Chapter 14 : Oscillations

Review : Key Equations

Equation of motion

\[ x(t) = A\cos{ ( \omega t + \phi ) } \]

Angular frequency

\[ \omega = \sqrt{\frac{k}{m}} \]

Frequency

\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } \]

Period

\[ T = 1/f = 2\pi \sqrt{ \frac{m}{k} } \]

Example: Grocery Store Scale

If we put a single 2 kg object on the scale and let it oscillate up and down, we find that it makes exactly 78 oscillations in one minute.

If we put two identical 2 kg objects on the scale, we find that it makes exactly 60 oscillations in one minute.

Determine:

the mass of the pan: m_pan
the spring constant: k

Aside : Modern Scales

Most scales today don't use mechanical springs, but rather crystals of various materials which create a voltage when under stress.

Referred to as the piezoelectric effect.

Energy in the Simple Harmonic Oscillator

Position equation

\[ x(t) = A\cos{ ( \omega t + \phi ) } \]

Velocity equation

\[ v(t) = dx/dt = -A\omega \sin{ ( \omega t + \phi ) } \]

Potential Energy U_s

\[ U(t) = \frac{1}{2}kx^2 = \frac{1}{2}kA^2 \cos^2(...) \]

Kinetic Energy K

\[ K(t) = \frac{1}{2}mv^2 = \frac{1}{2}mA^2 \omega^2 \sin^2(...) \]

ω² = k/m so:

\[ K(t) = \frac{1}{2}mv^2 = \frac{1}{2}kA^2 \sin^2(...) \]

Total Mechanical Energy

\[ E = U + K = \frac{1}{2}kA^2 \sin^2(...) + \frac{1}{2}kA^2 \cos^2(...) = \frac{1}{2}kA^2 \]

Kinetic Energy Potential Energy Total Energy

Important Side Effect

Quantity

Time Average

\[ U(t) = \frac{1}{2}kx^2 = \frac{1}{2}kA^2 \cos^2(...) \]

\[ \langle U(t) \rangle = \frac{1}{2}kA^2 \langle \cos^2(...) \rangle = ( \frac{1}{2}kA^2 )(\frac{1}{2} ) = \frac{E}{2} \]

\[ K(t) = \frac{1}{2}mv^2 = \frac{1}{2}kA^2 \sin^2(...) \]

\[ \langle K(t) \rangle = \frac{1}{2}kA^2 \langle \sin^2(...) \rangle = ( \frac{1}{2}kA^2 )(\frac{1}{2} ) = \frac{E}{2} \]

The average potential energy is the same as the average kinetic energy.
Each of those is exactly half the total mechanical energy.

Later: we'll be looking at energy and power carried by waves and it'll be much easier to calculate one of these.

Note: this is an example of a more general result called the Virial Theorem. If we have a potential energy function of the form U=αr^N, then: 2K_avg = N U_avg
For a linear restoring force like a spring, U=½kx² so N=2 and we just get K_avg = U_avg like we found here.
For a 1/r² force like gravity (a planet orbiting a star), or electricity (an electron orbiting a proton),
U_avg = -2K_avg meaning the total mechanical energy is negative (the objects are 'bound' together).

Initial Conditions

Allegedly we can fit any sinusoidal/cosinusoidal motion into the form:
\[ x(t) = A\cos{ ( \omega t + \phi ) } \]

Let's see how we would determine the amplitude and phase parameters if we happen to have measured the position and velocity of the object at t=0.
Suppose we attach a 0.300 kg mass to a k=19.2 N/m spring.
We compress the spring by 10 cm give it a shove in the +X direction with an initial velocity of +40 cm/s.
(a) Determine the amplitude of the motion, and the maximum velocity of the object as it oscillates back and forth on the spring.
(b) Write the motion of the mass in the form
\[ x(t) = A\cos{ ( \omega t + \phi ) } \]

(We know that x(0)=-0.1 m and v(0)=+0.4 m/s, so...)

Calculators and Inverse Trig Functions

Trig functions are periodic, which means there are an infinite number of solutions for things like \( \sin{\theta} = 0.5 \).
Your calculator just gives one value though, so which one will it be? (And is the 'right' answer for the particular scenario?)

SINE COSINE TANGENT

sin^-1 returns -90 to 90 deg cos^-1 returns 0 to 180 deg tan^-1 returns -90 to 90 deg

Review : Key Equations
Equation of motion	\[ x(t) = A\cos{ ( \omega t + \phi ) } \]
Angular frequency	\[ \omega = \sqrt{\frac{k}{m}} \]
Frequency	\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } \]
Period	\[ T = 1/f = 2\pi \sqrt{ \frac{m}{k} } \]

Energy in the Simple Harmonic Oscillator
Position equation	\[ x(t) = A\cos{ ( \omega t + \phi ) } \]
Velocity equation	\[ v(t) = dx/dt = -A\omega \sin{ ( \omega t + \phi ) } \]
Potential Energy U_s	\[ U(t) = \frac{1}{2}kx^2 = \frac{1}{2}kA^2 \cos^2(...) \]
Kinetic Energy K	\[ K(t) = \frac{1}{2}mv^2 = \frac{1}{2}mA^2 \omega^2 \sin^2(...) \]
ω² = k/m so:	\[ K(t) = \frac{1}{2}mv^2 = \frac{1}{2}kA^2 \sin^2(...) \]
Total Mechanical Energy	\[ E = U + K = \frac{1}{2}kA^2 \sin^2(...) + \frac{1}{2}kA^2 \cos^2(...) = \frac{1}{2}kA^2 \]

Kinetic Energy	Potential Energy	Total Energy

Important Side Effect
Quantity	Time Average
\[ U(t) = \frac{1}{2}kx^2 = \frac{1}{2}kA^2 \cos^2(...) \]	\[ \langle U(t) \rangle = \frac{1}{2}kA^2 \langle \cos^2(...) \rangle = ( \frac{1}{2}kA^2 )(\frac{1}{2} ) = \frac{E}{2} \]
\[ K(t) = \frac{1}{2}mv^2 = \frac{1}{2}kA^2 \sin^2(...) \]	\[ \langle K(t) \rangle = \frac{1}{2}kA^2 \langle \sin^2(...) \rangle = ( \frac{1}{2}kA^2 )(\frac{1}{2} ) = \frac{E}{2} \]
The average potential energy is the same as the average kinetic energy. Each of those is exactly half the total mechanical energy. Later: we'll be looking at energy and power carried by waves and it'll be much easier to calculate one of these. Note: this is an example of a more general result called the Virial Theorem. If we have a potential energy function of the form U=αr^N, then: 2K_avg = N U_avg For a linear restoring force like a spring, U=½kx² so N=2 and we just get K_avg = U_avg like we found here. For a 1/r² force like gravity (a planet orbiting a star), or electricity (an electron orbiting a proton), U_avg = -2K_avg meaning the total mechanical energy is negative (the objects are 'bound' together).

Calculators and Inverse Trig Functions
Trig functions are periodic, which means there are an infinite number of solutions for things like \( \sin{\theta} = 0.5 \). Your calculator just gives one value though, so which one will it be? (And is the 'right' answer for the particular scenario?)
SINE	COSINE	TANGENT

sin^-1 returns -90 to 90 deg	cos^-1 returns 0 to 180 deg	tan^-1 returns -90 to 90 deg