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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
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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 +xm and -xm • 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)
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 (w) -- w = 2pf = 2p/T • Unit = radians per second (360 degrees = 2p 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) = xmcos(wt + f) • Amplitude (xm) -- the maximum displacement from the center • Phase angle (f) -- offset due to not starting at x=xm (“start” means t=0) • Remember that (wt+f) is in radians
SHM in Action • Consider SHM with f=0: • x = xmcos(wt) • Remember w=2p/T • t=0, wt=0, cos (0) = 1 • x=xm • t=1/2T, wt=p, cos (p) = -1 • x=-xm • t=T, wt=2p, cos (2p) = 1 • x=xm
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/2p is the offset as a fraction of one period • Example: SHO’s f=p & f=0 are offset 1/2 period • They are phase shifted by 1/2 period
Velocity • If we differentiate the equation for displacement w.r.t. time, we get velocity: • v(t)=-wxmsin(wt + f) • Why is velocity negative? • Since the particle moves from +xm to -xm the velocity must be negative (and then positive in the other direction) • Velocity is proportional to w • 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)=-w2xmcos(wt + f) • This equation is similar to the equation for displacement • Making a substitution yields: • a(t)=-w2x(t)
x, v and a • Consider SMH with f=0: • x = xmcos(wt) • v = -wxmsin(wt) = -vmsin(wt) • a = -w2xmcos(wt) = -amcos(wt) • When displacement is greatest (cos(wt)=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(wt)=0), velocity is maximum and acceleration is zero • Mass coasts through the middle at high speed
Force • Remember that: a=-w2x • But, F=ma so, • F=-mw2x • Since m and w are constant we can write the expression for force as: • F=-kx • Where k=mw2 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=mw2 • We can thus find the angular frequency and the period as a function of m and k
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=2p(m/k)0.5 m/k=(T/2p)2 m=T2k/4p2
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
Potential Energy • Potential energy is the integral of force • From our expression for x U=½kxm2cos2(wt+f)
Kinetic Energy • Kinetic energy depends on the velocity, • K=½mv2 = ½mw2xm2 sin2(wt+f) • Since w2=k/m, • K = ½kxm2 sin2(wt+f) • The total energy E=U+K which will give: E= ½kxm2
Next Time • Read: 15.4-15.6