Simple Harmonic Motion Physics 202 Professor Vogel (Professor Carkner’s notes, ed) Lecture 2

Simple Harmonic Motion  Any motion that repeats itself in a sinusoidal fashion  e.g. a mass on a spring  A mass that moves between +x m and -x m  with period T  Properties vary from a positive maximum to a negative minimum  Position (x)  Velocity (v)  Acceleration (a)  The system undergoing simple harmonic motion (SHM) is a simple harmonic oscillator (SHO)

SHM Snapshots

Key Quantities  Frequency (f) -- number of complete oscillations per unit time  Unit=hertz (Hz) = 1 oscillation per second = s -1  Period (T) -- time for one complete oscillation  T=1/f  Angular frequency (  ) --  = 2  f = 2  /T  Unit = radians per second (360 degrees = 2  radians)  We use angular frequency because the motion cycles

Equation of Motion  What is the position (x) of the mass at time (t)?  The displacement from the origin of a particle undergoing simple harmonic motion is:  x(t) = x m cos(  t +  )  Amplitude (x m ) -- the maximum displacement from the center  Phase angle (  ) -- offset due to not starting at x=x m (“start” means t=0)  Remember that (  t+  ) is in radians

SHM Formula Reference

SHM in Action  Consider SHM with  =0:  x = x m cos(  t)  Remember  =2  /T  t=0,  t=0, cos (0) = 1  x=x m  t=1/2T,  t= , cos (  ) = -1  x=-x m  t=T,  t=2 , cos (2  ) = 1  x=x m

Phase  The phase of SHM is the quantity in parentheses, i.e. cos( phase )  The difference in phase between 2 SHM curves indicates how far out of phase the motion is  The difference/2  is the offset as a fraction of one period  Example: SHO’s  =  &  =0 are offset 1/2 period  They are phase shifted by 1/2 period

Amplitude, Period and Phase

Velocity If we differentiate the equation for displacement w.r.t. time, we get velocity: v(t)=-  x m sin(  t +  ) Why is velocity negative? Since the particle moves from +x m to -x m the velocity must be negative (and then positive in the other direction) Velocity is proportional to  High frequency (many cycles per second) means larger velocity

Acceleration If we differentiate the equation for velocity w.r.t. time, we get acceleration a(t)=-   x m cos(  t +  ) This equation is similar to the equation for displacement Making a substitution yields: a(t)=-  2 x(t)

x, v and a Consider SMH with  =0: x = x m cos(  t) v = -  x m sin(  t) = -v m sin(  t) a = -   x m cos(  t) = - a m cos(  t) When displacement is greatest (cos(  t)=1), velocity is zero and acceleration is maximum Mass is momentarily at rest, but being pulled hard in the other direction When displacement is zero (cos(  t)=0), velocity is maximum and acceleration is zero Mass coasts through the middle at high speed

Force  Remember that: a=-  2 x  But, F=ma so,  F=-m  2 x  Since m and  are constant we can write the expression for force as:  F=-kx  Where k=m  2 is the spring constant  This is Hooke’s Law  Simple harmonic motion is motion where force is proportional to displacement but opposite in sign  Why is the sign negative?

Linear Oscillator  A simple 1-dimensional SHM system is called a linear oscillator  Example: a mass on a spring  In such a system, k=m  2  We can thus find the angular frequency and the period as a function of m and k

Linear Oscillator

Application of the Linear Oscillator: Mass in Free Fall  A normal spring scale does not work in the absence of gravity  However, for a linear oscillator the mass depends only on the period and the spring constant: T=2  (m/k) 0.5 m/k=(T/2  ) 2 m=T 2 k/4  2

SHM and Energy A linear oscillator has a total energy E, which is the sum of the potential and kinetic energies (E=U+K) U and K change as the mass oscillates As one increases the other decreases Energy must be conserved

SHM Energy Conservation

Potential Energy Potential energy is the integral of force From our expression for x U=½kx m 2 cos 2 (  t+  )

Kinetic Energy Kinetic energy depends on the velocity, K=½mv 2 = ½m  2 x m 2 sin 2 (  t+  ) Since  2 =k/m, K = ½kx m 2 sin 2 (  t+  ) The total energy E=U+K which will give: E= ½kx m 2

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