Specification
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 = − ω 2 x a = -\omega^2 x a = − ω 2 x A particle undergoing SHM can be described using time period T T T f f f ω \omega ω The time period in terms of frequency and angular frequency: T = 1 f = 2 π ω T = \frac{1}{f} = \frac{2\pi}{\omega} T = f 1 = ω 2 π The time period of a mass-spring system: T = 2 π m k T = 2\pi \sqrt{\frac{m}{k}} T = 2 π k m The time period of a simple pendulum: T = 2 π L g T = 2\pi \sqrt{\frac{L}{g}} T = 2 π g L 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 ( ω t + Φ ) , v = ω x 0 cos ( ω t + Φ ) , v = ± ω { x 0 2 − x 2 } , E T = { 1 } { 2 } m ω 2 x 0 2 , E P = { 1 } { 2 } m ω 2 x 2 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 X = x 0 sin ( ω t + Φ ) , v = ω x 0 cos ( ω t + Φ ) , v = ± ω { x 0 2 − x 2 } , E T = 1 { } { 2 } m ω 2 x 0 2 , E P = 1 { } { 2 } m ω 2 x 2 Throughout this file, amplitude is A A A ϕ 0 \phi_0 ϕ 0 x 0 x_0 x 0 Φ \Phi Φ
Fundamental Principles Simple Harmonic Motion (SHM) is a periodic oscillation about a stable equilibrium position, Characterized by a restoring force F F F x x x
\begin`\{aligned}` F \propto -x \\ F_\{\mathrm\{net\}\} = -kx = m \frac\{d^2x\}\{dt^2\}, \end`\{aligned}` Where k > 0 k > 0 k > 0
{ d 2 x } { d t 2 } + ω 2 x = 0 , ω = { { k } { m } } .1 } ({) \frac\{d^2x\}\{dt^2\} + \omega^2 x = 0, \quad \omega = \sqrt\{\frac\{k\}\{m\}\}. \tag\{1\} d { 2 x } { d t 2 } + ω 2 x = 0 , ω = { k { } { m }} .1 } ( { ) Here, ω \omega ω − 1 ^{-1} − 1
Key Characteristics:
Equilibrium Position: Point where net force vanishes (F n e t = 0 F_{\mathrm{net}} = 0 F net = 0 Amplitude (A A A ∣ x ∣ m a x = A |x|_{\mathrm{max}} = A ∣ x ∣ max = A Isochrony: Period T T T Conditions for Ideal SHM:
Restoring force obeys Hooke’s law: F = − k x F = -kx F = − k x Zero dissipative forces (undamped motion). Constant total mechanical energy. Kinematic Relations The general solution to Equation (1) is:
X ( t ) = A cos ( ω t + ϕ 0 ) , 2 } ({) X(t) = A \cos(\omega t + \phi_0), \tag\{2\} X ( t ) = A cos ( ω t + ϕ 0 ) , 2 } ( { ) Where ϕ 0 \phi_0 ϕ 0 v v v a a a
V(t) = \frac`\{dx}, \{dt}` = -\omega A \sin(\omega t + \phi_0), \tag\{3\} A ( t ) = { d 2 x } { d t 2 } = − ω 2 A cos ( ω t + ϕ 0 ) = − ω 2 x .4 } ({) A(t) = \frac\{d^2x\}\{dt^2\} = -\omega^2 A \cos(\omega t + \phi_0) = -\omega^2 x. \tag\{4\} A ( t ) = d { 2 x } { d t 2 } = − ω 2 A cos ( ω t + ϕ 0 ) = − ω 2 x .4 } ( { ) Phase Relationships:
Displacement-Velocity: v = ± ω A 2 − x 2 v = \pm \omega \sqrt{A^2 - x^2} v = ± ω A 2 − x 2 Displacement-Acceleration: a = − ω 2 x a = -\omega^2 x a = − ω 2 x Extrema: ∣ v ∣ m a x = ω A |v|_{\mathrm{max}} = \omega A ∣ v ∣ max = ω A x = 0 x = 0 x = 0 ∣ a ∣ m a x = ω 2 A |a|_{\mathrm{max}} = \omega^2 A ∣ a ∣ max = ω 2 A x = ± A x = \pm A x = ± A Graphical Interpretation:
x ( t ) x(t) x ( t ) v ( t ) v(t) v ( t ) a ( t ) a(t) a ( t ) a ( t ) a(t) a ( t ) x ( t ) x(t) x ( t ) a ∝ − x a \propto -x a ∝ − x Energy Conservation Total mechanical energy E t o t a l E_{\mathrm{total}} E total
E { { t o t a l } } = K + U = { 1 } { 2 } m v 2 + { 1 } { 2 } k x 2 .5 } ({) E_\{\mathrm\{total\}\} = K + U = \frac\{1\}\{2\}mv^2 + \frac\{1\}\{2\}kx^2. \tag\{5\} E { { t o t a l }} = K + U = 1 { } { 2 } m v 2 + 1 { } { 2 } k x 2 .5 } ( { ) Substituting Equations (2)—(4) yields: Kinetic Energy (K K K
K = { 1 } { 2 } m ω 2 A 2 sin 2 ( ω t + ϕ 0 ) = { 1 } { 2 } m ω 2 ( A 2 − x 2 ) .6 } ({) 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\} K = 1 { } { 2 } m ω 2 A 2 sin 2 ( ω t + ϕ 0 ) = 1 { } { 2 } m ω 2 ( A 2 − x 2 ) .6 } ( { ) Potential Energy (U U U
U = { 1 } { 2 } k x 2 = { 1 } { 2 } m ω 2 x 2 .7 } ({) U = \frac\{1\}\{2\}kx^2 = \frac\{1\}\{2\}m\omega^2 x^2. \tag\{7\} U = 1 { } { 2 } k x 2 = 1 { } { 2 } m ω 2 x 2 .7 } ( { ) Total Energy:
E { { t o t a l } } = { 1 } { 2 } k A 2 = { 1 } { 2 } m ω 2 A 2 .8 } ({) E_\{\mathrm\{total\}\} = \frac\{1\}\{2\}kA^2 = \frac\{1\}\{2\}m\omega^2 A^2. \tag\{8\} E { { t o t a l }} = 1 { } { 2 } k A 2 = 1 { } { 2 } m ω 2 A 2 .8 } ( { ) Energy Oscillations:
K m a x = E t o t a l K_{\mathrm{max}} = E_{\mathrm{total}} K max = E total x = 0 x = 0 x = 0 U m a x = E t o t a l U_{\mathrm{max}} = E_{\mathrm{total}} U max = E total x = ± A x = \pm A x = ± A Example Systems Simple Pendulum Description: Point mass m m m L L L g g g θ \theta θ sin θ ≈ θ \sin\theta \approx \theta sin θ ≈ θ
{ d 2 θ } { d t 2 } + { g } { L } θ = 0.9 } ({) \frac\{d^2\theta\}\{dt^2\} + \frac\{g\}\{L\}\theta = 0. \tag\{9\} d { 2 θ } { d t 2 } + g { } { L } θ = 0.9 } ( { ) This matches Equation (1) with ω = g / L \omega = \sqrt{g/L} ω = g / L
T = { 2 π } { ω } = 2 π { { L } { g } } .10 } ({) T = \frac\{2\pi\}\{\omega\} = 2\pi \sqrt\{\frac\{L\}\{g\}\}. \tag\{10\} T = 2 { π } { ω } = 2 π { L { } { g }} .10 } ( { ) Properties:
T ∝ L T \propto \sqrt{L} T ∝ L T ∝ 1 / g T \propto 1/\sqrt{g} T ∝ 1/ g m m m A A A θ ≪ 1 \theta \ll 1 θ ≪ 1 Mass-Spring System 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 m m k k k
M { d 2 x } { d t 2 } = − k x ⟹ { d 2 x } { d t 2 } + { k } { m } x = 0.11 } ({) M\frac\{d^2x\}\{dt^2\} = -kx \implies \frac\{d^2x\}\{dt^2\} + \frac\{k\}\{m\}x = 0. \tag\{11\} M d { 2 x } { d t 2 } = − k x ⟹ d { 2 x } { d t 2 } + k { } { m } x = 0.11 } ( { ) Period:
T = 2 π { { m } { k } } .12 } ({) T = 2\pi \sqrt\{\frac\{m\}\{k\}\}. \tag\{12\} T = 2 π { m { } { k }} .12 } ( { ) Properties:
T ∝ m T \propto \sqrt{m} T ∝ m T ∝ 1 / k T \propto 1/\sqrt{k} T ∝ 1/ k A A A Angular Frequency and Phase Angular Frequency (ω \omega ω
ω = 2 π f = { 2 π } { T } , 13 } ({) \omega = 2\pi f = \frac\{2\pi\}\{T\}, \tag\{13\} ω = 2 π f = 2 { π } { T } , 13 } ( { ) Where f f f
Phase Angle (ϕ \phi ϕ
X ( t ) = A cos ( ω t + ϕ 0 ) . X(t) = A \cos(\omega t + \phi_0). X ( t ) = A cos ( ω t + ϕ 0 ) . Phase Difference (Δ ϕ \Delta\phi Δ ϕ Δ ϕ = ω Δ t = { 2 π Δ t } { T } .14 } ({) \Delta\phi = \omega \Delta t = \frac\{2\pi \Delta t\}\{T\}. \tag\{14\} Δ ϕ = ω Δ t = 2 { π Δ t } { T } .14 } ( { ) Measured in radians (1 rad ≈ \approx ≈ Summary of Key Equations Quantity Expression Displacement x = A cos ( ω t + ϕ 0 ) x = A \cos(\omega t + \phi_0) x = A cos ( ω t + ϕ 0 ) Velocity v = − ω A sin ( ω t + ϕ 0 ) v = -\omega A \sin(\omega t + \phi_0) v = − ω A sin ( ω t + ϕ 0 ) Acceleration a = − ω 2 x a = -\omega^2 x a = − ω 2 x Angular Frequency ω = k / m \omega = \sqrt{k/m} ω = k / m ω = g / L \omega = \sqrt{g/L} ω = g / L Period T = 2 π / ω T = 2\pi / \omega T = 2 π / ω Kinetic Energy K = 1 2 m ω 2 ( A 2 − x 2 ) K = \frac{1}{2}m\omega^2(A^2 - x^2) K = 2 1 m ω 2 ( A 2 − x 2 ) Potential Energy U = 1 2 m ω 2 x 2 U = \frac{1}{2}m\omega^2 x^2 U = 2 1 m ω 2 x 2 Total Energy E t o t a l = 1 2 m ω 2 A 2 E_{\mathrm{total}} = \frac{1}{2}m\omega^2 A^2 E total = 2 1 m ω 2 A 2
Derivation of the SHM Solution Verification by Direct Substitution Claim: x ( t ) = A cos ( ω t + ϕ 0 ) x(t) = A\cos(\omega t + \phi_0) x ( t ) = A cos ( ω t + ϕ 0 ) d 2 x d t 2 + ω 2 x = 0 \frac{d^2x}{dt^2} + \omega^2 x = 0 d t 2 d 2 x + ω 2 x = 0 A A A ω \omega ω ϕ 0 \phi_0 ϕ 0
Proof. First derivative:
\frac`\{dx}, \{dt}` = -\omega A \sin(\omega t + \phi_0) Second derivative:
{ d 2 x } { d t 2 } = − ω 2 A cos ( ω t + ϕ 0 ) = − ω 2 x \frac\{d^2x\}\{dt^2\} = -\omega^2 A \cos(\omega t + \phi_0) = -\omega^2 x d { 2 x } { d t 2 } = − ω 2 A cos ( ω t + ϕ 0 ) = − ω 2 x Substituting into the equation of motion:
{ d 2 x } { d t 2 } + ω 2 x = − ω 2 x + ω 2 x = 0 ■ \frac\{d^2x\}\{dt^2\} + \omega^2 x = -\omega^2 x + \omega^2 x = 0 \quad \blacksquare d { 2 x } { d t 2 } + ω 2 x = − ω 2 x + ω 2 x = 0 ■ Claim: x ( t ) = A sin ( ω t + ϕ 0 ) x(t) = A\sin(\omega t + \phi_0) x ( t ) = A sin ( ω t + ϕ 0 )
Using the identity sin θ = cos ( θ − π 2 ) \sin\theta = \cos\!\left(\theta - \frac{\pi}{2}\right) sin θ = cos ( θ − 2 π )
A sin ( ω t + ϕ 0 ) = A cos ( ω t + ϕ 0 − { π } { 2 } ) A\sin(\omega t + \phi_0) = A\cos\!\left(\omega t + \phi_0 - \frac\{\pi\}\{2\}\right) A sin ( ω t + ϕ 0 ) = A cos ( ω t + ϕ 0 − π { } { 2 } ) This is the cosine form with a shifted phase. Since ϕ 0 \phi_0 ϕ 0 − π / 2 -\pi/2 − π /2 ( A , ϕ 0 ) (A, \phi_0) ( A , ϕ 0 )
Direct substitution confirms:
{ d 2 } { d t 2 } [ A sin ( ω t + ϕ 0 ) ] = − ω 2 A sin ( ω t + ϕ 0 ) = − ω 2 x ■ \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 d { 2 } { d t 2 } [ A sin ( ω t + ϕ 0 ) ] = − ω 2 A sin ( ω t + ϕ 0 ) = − ω 2 x ■ 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 x = A x = A v = 0 v = 0 v = 0 x = A cos ( ω t ) x = A\cos(\omega t) x = A cos ( ω t ) cos ( 0 ) = 1 \cos(0) = 1 cos ( 0 ) = 1 sin ( 0 ) = 0 \sin(0) = 0 sin ( 0 ) = 0 Released from x = 0 x = 0 x = 0 v = v max v = v_{\max} v = v m a x x = A sin ( ω t ) x = A\sin(\omega t) x = A sin ( ω t ) sin ( 0 ) = 0 \sin(0) = 0 sin ( 0 ) = 0 cos ( 0 ) = 1 \cos(0) = 1 cos ( 0 ) = 1 General initial state Either with appropriate ϕ 0 \phi_0 ϕ 0 Phase angle absorbs the offset
The IB data booklet uses the sine convention (x = x 0 sin ( ω t + Φ ) x = x_0 \sin(\omega t + \Phi) x = x 0 sin ( ω t + Φ ) ± π / 2 \pm\pi/2 ± π /2
Worked Examples, SL Level Example 1: Period and Frequency of a Mass-Spring System A spring of stiffness k = 200 N / m k = 200\mathrm{ N/m} k = 200 N/m 0.50 k g 0.50\mathrm{ kg} 0.50 kg T T T f f f
T = 2 π { { m } { k } } = 2 π { { 0 .50 } { 200 } } = 2 π ( 0.0500 ) = 0.314 { s } T = 2\pi \sqrt\{\frac\{m\}\{k\}\} = 2\pi \sqrt\{\frac\{0.50\}\{200\}\} = 2\pi(0.0500) = 0.314\mathrm\{ s\} T = 2 π { m { } { k }} = 2 π { 0 { .50 } { 200 }} = 2 π ( 0.0500 ) = 0.314 { s } F = { 1 } { T } = { 1 } { 0.314 } = 3.18 { H z } F = \frac\{1\}\{T\} = \frac\{1\}\{0.314\} = 3.18\mathrm\{ Hz\} F = 1 { } { T } = 1 { } { 0.314 } = 3.18 { H z } Example 2: Displacement and Velocity at a Given Time A mass-spring system has amplitude A = 0.10 m A = 0.10\mathrm{ m} A = 0.10 m T = 0.50 s T = 0.50\mathrm{ s} T = 0.50 s t = 0 t = 0 t = 0 x x x v v v t = 0.125 s t = 0.125\mathrm{ s} t = 0.125 s
Angular frequency:
ω = { 2 π } { T } = { 2 π } { 0.50 } = 4 π { r a d / s } \omega = \frac\{2\pi\}\{T\} = \frac\{2\pi\}\{0.50\} = 4\pi\mathrm\{ rad/s\} ω = 2 { π } { T } = 2 { π } { 0.50 } = 4 π { r a d / s } With x ( 0 ) = A x(0) = A x ( 0 ) = A x = A cos ( ω t ) x = A\cos(\omega t) x = A cos ( ω t )
X ( 0.125 ) = 0.10 cos ( 4 π × 0.125 ) = 0.10 cos ( { π } { 2 } ) = 0 { m } X(0.125) = 0.10\cos(4\pi \times 0.125) = 0.10\cos\!\left(\frac\{\pi\}\{2\}\right) = 0\mathrm\{ m\} X ( 0.125 ) = 0.10 cos ( 4 π × 0.125 ) = 0.10 cos ( π { } { 2 } ) = 0 { m } V ( 0.125 ) = − ω A sin ( ω t ) = − 4 π ( 0.10 ) sin ( { π } { 2 } ) = − 1.26 { m / s } V(0.125) = -\omega A \sin(\omega t) = -4\pi(0.10)\sin\!\left(\frac\{\pi\}\{2\}\right) = -1.26\mathrm\{ m/s\} V ( 0.125 ) = − ω A sin ( ω t ) = − 4 π ( 0.10 ) sin ( π { } { 2 } ) = − 1.26 { m / s } The mass passes through equilibrium at t = T / 4 = 0.125 s t = T/4 = 0.125\mathrm{ s} t = T /4 = 0.125 s ∣ v ∣ = ω A = 1.26 m / s |v| = \omega A = 1.26\mathrm{ m/s} ∣ v ∣ = ω A = 1.26 m/s
Example 3: Amplitude from Energy A 0.30 k g 0.30\mathrm{ kg} 0.30 kg k = 120 N / m k = 120\mathrm{ N/m} k = 120 N/m 1.6 m / s 1.6\mathrm{ m/s} 1.6 m/s
At x = 0 x = 0 x = 0
E { { t o t a l } } = { 1 } { 2 } m v { max } 2 = { 1 } { 2 } ( 0.30 ) ( 1.6 ) 2 = 0.384 { J } E_\{\mathrm\{total\}\} = \frac\{1\}\{2\}mv_\{\max\}^2 = \frac\{1\}\{2\}(0.30)(1.6)^2 = 0.384\mathrm\{ J\} E { { t o t a l }} = 1 { } { 2 } m v { max } 2 = 1 { } { 2 } ( 0.30 ) ( 1.6 ) 2 = 0.384 { J } At x = A x = A x = A
E { { t o t a l } } = { 1 } { 2 } k A 2 ⟹ A = { { 2 E { { t o t a l } } } { k } } = { { 2 ( 0.384 ) } { 120 } } = { 0.00640 } = 0.0800 { m } 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\} E { { t o t a l }} = 1 { } { 2 } k A 2 ⟹ A = { 2 { E { { t o t a l }}} { k }} = { 2 { ( 0.384 )} { 120 }} = { 0.00640 } = 0.0800 { m } Example 4: Simple Pendulum Period A simple pendulum on Earth has a period of 2.00 s 2.00\mathrm{ s} 2.00 s g = 9.81 m / s 2 g = 9.81\mathrm{ m/s}^2 g = 9.81 m/s 2
L = { g T 2 } { 4 π 2 } = { ( 9.81 ) ( 2.00 ) 2 } { 4 π 2 } = { 3 9.2 } { 39.5 } = 0.993 { m } 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\} L = g { T 2 } { 4 π 2 } = ( { 9.81 ) ( 2.00 ) 2 } { 4 π 2 } = 3 { 9.2 } { 39.5 } = 0.993 { m } This is close to 1.00 m 1.00\mathrm{ m} 1.00 m T = 2 s T = 2\mathrm{ s} T = 2 s 1 m 1\mathrm{ m} 1 m
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 t = 0 t = 0 t = 0 t = 0 t = 0 Δ ϕ \Delta\phi Δ ϕ
System A (cosine form): x A = A cos ( ω t ) x_A = A\cos(\omega t) x A = A cos ( ω t )
System B (starts at x = 0 x = 0 x = 0 x B = A sin ( ω t ) = A cos ( ω t − π 2 ) x_B = A\sin(\omega t) = A\cos\!\left(\omega t - \frac{\pi}{2}\right) x B = A sin ( ω t ) = A cos ( ω t − 2 π )
Comparing phases: ϕ A = 0 \phi_A = 0 ϕ A = 0 ϕ B = − π / 2 \phi_B = -\pi/2 ϕ B = − π /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 ∘ 90^\circ 9 0 ∘
Example 6: Energy Partition at a Given Displacement A mass-spring system has A = 0.15 m A = 0.15\mathrm{ m} A = 0.15 m k = 80.0 N / m k = 80.0\mathrm{ N/m} k = 80.0 N/m m = 0.20 k g m = 0.20\mathrm{ kg} m = 0.20 kg x = 0.090 m x = 0.090\mathrm{ m} x = 0.090 m
E { { t o t a l } } = { 1 } { 2 } k A 2 = { 1 } { 2 } ( 80.0 ) ( 0.15 ) 2 = 0.900 { J } E_\{\mathrm\{total\}\} = \frac\{1\}\{2\}kA^2 = \frac\{1\}\{2\}(80.0)(0.15)^2 = 0.900\mathrm\{ J\} E { { t o t a l }} = 1 { } { 2 } k A 2 = 1 { } { 2 } ( 80.0 ) ( 0.15 ) 2 = 0.900 { J } U = { 1 } { 2 } k x 2 = { 1 } { 2 } ( 80.0 ) ( 0.090 ) 2 = 0.324 { J } U = \frac\{1\}\{2\}kx^2 = \frac\{1\}\{2\}(80.0)(0.090)^2 = 0.324\mathrm\{ J\} U = 1 { } { 2 } k x 2 = 1 { } { 2 } ( 80.0 ) ( 0.090 ) 2 = 0.324 { J } K = E { { t o t a l } } − U = 0.900 − 0.324 = 0.576 { J } K = E_\{\mathrm\{total\}\} - U = 0.900 - 0.324 = 0.576\mathrm\{ J\} K = E { { t o t a l }} − U = 0.900 − 0.324 = 0.576 { J } V = { { 2 K } { m } } = { { 2 ( 0.576 ) } { 0.20 } } = { 5.76 } = 2.40 { m / s } V = \sqrt\{\frac\{2K\}\{m\}\} = \sqrt\{\frac\{2(0.576)\}\{0.20\}\} = \sqrt\{5.76\} = 2.40\mathrm\{ m/s\} V = { 2 { K } { m }} = { 2 { ( 0.576 )} { 0.20 }} = { 5.76 } = 2.40 { m / s } HL Extension: Phase and Advanced Problems Determining Phase Angle from Initial Conditions Given initial displacement x ( 0 ) = x 0 x(0) = x_0 x ( 0 ) = x 0 v ( 0 ) = v 0 v(0) = v_0 v ( 0 ) = v 0 A A A ϕ 0 \phi_0 ϕ 0
Using the cosine form x = A cos ( ω t + ϕ 0 ) x = A\cos(\omega t + \phi_0) x = A cos ( ω t + ϕ 0 )
X ( 0 ) = A cos ϕ 0 = x 0 i } ({) X(0) = A\cos\phi_0 = x_0 \tag\{i\} X ( 0 ) = A cos ϕ 0 = x 0 i } ( { ) V(0) = -\omega A\sin\phi_0 = v_0 \tag`\{ii}` Squaring and adding (i) and (ii):
A 2 cos 2 ϕ 0 + A 2 sin 2 ϕ 0 = x 0 2 + { v 0 2 } { ω 2 } A^2\cos^2\phi_0 + A^2\sin^2\phi_0 = x_0^2 + \frac\{v_0^2\}\{\omega^2\} A 2 cos 2 ϕ 0 + A 2 sin 2 ϕ 0 = x 0 2 + v { 0 2 } { ω 2 } A = { x 0 2 + { v 0 2 } { ω 2 } } 15 } ({) A = \sqrt\{x_0^2 + \frac\{v_0^2\}\{\omega^2\}\} \tag\{15\} A = { x 0 2 + v { 0 2 } { ω 2 }} 15 } ( { ) Dividing (ii) by (i):
tan ϕ 0 = − { v 0 } { ω x 0 } 16 } ({) \tan\phi_0 = -\frac\{v_0\}\{\omega x_0\} \tag\{16\} tan ϕ 0 = − v { 0 } { ω x 0 } 16 } ( { )
Equation (16) alone is ambiguous: arctan \arctan arctan ( − π / 2 , π / 2 ) (-\pi/2,\, \pi/2) ( − π /2 , π /2 )
Phase could be in any quadrant. Always verify the signs of cos ϕ 0 = x 0 / A \cos\phi_0 = x_0/A cos ϕ 0 = x 0 / A sin ϕ 0 = − v 0 / ( ω A ) \sin\phi_0 = -v_0/(\omega A) sin ϕ 0 = − v 0 / ( ω A )
HL Example 1: Phase Angle Determination A 0.40 k g 0.40\mathrm{ kg} 0.40 kg k = 160 N / m k = 160\mathrm{ N/m} k = 160 N/m x ( 0 ) = 0.060 m x(0) = 0.060\mathrm{ m} x ( 0 ) = 0.060 m v ( 0 ) = 1.20 m / s v(0) = 1.20\mathrm{ m/s} v ( 0 ) = 1.20 m/s
ω = { { k } { m } } = { { 1 60 } { 0.40 } } = 20.0 { r a d / s } \omega = \sqrt\{\frac\{k\}\{m\}\} = \sqrt\{\frac\{160\}\{0.40\}\} = 20.0\mathrm\{ rad/s\} ω = { k { } { m }} = { 1 { 60 } { 0.40 }} = 20.0 { r a d / s } A = { ( 0.060 ) 2 + { ( 1.20 ) 2 } { ( 20.0 ) 2 } } = { 0.00360 + 0.00360 } = { 0.00720 } = 0.0849 { m } A = \sqrt\{(0.060)^2 + \frac\{(1.20)^2\}\{(20.0)^2\}\} = \sqrt\{0.00360 + 0.00360\} = \sqrt\{0.00720\} = 0.0849\mathrm\{ m\} A = { ( 0.060 ) 2 + ( { 1.20 ) 2 } {( 20.0 ) 2 }} = { 0.00360 + 0.00360 } = { 0.00720 } = 0.0849 { m } tan ϕ 0 = − { v 0 } { ω x 0 } = − { 1 .20 } { ( 20.0 ) ( 0.060 ) } = − 1.00 \tan\phi_0 = -\frac\{v_0\}\{\omega x_0\} = -\frac\{1.20\}\{(20.0)(0.060)\} = -1.00 tan ϕ 0 = − v { 0 } { ω x 0 } = − 1 { .20 } {( 20.0 ) ( 0.060 )} = − 1.00 So ϕ 0 = − π / 4 \phi_0 = -\pi/4 ϕ 0 = − π /4 ϕ 0 = 3 π / 4 \phi_0 = 3\pi/4 ϕ 0 = 3 π /4 cos ϕ 0 = x 0 / A = 0.707 > 0 \cos\phi_0 = x_0/A = 0.707 > 0 cos ϕ 0 = x 0 / A = 0.707 > 0 sin ϕ 0 = − v 0 / ( ω A ) = − 0.707 < 0 \sin\phi_0 = -v_0/(\omega A) = -0.707 < 0 sin ϕ 0 = − v 0 / ( ω A ) = − 0.707 < 0 ϕ 0 = − π / 4 = − 0.785 r a d \phi_0 = -\pi/4 = -0.785\mathrm{ rad} ϕ 0 = − π /4 = − 0.785 rad
HL Example 2: Velocity at Given Displacement A pendulum oscillates with amplitude A = 0.120 m A = 0.120\mathrm{ m} A = 0.120 m ω = 3.50 r a d / s \omega = 3.50\mathrm{ rad/s} ω = 3.50 rad/s x = 0.050 m x = 0.050\mathrm{ m} x = 0.050 m
V = ± ω { A 2 − x 2 } = ± 3.50 { ( 0.120 ) 2 − ( 0.050 ) 2 } = ± 3.50 { 0.0119 } = ± 0.382 { m / s } 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\} V = ± ω { A 2 − x 2 } = ± 3.50 { ( 0.120 ) 2 − ( 0.050 ) 2 } = ± 3.50 { 0.0119 } = ± 0.382 { m / s } The ± \pm ± x = 0.050 m x = 0.050\mathrm{ m} x = 0.050 m
HL Example 3: Energy at Arbitrary Displacement A 0.50 k g 0.50\mathrm{ kg} 0.50 kg k = 200 N / m k = 200\mathrm{ N/m} k = 200 N/m 0.10 m 0.10\mathrm{ m} 0.10 m x = 0.060 m x = 0.060\mathrm{ m} x = 0.060 m
E T = { 1 } { 2 } m ω 2 A 2 = { 1 } { 2 } ( 200 ) ( 0.10 ) 2 = 1.00 { J } E_T = \frac\{1\}\{2\}m\omega^2 A^2 = \frac\{1\}\{2\}(200)(0.10)^2 = 1.00\mathrm\{ J\} E T = 1 { } { 2 } m ω 2 A 2 = 1 { } { 2 } ( 200 ) ( 0.10 ) 2 = 1.00 { J } E P = { 1 } { 2 } m ω 2 x 2 = { 1 } { 2 } ( 200 ) ( 0.060 ) 2 = 0.360 { J } E_P = \frac\{1\}\{2\}m\omega^2 x^2 = \frac\{1\}\{2\}(200)(0.060)^2 = 0.360\mathrm\{ J\} E P = 1 { } { 2 } m ω 2 x 2 = 1 { } { 2 } ( 200 ) ( 0.060 ) 2 = 0.360 { J } K = E T − E P = 1.00 − 0.360 = 0.640 { J } K = E_T - E_P = 1.00 - 0.360 = 0.640\mathrm\{ J\} K = E T − E P = 1.00 − 0.360 = 0.640 { J } V = { { 2 K } { m } } = { { 2 ( 0.640 ) } { 0.50 } } = 1.60 { m / s } V = \sqrt\{\frac\{2K\}\{m\}\} = \sqrt\{\frac\{2(0.640)\}\{0.50\}\} = 1.60\mathrm\{ m/s\} V = { 2 { K } { m }} = { 2 { ( 0.640 )} { 0.50 }} = 1.60 { m / s } Verification: ω = k / m = 20.0 r a d / s \omega = \sqrt{k/m} = 20.0\mathrm{ rad/s} ω = k / m = 20.0 rad/s v = 20.0 ( 0.10 ) 2 − ( 0.060 ) 2 = 1.60 m / s v = 20.0\sqrt{(0.10)^2 - (0.060)^2} = 1.60\mathrm{ m/s} v = 20.0 ( 0.10 ) 2 − ( 0.060 ) 2 = 1.60 m/s ✓ \checkmark ✓
HL Example 4: Period, Amplitude, and Total Energy A 0.50 k g 0.50\mathrm{ kg} 0.50 kg 2.0 J 2.0\mathrm{ J} 2.0 J 0.20 m 0.20\mathrm{ m} 0.20 m ω \omega ω k k k x = 0.10 m x = 0.10\mathrm{ m} x = 0.10 m
E T = { 1 } { 2 } m ω 2 A 2 ⟹ ω 2 = { 2 E T } { m A 2 } = { 2 ( 2.0 ) } { ( 0.50 ) ( 0.20 ) 2 } = 200 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 E T = 1 { } { 2 } m ω 2 A 2 ⟹ ω 2 = 2 { E T } { m A 2 } = 2 { ( 2.0 )} {( 0.50 ) ( 0.20 ) 2 } = 200 ω = { 200 } = 14.1 { r a d / s } , k = m ω 2 = ( 0.50 ) ( 200 ) = 100 { N / m } \omega = \sqrt\{200\} = 14.1\mathrm\{ rad/s\}, \quad k = m\omega^2 = (0.50)(200) = 100\mathrm\{ N/m\} ω = { 200 } = 14.1 { r a d / s } , k = m ω 2 = ( 0.50 ) ( 200 ) = 100 { N / m } V = ω { A 2 − x 2 } = 14.1 { ( 0.20 ) 2 − ( 0.10 ) 2 } = 14.1 ( 0.173 ) = 2.44 { m / s } 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\} V = ω { A 2 − x 2 } = 14.1 { ( 0.20 ) 2 − ( 0.10 ) 2 } = 14.1 ( 0.173 ) = 2.44 { m / s } 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 − b t A(t) = A_0 e^{-bt} A ( t ) = A 0 e − b t b b b
The system oscillates with gradually decreasing amplitude. The period is slightly longer than the undamped period: T d a m p e d > T 0 T_{\mathrm{damped}} \gt T_0 T damped > T 0 Energy is dissipated each cycle, primarily as heat via friction or air resistance. The quality factor Q = 2 π × e n e r g y s t o r e d e n e r g y l o s t p e r c y c l e Q = 2\pi \times \frac{\mathrm{energy stored}}{\mathrm{energy lost per cycle}} Q = 2 π × energylostpercycle energystored Q Q Q 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 − α t x(t) = (c_1 + c_2 t)e^{-\alpha t} x ( t ) = ( c 1 + c 2 t ) e − α t c 1 c_1 c 1 c 2 c_2 c 2 α \alpha α 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 A 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 v^2 v 2 E ( t ) ≈ E 0 e − 2 b t E(t) \approx E_0 e^{-2bt} E ( t ) ≈ E 0 e − 2 b t E ∝ A 2 E \propto A^2 E ∝ A 2 A ( t ) ∝ e − b t A(t) \propto e^{-bt} A ( t ) ∝ e − b t
Phase Space (Qualitative) An undamped SHM traces a closed ellipse in the phase plane (x x x v v v
Intuition Simple harmonic motion is the natural dance of any system displaced from equilibrium. The restoring force is proportional to displacement, like a spring that pulls harder the more you stretch it. Energy continuously converts between kinetic and potential, and the total energy remains constant. A pendulum and a mass on a spring both exhibit SHM, but their periods depend on different properties. The phase angle tells you where in the cycle the oscillation starts, and damping gradually drains energy until the motion stops.
Common Pitfalls 1. Confusing ω \omega ω f f f ω = 2 π f \omega = 2\pi f ω = 2 π f ω \omega ω 2 π 2\pi 2 π f f f f f f ω \omega ω v max = ω A v_{\max} = \omega A v m a x = ω A ω \omega ω
2. Pendulum Period is Independent of Mass T = 2 π L / g T = 2\pi\sqrt{L/g} T = 2 π L / g m m m T ∝ m T \propto \sqrt{m} T ∝ m m g mg m g m m m
3. Sign Errors in the Restoring Force The defining characteristic of SHM is F = − k x F = -kx F = − k x a = − ω 2 x a = -\omega^2 x a = − ω 2 x
4. Mixing Up Phase Angle Conventions The cosine form x = A cos ( ω t + ϕ 0 ) x = A\cos(\omega t + \phi_0) x = A cos ( ω t + ϕ 0 ) x = A sin ( ω t + Φ ) x = A\sin(\omega t + \Phi) x = A sin ( ω t + Φ ) x ( 0 ) = 0 x(0) = 0 x ( 0 ) = 0 ϕ 0 = ± π / 2 \phi_0 = \pm\pi/2 ϕ 0 = ± π /2 0 0 0
5. Small-Angle Approximation Validity T = 2 π L / g T = 2\pi\sqrt{L/g} T = 2 π L / g sin θ ≈ θ \sin\theta \approx \theta sin θ ≈ θ θ < 10 ∘ \theta \lt 10^\circ θ < 1 0 ∘ θ < 5 ∘ \theta \lt 5^\circ θ < 5 ∘ 15 ∘ 15^\circ 1 5 ∘
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 m g / k mg/k m g / k
Problem Set Problem 1 (SL) A 0.25 k g 0.25\mathrm{ kg} 0.25 kg k = 400 N / m k = 400\mathrm{ N/m} k = 400 N/m
Solution (a)
T = 2 π { { m } { k } } = 2 π { { 0 .25 } { 400 } } = 2 π ( 0.0250 ) = 0.157 { s } T = 2\pi\sqrt\{\frac\{m\}\{k\}\} = 2\pi\sqrt\{\frac\{0.25\}\{400\}\} = 2\pi(0.0250) = 0.157\mathrm\{ s\} T = 2 π { m { } { k }} = 2 π { 0 { .25 } { 400 }} = 2 π ( 0.0250 ) = 0.157 { s } (b)
T { { E a r t h } } = 2 π { { 1 .50 } { 9.81 } } = 2.46 { s } T_\{\mathrm\{Earth\}\} = 2\pi\sqrt\{\frac\{1.50\}\{9.81\}\} = 2.46\mathrm\{ s\} T { { E a r t h }} = 2 π { 1 { .50 } { 9.81 }} = 2.46 { s } On Mars:
T { { M a r s } } = 2 π { { 1 .50 } { 3.71 } } = 4.00 { s } T_\{\mathrm\{Mars\}\} = 2\pi\sqrt\{\frac\{1.50\}\{3.71\}\} = 4.00\mathrm\{ s\} T { { M a r s }} = 2 π { 1 { .50 } { 3.71 }} = 4.00 { s } The period is longer on Mars due to the weaker gravitational field.
Problem 3 (SL) A mass-spring system has amplitude 0.080 m 0.080\mathrm{ m} 0.080 m k = 500 N / m k = 500\mathrm{ N/m} k = 500 N/m 0.50 k g 0.50\mathrm{ kg} 0.50 kg 0.040 m 0.040\mathrm{ m} 0.040 m
Solution E T = { 1 } { 2 } k A 2 = { 1 } { 2 } ( 500 ) ( 0.080 ) 2 = 1.60 { J } E_T = \frac\{1\}\{2\}kA^2 = \frac\{1\}\{2\}(500)(0.080)^2 = 1.60\mathrm\{ J\} E T = 1 { } { 2 } k A 2 = 1 { } { 2 } ( 500 ) ( 0.080 ) 2 = 1.60 { J } U = { 1 } { 2 } k x 2 = { 1 } { 2 } ( 500 ) ( 0.040 ) 2 = 0.400 { J } U = \frac\{1\}\{2\}kx^2 = \frac\{1\}\{2\}(500)(0.040)^2 = 0.400\mathrm\{ J\} U = 1 { } { 2 } k x 2 = 1 { } { 2 } ( 500 ) ( 0.040 ) 2 = 0.400 { J } K = E T − U = 1.60 − 0.400 = 1.20 { J } K = E_T - U = 1.60 - 0.400 = 1.20\mathrm\{ J\} K = E T − U = 1.60 − 0.400 = 1.20 { J } At half the amplitude, the split is 75% kinetic, 25% potential (energy scales as x 2 x^2 x 2
Problem 4 (SL) Two identical pendulums are released simultaneously. Pendulum P is released from an angle of 5 ∘ 5^\circ 5 ∘ 10 ∘ 10^\circ 1 0 ∘
Solution For ideal SHM, the period is independent of amplitude:
T = 2 π { { L } { g } } T = 2\pi\sqrt\{\frac\{L\}\{g\}\} T = 2 π { L { } { g }} Both pendulums have the same length and gravitational field, so T P = T Q T_P = T_Q T P = T Q isochrony . In practice, Q’s period is very slightly longer because the small-angle approximation is less accurate at 10 ∘ 10^\circ 1 0 ∘
Problem 5 (SL) A 2.0 k g 2.0\mathrm{ kg} 2.0 kg 0.30 m 0.30\mathrm{ m} 0.30 m x = 0.20 m x = 0.20\mathrm{ m} x = 0.20 m 2.0 m / s 2.0\mathrm{ m/s} 2.0 m/s
Solution From v 2 = ω 2 ( A 2 − x 2 ) = k m ( A 2 − x 2 ) v^2 = \omega^2(A^2 - x^2) = \frac{k}{m}(A^2 - x^2) v 2 = ω 2 ( A 2 − x 2 ) = m k ( A 2 − x 2 )
( 2.0 ) 2 = { k } { 2.0 } ( ( 0.30 ) 2 − ( 0.20 ) 2 ) = { k } { 2.0 } ( 0.050 ) (2.0)^2 = \frac\{k\}\{2.0\}\!\left((0.30)^2 - (0.20)^2\right) = \frac\{k\}\{2.0\}(0.050) ( 2.0 ) 2 = k { } { 2.0 } ( ( 0.30 ) 2 − ( 0.20 ) 2 ) = k { } { 2.0 } ( 0.050 ) 4.0 = 0.025 k ⟹ k = 160 { N / m } 4.0 = 0.025k \implies k = 160\mathrm\{ N/m\} 4.0 = 0.025 k ⟹ k = 160 { N / m } Total energy:
E T = { 1 } { 2 } k A 2 = { 1 } { 2 } ( 160 ) ( 0.30 ) 2 = 7.2 { J } E_T = \frac\{1\}\{2\}kA^2 = \frac\{1\}\{2\}(160)(0.30)^2 = 7.2\mathrm\{ J\} E T = 1 { } { 2 } k A 2 = 1 { } { 2 } ( 160 ) ( 0.30 ) 2 = 7.2 { J } Problem 6 (HL) A particle undergoes SHM with ω = 8.0 r a d / s \omega = 8.0\mathrm{ rad/s} ω = 8.0 rad/s t = 0 t = 0 t = 0 x = 0.030 m x = 0.030\mathrm{ m} x = 0.030 m v = − 0.20 m / s v = -0.20\mathrm{ m/s} v = − 0.20 m/s t = 0.50 s t = 0.50\mathrm{ s} t = 0.50 s
Solution Amplitude:
A = { x 0 2 + { v 0 2 } { ω 2 } } = { ( 0.030 ) 2 + { ( 0.20 ) 2 } { ( 8.0 ) 2 } } = { 0.000900 + 0.000625 } = { 0.001525 } = 0.0391 { m } 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\} A = { x 0 2 + v { 0 2 } { ω 2 }} = { ( 0.030 ) 2 + ( { 0.20 ) 2 } {( 8.0 ) 2 }} = { 0.000900 + 0.000625 } = { 0.001525 } = 0.0391 { m } Phase angle:
tan ϕ 0 = − { v 0 } { ω x 0 } = − { − 0.20 } { ( 8.0 ) ( 0.030 ) } = 0.833 \tan\phi_0 = -\frac\{v_0\}\{\omega x_0\} = -\frac\{-0.20\}\{(8.0)(0.030)\} = 0.833 tan ϕ 0 = − v { 0 } { ω x 0 } = − − { 0.20 } {( 8.0 ) ( 0.030 )} = 0.833 ϕ 0 = arctan ( 0.833 ) = 0.694 r a d \phi_0 = \arctan(0.833) = 0.694\mathrm{ rad} ϕ 0 = arctan ( 0.833 ) = 0.694 rad cos ϕ 0 = x 0 / A = 0.767 > 0 \cos\phi_0 = x_0/A = 0.767 > 0 cos ϕ 0 = x 0 / A = 0.767 > 0 sin ϕ 0 = − v 0 / ( ω A ) = 0.639 > 0 \sin\phi_0 = -v_0/(\omega A) = 0.639 > 0 sin ϕ 0 = − v 0 / ( ω A ) = 0.639 > 0
Displacement at t = 0.50 s t = 0.50\mathrm{ s} t = 0.50 s
X = 0.0391 cos ( 8.0 × 0.50 + 0.694 ) = 0.0391 cos ( 4.694 ) = 0.0391 ( − 0.0296 ) = − 0.00116 { m } X = 0.0391\cos(8.0 \times 0.50 + 0.694) = 0.0391\cos(4.694) = 0.0391(-0.0296) = -0.00116\mathrm\{ m\} X = 0.0391 cos ( 8.0 × 0.50 + 0.694 ) = 0.0391 cos ( 4.694 ) = 0.0391 ( − 0.0296 ) = − 0.00116 { m } Problem 7 (HL) A 0.60 k g 0.60\mathrm{ kg} 0.60 kg 0.48 J 0.48\mathrm{ J} 0.48 J 0.040 m 0.040\mathrm{ m} 0.040 m x = 0.020 m x = 0.020\mathrm{ m} x = 0.020 m K = U K = U K = U
Solution (a)
K = { 2 E T } { A 2 } = { 2 ( 0.48 ) } { ( 0.040 ) 2 } = 600 { N / m } K = \frac\{2E_T\}\{A^2\} = \frac\{2(0.48)\}\{(0.040)^2\} = 600\mathrm\{ N/m\} K = 2 { E T } { A 2 } = 2 { ( 0.48 )} {( 0.040 ) 2 } = 600 { N / m } (b)
V { max } = ω A = A { { k } { m } } = 0.040 { { 6 00 } { 0.60 } } = 0.040 ( 31.6 ) = 1.26 { m / s } 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\} V { max } = ω A = A { k { } { m }} = 0.040 { 6 { 00 } { 0.60 }} = 0.040 ( 31.6 ) = 1.26 { m / s } (c)
V = ω { A 2 − x 2 } = 31.6 { ( 0.040 ) 2 − ( 0.020 ) 2 } = 31.6 { 0.00120 } = 1.10 { m / s } 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\} V = ω { A 2 − x 2 } = 31.6 { ( 0.040 ) 2 − ( 0.020 ) 2 } = 31.6 { 0.00120 } = 1.10 { m / s } (d) K = U ⟹ E T = 2 U K = U \implies E_T = 2U K = U ⟹ E T = 2 U
{ 1 } { 2 } k A 2 = 2 × { 1 } { 2 } k x 2 ⟹ x = { A } { { 2 } } = { 0 .040 } { { 2 } } = 0.0283 { m } \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\} 1 { } { 2 } k A 2 = 2 × 1 { } { 2 } k x 2 ⟹ x = A { } { { 2 }} = 0 { .040 } { { 2 }} = 0.0283 { m } This holds for any SHM regardless of system parameters.
Problem 8 (HL) A simple pendulum of length 2.00 m 2.00\mathrm{ m} 2.00 m 3.13 m / s 3.13\mathrm{ m/s} 3.13 m/s 0.50 m 0.50\mathrm{ m} 0.50 m
Solution A = { v { max } } { ω } = { 3 .13 } { 2.21 } = 1.42 { m } A = \frac\{v_\{\max\}\}\{\omega\} = \frac\{3.13\}\{2.21\} = 1.42\mathrm\{ m\} A = v { { max }} { ω } = 3 { .13 } { 2.21 } = 1.42 { m } This corresponds to an angular amplitude of 0.71 r a d ≈ 41 ∘ 0.71\mathrm{ rad} \approx 41^\circ 0.71 rad ≈ 4 1 ∘
(b)
A { max } = ω 2 A = ( 4.905 ) ( 1.42 ) = 6.97 { m / s } 2 A_\{\max\} = \omega^2 A = (4.905)(1.42) = 6.97\mathrm\{ m/s\}^2 A { max } = ω 2 A = ( 4.905 ) ( 1.42 ) = 6.97 { m / s } 2 (c) The release point is at x = A = 1.42 m x = A = 1.42\mathrm{ m} x = A = 1.42 m 0.50 m 0.50\mathrm{ m} 0.50 m x = A − 0.50 = 0.92 m x = A - 0.50 = 0.92\mathrm{ m} x = A − 0.50 = 0.92 m
V = ω { A 2 − x 2 } = 2.21 { ( 1.42 ) 2 − ( 0.92 ) 2 } = 2.21 { 1.170 } = 2.21 ( 1.082 ) = 2.39 { m / s } 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\} V = ω { A 2 − x 2 } = 2.21 { ( 1.42 ) 2 − ( 0.92 ) 2 } = 2.21 { 1.170 } = 2.21 ( 1.082 ) = 2.39 { m / s } For the A-Level treatment of this topic, see Oscillations .
Diagnostic Test Ready to test your understanding of Simple Harmonic Motion ? The contains the hardest questions within the
IB specification for this topic, each with a full worked solution.
Unit tests probe edge cases and common misconceptions. Integration tests combine Simple Harmonic Motion with other physics topics to test synthesis under exam conditions.
See for instructions on self-marking and building a personal test matrix.
Summary SHM defined by a = − ω 2 x a = -\omega^2 x a = − ω 2 x T = 2 π / ω T = 2\pi/\omega T = 2 π / ω T = 2 π m / k T = 2\pi\sqrt{m/k} T = 2 π m / k T = 2 π L / g T = 2\pi\sqrt{L/g} T = 2 π L / g At equilibrium: v = v max v = v_{\max} v = v m a x a = 0 a = 0 a = 0 v = 0 v = 0 v = 0 a = a max a = a_{\max} a = a m a x Total energy = 1 2 k A 2 = 1 2 m ω 2 A 2 = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2 = 2 1 k A 2 = 2 1 m ω 2 A 2 ↔ \leftrightarrow ↔ Cross-References Topic Site Link [Oscillations and SHM] A-Level View [Oscillations and SHM] IB View
Common Mistakes Confusing SHM with uniform circular motion: SHM is one-dimensional oscillation; uniform circular motion is two-dimensional rotation. They’re related (SHM is the projection of uniform circular motion onto a diameter), but students often confuse the equations.
Forgetting that acceleration is maximum at maximum displacement: In SHM, a = − ω 2 x a = -\omega^2 x a = − ω 2 x x = A x = A x = A x = 0 x = 0 x = 0
Misidentifying the phase angle: The phase angle ϕ \phi ϕ x = A cos ( ω t + ϕ ) x = A\cos(\omega t + \phi) x = A cos ( ω t + ϕ )
Confusing the period formulas: For a spring: T = 2 π m / k T = 2\pi\sqrt{m/k} T = 2 π m / k T = 2 π L / g T = 2\pi\sqrt{L/g} T = 2 π L / g
Forgetting that total energy is constant: In SHM, kinetic and potential energy exchange, but the total energy E = 1 2 k A 2 E = \frac{1}{2}kA^2 E = 2 1 k A 2
See Also 