Simple Harmonic Motion

Differential equation approach to SHM, springs, and pendulums

🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Start Interactive Lesson →

Simple Harmonic Motion

Differential Equation

For restoring force F=−kxF = -kx:

md2xdt2=−kxm\frac{d^2x}{dt^2} = -kx

d2xdt2+ω02x=0\frac{d^2x}{dt^2} + \omega_0^2x = 0

where ω0=k/m\omega_0 = \sqrt{k/m} is natural angular frequency.

General Solution

x(t)=Acos⁥(ω0t+ϕ)x(t) = A\cos(\omega_0 t + \phi)

or equivalently:

x(t)=C1cos⁥(ω0t)+C2sin⁥(ω0t)x(t) = C_1\cos(\omega_0 t) + C_2\sin(\omega_0 t)

where AA is amplitude and ϕ\phi is phase constant.

Velocity and Acceleration

v(t)=dxdt=−Aω0sin⁥(ω0t+ϕ)v(t) = \frac{dx}{dt} = -A\omega_0\sin(\omega_0 t + \phi)

a(t)=dvdt=−Aω02cos⁥(ω0t+ϕ)=−ω02xa(t) = \frac{dv}{dt} = -A\omega_0^2\cos(\omega_0 t + \phi) = -\omega_0^2 x

Maximum velocity: vmax=Aω0v_{max} = A\omega_0

Maximum acceleration: amax=Aω02a_{max} = A\omega_0^2

Energy in SHM

Kinetic energy: KE=12mv2=12mA2ω02sin⁥2(ω0t+ϕ)KE = \frac{1}{2}mv^2 = \frac{1}{2}mA^2\omega_0^2\sin^2(\omega_0 t + \phi)

Potential energy: PE=12kx2=12kA2cos⁥2(ω0t+ϕ)PE = \frac{1}{2}kx^2 = \frac{1}{2}kA^2\cos^2(\omega_0 t + \phi)

Total energy: E=KE+PE=12kA2=constantE = KE + PE = \frac{1}{2}kA^2 = \text{constant}

(Using k=mω02k = m\omega_0^2)

Initial Conditions

Given x0=x(0)x_0 = x(0) and v0=v(0)v_0 = v(0):

A=x02+v02ω02A = \sqrt{x_0^2 + \frac{v_0^2}{\omega_0^2}}

tanâĄÏ•=−v0ω0x0\tan\phi = -\frac{v_0}{\omega_0 x_0}

Period and Frequency

Period: T=2πω0=2πmkT = \frac{2\pi}{\omega_0} = 2\pi\sqrt{\frac{m}{k}}

Frequency: f=1T=ω02π=12πkmf = \frac{1}{T} = \frac{\omega_0}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

Simple Pendulum

For small angles (sin⁡ξ≈ξ\sin\theta \approx \theta):

d2Ξdt2+gLΞ=0\frac{d^2\theta}{dt^2} + \frac{g}{L}\theta = 0

ω0=gL\omega_0 = \sqrt{\frac{g}{L}}

T=2πLgT = 2\pi\sqrt{\frac{L}{g}}

Physical Pendulum

Extended object rotating about pivot:

Torque: τ=−mgdsin⁡ξ≈−mgdΞ\tau = -mgd\sin\theta \approx -mgd\theta

where dd is distance from pivot to center of mass.

Id2ξdt2=−mgdξI\frac{d^2\theta}{dt^2} = -mgd\theta

ω0=mgdI\omega_0 = \sqrt{\frac{mgd}{I}}

T=2πImgdT = 2\pi\sqrt{\frac{I}{mgd}}

Torsional Pendulum

Restoring torque: τ=−ÎșΞ\tau = -\kappa\theta

where Îș\kappa is torsional constant.

Id2Ξdt2=−ÎșΞI\frac{d^2\theta}{dt^2} = -\kappa\theta

ω0=ÎșI\omega_0 = \sqrt{\frac{\kappa}{I}}

Two-Body Oscillator

Two masses m1m_1 and m2m_2 connected by spring with constant kk:

Reduced mass: Ό=m1m2m1+m2\mu = \frac{m_1m_2}{m_1 + m_2}

ω0=kÎŒ\omega_0 = \sqrt{\frac{k}{\mu}}

System oscillates as if single mass Ό\mu attached to spring kk.

Vertical Spring

Mass hanging from spring:

Equilibrium: kxeq=mgkx_{eq} = mg

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

md2ydt2=−kym\frac{d^2y}{dt^2} = -ky

Same SHM equation! Period independent of gravity:

T=2πmkT = 2\pi\sqrt{\frac{m}{k}}

📚 Practice Problems

No example problems available yet.