PHYS 218 Chapter 13

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Periodic Motion

Motion around a stable point (energy trough)

If there is no work being done by any forces to remove energy, then the motion continues forever.

Terms

Amplitude

Maximum displacement from the equilibrium position

Period

Time needed to complete a full cycle (measured in seconds)

Frequency

Number of cycles that can be done in a given time period ($f={\frac {1}{t}}$)

Angular Frequency

$\omega =2\pi f$ (in radians/second)

Harmonic Oscillation Body

Simple Harmonic Motion (SHM)

Restoring force is directly proportional to the displacement from the equilibrium point: $F=-k\Delta x$

Equation of simple harmonic motion

Linear acceleration
${\frac {d^{2}x(t)}{dt^{2}}}={\frac {-k}{m}}x(t)$

General Solution for form above

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

• A is the amplitude
• ω is the angular frequency: square root of "stuff" in front of the x(t) in Equation 1 = $\omega ={\sqrt {\tfrac {-k}{m}}}$
• φ is the phase

A and φ can be obtained from initial conditions: Given:

$x(t=0)=x_{0}$ (e.g. 3 m)

$v(t=0)=v_{0}$ (e.g. 2 m/s)

We can find:

$x_{0}=Acos{\varphi }$

$v_{0}=-A\omega \sin {\varphi }$

Energy of SHM

For a Spring

$E={\frac {1}{2}}kA^{2}$

Proof:
${\begin{aligned}E&={\frac {1}{2}}k(l_{0}-x)^{2}+{\frac {1}{2}}m\left({\frac {dx}{dt}}\right)^{2}\\x(t)&=A\cos(\omega t+\varphi )\\E&={\frac {1}{2}}kA^{2}\cos ^{2}(\omega t+\varphi )+{\frac {1}{2}}mA^{2}+\omega ^{2}\sin ^{2}(\omega t+\varphi )\\&={\frac {1}{2}}kA^{2}\end{aligned}}$

For a Simple Pendulum

If pendulum has length of $L$ Only works for very small angles where $\sin(\theta )\approx \theta$,

$s(t)=A\cos \left({\sqrt {\frac {g}{L}}}t+\varphi \right)$

For a Physical Pendulum

An extended body that is fixed around an axis (not at center of gravity) Distance from fixed axis to center of gravity is $L$

$\theta (t)=A\cos \left({\sqrt {\frac {Lmg}{I}}}t+\varphi \right)$

Damped Oscillation

In SHM, we had $F=-Cx$. What happens if we add another force... proportional to velocity, per se. $F=-bv$

Now our forces in the system:
$\sum F_{x}=\underbrace {-kx} _{\mbox{restoring force}}+\underbrace {-bv_{x}} _{\mbox{friction}}=m{\frac {d^{2}x}{dt^{2}}}$

General Solution:
$x(t)=A\mathrm {e} ^{-{\frac {b}{2m}}t}\cos(\omega t+\varphi )$

Solve for ω by taking derivative and plugging into the first equation:
$\omega ={\sqrt {{\frac {k}{m}}-{\frac {b^{2}}{4m^{2}}}}}$

Determine Damping Case

Note if ${\frac {k}{m}}-{\frac {b^{2}}{4m^{2}}}=0$, then $b=2{\sqrt {km}}$

3 Cases:

1. $b<2{\sqrt {km}}$ — ω is real (underdamping)
2. $b=2{\sqrt {km}}$ — ω = 0 (critical damping)
3. $b>2{\sqrt {km}}$ — ω is imaginary (overdamping)

Underdamping

System oscillates less and less every time until it completely stops (sine curve tapers as $t$ increases)

Envelope is approximately e^−t

Critical Damping

ω = 0, so system won't even oscillate; asymptotically approaches lowest point of equilibrium $x(t)=A\mathrm {e} ^{-{\frac {b^{2}}{4m^{2}}}t}\cos {\varphi }$

Overdamping

Same as critical damping, but approaches 0 faster.