Simple Harmonic Motion

Differential equation approach to SHM, springs, and pendulums

Simple Harmonic Motion

Differential Equation

For restoring force $F = -kx$ :

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

$\frac{d^2x}{dt^2} + \omega_0^2x = 0$

where $\omega_0 = \sqrt{k/m}$ is natural angular frequency.

General Solution

$x(t) = A\cos(\omega_0 t + \phi)$

or equivalently:

$x(t) = C_1\cos(\omega_0 t) + C_2\sin(\omega_0 t)$

where $A$ is amplitude and $\phi$ is phase constant.

Velocity and Acceleration

$v(t) = \frac{dx}{dt} = -A\omega_0\sin(\omega_0 t + \phi)$

$a(t) = \frac{dv}{dt} = -A\omega_0^2\cos(\omega_0 t + \phi) = -\omega_0^2 x$

Maximum velocity: $v_{max} = A\omega_0$

Maximum acceleration: $a_{max} = A\omega_0^2$

Energy in SHM

Kinetic energy: $KE = \frac{1}{2}mv^2 = \frac{1}{2}mA^2\omega_0^2\sin^2(\omega_0 t + \phi)$

Potential energy: $PE = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\cos^2(\omega_0 t + \phi)$

Total energy: $E = KE + PE = \frac{1}{2}kA^2 = \text{constant}$

(Using $k = m\omega_0^2$ )

Initial Conditions

Given $x_0 = x(0)$ and $v_0 = v(0)$ :

$A = \sqrt{x_0^2 + \frac{v_0^2}{\omega_0^2}}$

$\tan\phi = -\frac{v_0}{\omega_0 x_0}$

Period and Frequency

Period: $T = \frac{2\pi}{\omega_0} = 2\pi\sqrt{\frac{m}{k}}$

Frequency: $f = \frac{1}{T} = \frac{\omega_0}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Simple Pendulum

For small angles ( $\sin\theta \approx \theta$ ):

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

$\omega_0 = \sqrt{\frac{g}{L}}$

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

Physical Pendulum

Extended object rotating about pivot:

Torque: $\tau = -mgd\sin\theta \approx -mgd\theta$

where $d$ is distance from pivot to center of mass.

$I\frac{d^2\theta}{dt^2} = -mgd\theta$

$\omega_0 = \sqrt{\frac{mgd}{I}}$

$T = 2\pi\sqrt{\frac{I}{mgd}}$

Torsional Pendulum

Restoring torque: $\tau = -\kappa\theta$

where $\kappa$ is torsional constant.

$I\frac{d^2\theta}{dt^2} = -\kappa\theta$

$\omega_0 = \sqrt{\frac{\kappa}{I}}$

Two-Body Oscillator

Two masses $m_1$ and $m_2$ connected by spring with constant $k$ :

Reduced mass: $\mu = \frac{m_1m_2}{m_1 + m_2}$

$\omega_0 = \sqrt{\frac{k}{\mu}}$

System oscillates as if single mass $\mu$ attached to spring $k$ .

Vertical Spring

Mass hanging from spring:

Equilibrium: $kx_{eq} = mg$

Displacement from equilibrium: $y = x - x_{eq}$

$m\frac{d^2y}{dt^2} = -ky$

Same SHM equation! Period independent of gravity:

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

📚 Practice Problems

1Problem 1medium

❓ Question:

A 0.5 kg mass attached to a spring (k = 200 N/m) oscillates with amplitude A = 0.1 m. Find: (a) the angular frequency and period, (b) the maximum velocity and acceleration, and (c) the velocity when x = 0.05 m.

💡 Show Solution

Given:

m = 0.5 kg
k = 200 N/m
A = 0.1 m

(a) Angular frequency and period:

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400}$

$\boxed{\omega = 20 \text{ rad/s}}$

$T = \frac{2\pi}{\omega} = \frac{2\pi}{20}$

$\boxed{T = 0.314 \text{ s}}$

(b) Maximum velocity and acceleration:

$v_{max} = \omega A = (20)(0.1)$

$\boxed{v_{max} = 2.0 \text{ m/s}}$

(Occurs at x = 0)

$a_{max} = \omega^2 A = (20)^2(0.1) = 400(0.1)$

$\boxed{a_{max} = 40 \text{ m/s}^2}$

(Occurs at x = ±A)

(c) Velocity at x = 0.05 m:

Energy method: $\frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$

$kA^2 = kx^2 + mv^2$

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

$v^2 = (20)^2[(0.1)^2 - (0.05)^2] = 400[0.01 - 0.0025]$

$v^2 = 400(0.0075) = 3.0$

$\boxed{v = 1.73 \text{ m/s}}$

2Problem 2hard

❓ Question:

A physical pendulum consists of a uniform rod (length L = 1.0 m, mass M = 2.0 kg) pivoted at one end. Find: (a) the period of small oscillations, (b) the angular frequency, and (c) compare to a simple pendulum of the same length.

💡 Show Solution

Given:

L = 1.0 m
M = 2.0 kg
Rod pivoted at end

(a) Period:

For physical pendulum: $T = 2\pi\sqrt{\frac{I}{mgd}}$

where:

I = $\frac{1}{3}ML^2$ = moment about pivot
d = L/2 = distance to CM

$T = 2\pi\sqrt{\frac{ML^2/3}{Mg(L/2)}} = 2\pi\sqrt{\frac{2L}{3g}}$

$T = 2\pi\sqrt{\frac{2(1.0)}{3(9.8)}} = 2\pi\sqrt{\frac{2}{29.4}}$

$T = 2\pi\sqrt{0.0680} = 2\pi(0.261)$

$\boxed{T = 1.64 \text{ s}}$

(b) Angular frequency:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{1.64}$

$\boxed{\omega = 3.83 \text{ rad/s}}$

(c) Comparison to simple pendulum:

Simple pendulum with length L: $T_{simple} = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{1.0}{9.8}}$

$T_{simple} = 2.01 \text{ s}$

$\frac{T_{rod}}{T_{simple}} = \frac{1.64}{2.01} = 0.816$

$\boxed{\text{Rod oscillates } 23\% \text{ faster}}$

Reason: Rod's effective length is $L_{eff} = \frac{2L}{3} = 0.67$ m

3Problem 3medium

❓ Question:

Derive the differential equation for a mass-spring system and solve it for initial conditions: at t = 0, x = 0.08 m and v = 0 (given: m = 2.0 kg, k = 50 N/m). Find the equation of motion x(t).

💡 Show Solution

Given:

m = 2.0 kg
k = 50 N/m
At t = 0: x₀ = 0.08 m, v₀ = 0

Differential equation:

Newton's second law: $F = ma$ $-kx = m\frac{d^2x}{dt^2}$

$\boxed{\frac{d^2x}{dt^2} + \frac{k}{m}x = 0}$

Or: $\frac{d^2x}{dt^2} + \omega^2 x = 0$ where $\omega = \sqrt{k/m}$

General solution:

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

Find constants:

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{2.0}} = \sqrt{25} = 5 \text{ rad/s}$

At t = 0: $x(0) = A\cos(\phi) = 0.08$

$v(0) = -A\omega\sin(\phi) = 0$

From v(0) = 0: $\sin(\phi) = 0$ → $\phi = 0$ or $\pi$

Since x(0) > 0 and $\cos(0) = 1$ : choose $\phi = 0$

$A = 0.08 \text{ m}$

Final answer:

$\boxed{x(t) = 0.08\cos(5t) \text{ m}}$

where t is in seconds.

Velocity: $v(t) = -0.4\sin(5t) \text{ m/s}$

Acceleration: $a(t) = -2.0\cos(5t) \text{ m/s}^2$

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