longer than critical damping.

Energy Dissipation in Damped Oscillations

The total mechanical energy of a lightly damped oscillator decays as:

$E(t) = E_0 \, e^{-t/\tau_E}$

where $\tau_E = \tau/2 = m/b$. Energy is lost twice as fast as the amplitude decays, since $E \propto A^2$.

warning

warning fastest overall. A lightly damped system passes through equilibrium sooner but overshoots. Heavy damping is slower than critical — adding more damping beyond the critical value makes the system slower, not faster.

Application. Car shock absorbers are designed to be critically damped: after hitting a bump, the car body returns to equilibrium as quickly as possible without bouncing up and down.

11. Forced Oscillations and Resonance — Detailed Treatment

Natural Frequency and Driving Frequency

Every oscillating system has a natural frequency $f_0 = \omega_0/(2\pi)$ — the frequency at which it oscillates when free from external forces.

When a periodic driving force $F = F_0\cos(\omega_d t)$ is applied at driving frequency $f_d = \omega_d/(2\pi)$, the system is compelled to oscillate at $f_d$.

Resonance

Definition. Resonance occurs when the driving frequency equals the natural frequency: $f_d = f_0$. At resonance, the system absorbs maximum energy per cycle from the driving force and the amplitude reaches its maximum value.

Amplitude Response

The steady-state amplitude of a driven, damped oscillator:

$A(f_d) = \frac◆LB◆F_0 / m◆RB◆◆LB◆\sqrt◆LB◆(\omega_0^2 - \omega_d^2)^2 + (b\omega_d/m)^2◆RB◆◆RB◆$

At resonance ($\omega_d = \omega_0$):

$A_{\max} = \frac◆LB◆F_0◆RB◆◆LB◆b\omega_0◆RB◆$

Key features of the resonance curve:

• The peak amplitude occurs at $f_d \approx f_0$, shifting slightly below $f_0$ as damping increases.
• Light damping produces a sharp, tall peak (high $Q$-factor).
• Heavy damping produces a broad, low peak.
• At resonance, the driving force is in phase with velocity ($90^\circ$ ahead of displacement), maximising power transfer: $P = F \cdot v$.

Applications of Resonance

Musical instruments: Wind instruments use resonating air columns; strings vibrate at natural frequencies (harmonics) to produce sustained, rich tones.

Tuning forks: Striking a tuning fork at $f_0$ causes a second identical fork to vibrate sympathetically — a classic demonstration of resonance.

Tacoma Narrows Bridge (1940): Wind-induced aeroelastic forces matched the bridge's natural torsional frequency, causing violent resonance and catastrophic collapse.

Microwave ovens: Microwaves at 2.45 GHz resonate with the rotational frequency of water molecules, efficiently transferring energy and heating food.

MRI scanners: Protons in body tissue resonate at specific radio frequencies in a strong magnetic field, enabling detailed internal imaging without ionising radiation.

tip

tip slightly below $f_0$. Less damping $\to$ higher, sharper peak closer to $f_0$. When sketching resonance curves for different damping levels, always label the natural frequency $f_0$ and show the amplitude axis clearly.

12. Phasor Diagrams for SHM

What Is a Phasor?

A phasor is a rotating vector whose projection onto a fixed axis gives the instantaneous value of a sinusoidal quantity. For SHM, a phasor of length $A$ rotates anticlockwise at angular speed $\omega$.

Constructing the Diagram

Three phasors originate from the same point, all rotating at $\omega$:

• Displacement phasor (length $A$): makes angle $\theta = \omega t + \phi$ with the horizontal. Its horizontal projection gives $x = A\cos(\omega t + \phi)$.
• Velocity phasor (length $A\omega$): leads the displacement phasor by $90^\circ$ (perpendicular, in the direction of rotation). Its horizontal projection gives $v = -A\omega\sin(\omega t + \phi)$.
• Acceleration phasor (length $A\omega^2$): leads the displacement phasor by $180^\circ$ (antiparallel). Its horizontal projection gives $a = -A\omega^2\cos(\omega t + \phi)$.

Phase Relationships

Quantity	Phase relative to $x$	Relationship
Displacement $x$	$0^\circ$	Reference
Velocity $v$	$+90^\circ$	$v$ leads $x$ by $\pi/2$
Acceleration $a$	$+180^\circ$	$a$ leads $x$ by $\pi$ (antiphase)

Equivalently: displacement lags velocity by $90^\circ$, and displacement lags acceleration by $180^\circ$.

Intuition. The phasor diagram makes phase relationships visually obvious. When the displacement phasor is horizontal (turning point, maximum $x$), the velocity phasor is vertical ($v = 0$) and the acceleration phasor points left ($a = -\omega^2 A$). When the displacement phasor is vertical (equilibrium, $x = 0$), the velocity phasor is horizontal ($v = \pm A\omega$) and the acceleration phasor is vertical ($a = 0$).

Adding Oscillations with Phasors

When two SHM oscillations of the same frequency are combined, the resultant is also SHM with the same frequency. The resultant phasor is the vector sum of the individual phasors.

For $x_1 = A_1\cos(\omega t)$ and $x_2 = A_2\cos(\omega t + \delta)$:

$\boxed{A_{\mathrm{res}} = \sqrt◆LB◆A_1^2 + A_2^2 + 2A_1 A_2 \cos\delta◆RB◆}$

Proof. The cosine rule applied to the triangle formed by the two phasors and their resultant, with included angle $\delta$. $\square$

The resultant phase is given by $\tan\phi_{\mathrm{res}} = \frac◆LB◆A_2 \sin\delta◆RB◆◆LB◆A_1 + A_2 \cos\delta◆RB◆$.

info

Board Coverage AQA Paper 2 | Edexcel CP2 | OCR (A) Paper 2 | CIE P2

Details

Problem 11

A mass-spring system has $m = 0.25$ kg and $k = 100$ N m$^{-1}$, oscillating with amplitude $0.08$ m. Calculate: (a) the total energy using $E = \frac{1}{2}kA^2$, (b) the KE and PE when $x = 0.04$ m, (c) show that KE and PE each oscillate at twice the displacement frequency.

Answer. (a) $E = \frac{1}{2}(100)(0.08)^2 = 0.320$ J.

(b) $E_p = \frac{1}{2}(100)(0.04)^2 = 0.080$ J. $E_k = E - E_p = 0.320 - 0.080 = 0.240$ J.

(c) $E_k(t) = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t) = \frac{E}{2}(1 - \cos 2\omega t)$. The argument $2\omega t$ means the angular frequency is $2\omega$, twice the displacement frequency $\omega$. Similarly $E_p(t) = \frac{E}{2}(1 + \cos 2\omega t)$. $\square$

If you get this wrong, revise: Energy in SHM — Alternative Forms and Frequency Doubling

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Problem 12

A damped oscillator has $m = 0.50$ kg and $b = 0.80$ kg s$^{-1}$. Calculate: (a) the time constant, (b) the fraction of initial amplitude remaining after 3.0 s, (c) the time for amplitude to fall to 10% of its initial value.

Answer. (a) $\tau = 2m/b = 2(0.50)/0.80 = 1.25$ s.

(b) $A(3.0)/A_0 = e^{-3.0/1.25} = e^{-2.4} = 0.0907$, so approximately 9.1%.

(c) $0.10 = e^{-t/1.25} \implies t = -1.25 \ln(0.10) = 1.25 \times 2.303 = 2.88$ s.

If you get this wrong, revise: Damping — Time Constant and Displacement-Time Graphs

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Problem 13

A system has natural frequency 440 Hz and is lightly damped. It is driven at: (a) 220 Hz, (b) 440 Hz, (c) 880 Hz. Rank the amplitudes from largest to smallest and explain your reasoning.

Answer. Largest at 440 Hz: this is resonance ($f_d = f_0$), so the amplitude peaks. The amplitudes at 220 Hz and 880 Hz are both smaller. On a linear frequency scale, 220 Hz is 220 Hz below resonance and 880 Hz is 440 Hz above, so the response at 220 Hz is larger than at 880 Hz. Ranking: 440 Hz (largest) > 220 Hz > 880 Hz (smallest).

If you get this wrong, revise: Forced Oscillations and Resonance — Detailed Treatment

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Problem 14

Two SHM oscillations of the same frequency have amplitudes 3.0 cm and 4.0 cm, with a phase difference of $60^\circ$. Find the amplitude of the resultant oscillation using the phasor method.

Answer. $A_{\mathrm{res}} = \sqrt◆LB◆A_1^2 + A_2^2 + 2A_1 A_2 \cos\delta◆RB◆ = \sqrt◆LB◆9 + 16 + 2(3)(4)\cos 60°◆RB◆ = \sqrt{25 + 12} = \sqrt{37} = 6.08$ cm.

If you get this wrong, revise: Phasor Diagrams for SHM

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Problem 15

A 200 g mass on a spring ($k = 50$ N m$^{-1}$) oscillates with amplitude 5.0 cm. A damping force with $b = 0.20$ kg s$^{-1}$ is present. Calculate: (a) the time constant, (b) the total energy at $t = 0$, (c) the total energy at $t = 2\tau$.

Answer. (a) $\tau = 2m/b = 2(0.200)/0.20 = 2.00$ s.

(b) $E_0 = \frac{1}{2}kA^2 = \frac{1}{2}(50)(0.05)^2 = 0.0625$ J.

(c) Since $E \propto A^2 \propto e^{-2t/\tau}$: $E(2\tau) = E_0 \, e^{-4} = 0.0625 \times 0.0183 = 1.14 \times 10^{-3}$ J.

If you get this wrong, revise: Damping — Time Constant and Displacement-Time Graphs

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Problem 16

Using a phasor diagram, explain why the velocity of an SHM oscillator is zero at the turning points and maximum at the equilibrium position.

Answer. The velocity phasor (length $A\omega$) leads the displacement phasor by $90^\circ$. The horizontal projection of the velocity phasor gives the instantaneous velocity. When the displacement phasor is horizontal (turning points, $x = \pm A$), the velocity phasor is vertical, giving a horizontal projection of zero ($v = 0$). When the displacement phasor is vertical (equilibrium, $x = 0$), the velocity phasor is horizontal, giving a maximum projection of $\pm A\omega$. $\square$

If you get this wrong, revise: Phasor Diagrams for SHM

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tip

Diagnostic Test Ready to test your understanding of Oscillations? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Oscillations with other physics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

danger

Common Pitfalls

• Confusing angular frequency omega with frequency f: Angular frequency omega = 2 * pi * f = 2 * pi / T. They are different quantities with different units (rad/s vs Hz). Substituting f where omega is needed (or vice versa) in equations like x = A cos(omega * t) will give completely wrong answers.

• Assuming all oscillations are simple harmonic: Simple harmonic motion requires the restoring force to be proportional to displacement (F = -kx). A pendulum only approximates SHM for small angles (less than about 10 degrees). A bouncing ball or a swing with large amplitude is NOT SHM.

• Forgetting that the maximum velocity occurs at equilibrium: In SHM, the velocity is maximum at the equilibrium position (x = 0) and zero at the maximum displacement (x = A). The acceleration is the opposite: zero at equilibrium and maximum at the extremes. Students often confuse these two.

• Misidentifying nodes and antinodes in stationary waves: Nodes are points of ZERO amplitude (always at rest). Antinodes are points of MAXIMUM amplitude. The distance between adjacent nodes is always half the wavelength (lambda/2). The distance between a node and the nearest antinode is lambda/4.