DRUG CLASSIFICATION 1

Simple Harmonic Motion (SHM)

• SHM represents the most basic form of oscillation.
• Stable equilibrium is necessary for SHM.
• A restoring force must act on the object, directing it back to the equilibrium position.
• The strength of the restoring force is proportionate to the object's displacement from equilibrium, expressed as $F \propto -x$.

Terminology

• Amplitude ($A$) is the maximum displacement from equilibrium.
• Period ($T$) denotes the time for one complete motion cycle.
• Frequency ($f$) measures cycles per unit time; $f = \frac{1}{T}$.
• Angular frequency ($\omega$) is calculated as $\omega = 2\pi f = \frac{2\pi}{T}$, measured in radians per second.

Mass on a Spring Example

Force

• Hooke's Law defines the restoring force: $F_x = -kx$.
• The spring constant, $k$, indicates stiffness.

Acceleration

• $a_x = \frac{F_x}{m} = -\frac{k}{m}x$
• Acceleration aligns with displacement, confirming SHM.

Angular Frequency

• $\omega = \sqrt{\frac{k}{m}}$

Period

• $T = 2\pi \sqrt{\frac{m}{k}}$

Frequency

• $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

SHM and Circular Motion

• SHM can be visualized as uniform circular motion projected onto a line.

Equations

• Displacement: $x = A\cos(\omega t)$
• Velocity: $v_x = -A\omega \sin(\omega t)$
• Acceleration: $a_x = -A\omega^2 \cos(\omega t)$
• $A$ represents amplitude, while $\omega$ is the angular frequency.

Maximum Values

• Max velocity: $v_{max} = A\omega$
• Max acceleration: $a_{max} = A\omega^2$

Block on a Spring Example

• A 2.00 kg block on a frictionless surface is attached to a spring ($k = 196 N/m$).
• It's pulled 20.0 cm from equilibrium and released.

Solution (a)

• Angular frequency: $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{196 N/m}{2.00 kg}} = 9.90 rad/s$
• Frequency: $f = \frac{\omega}{2\pi} = \frac{9.90 rad/s}{2\pi} = 1.58 Hz$
• Period: $T = \frac{1}{f} = \frac{1}{1.58 Hz} = 0.633 s$

Solution (b)

• Max velocity: $v_{max} = A\omega = (0.200 m)(9.90 rad/s) = 1.98 m/s$
• Max acceleration: $a_{max} = A\omega^2 = (0.200 m)(9.90 rad/s)^2 = 19.6 m/s^2$

Solution (c)

• Displacement vs. time: $x = A\cos(\omega t) = (0.200 m) \cos((9.90 rad/s)t)$
• Velocity vs. time: $v_x = -A\omega \sin(\omega t) = -(1.98 m/s) \sin((9.90 rad/s)t)$
• Acceleration vs. time: $a_x = -A\omega^2 \cos(\omega t) = -(19.6 m/s^2) \cos((9.90 rad/s)t)$