PHYSICS 231 Lecture 33: Oscillations Remco Zegers Question hours:Monday 9:15-10:15 Helproom PHY 231

The (loss of) ability to do work: entropy entropy: S=QR/T R refers to a reversible process The equation ONLY holds for a reversible process. example: Carnot engine: Hot reservoir: Shot=-Qhot/Thot (entropy is decreased) Cold reservoir: Scold=Qcold/Tcold We saw: efficiency for a general engine: e=1-Qcold/Qhot efficiency for a Carnot engine: e=1-Tcold/Thot So for a Carnot engine: Tcold/Thot=Qcold/Qhot and thus: Qhot/Thot=Qcold/Thot Total change in entropy: Shot+Scold=0 For a Carnot engine, there is no change in entropy PHY 231

The loss of ability to do work: entropy Now, consider the following irreversible case: entropy: S=QR/T This equation only holds for reversible processes. T=300 K conducting copper wire Qtransfer=1200 J We cut the irreversible process up into 2 reversible processes T=650 K Shot+Scold=Qhot/Thot+Qcold/Tcold=-1200/650+1200/300= =+1.6 J/K The entropy has increased! PHY 231

Hooke’s law Fs=-kx Hooke’s law If there is no friction, the mass continues to oscillate back and forth. If a force is proportional to the displacement x, but opposite in direction, the resulting motion of the object is called: simple harmonic oscillation PHY 231

Simple harmonic motion displacement x A a) b) time (s) c) Amplitude (A): maximum distance from equilibrium (unit: m) Period (T): Time to complete one full oscillation (unit: s) Frequency (f): Number of completed oscillations per second (unit: 1/s = 1 Herz [Hz]) f=1/T PHY 231

Simple harmonic motion displacement x 5cm 2 4 6 8 10 time (s) -5cm what is the amplitude of the harmonic oscillation? what is the period of the harmonic oscillation? what is the frequency of the harmonic oscillation? Amplitude: 5cm (0.05 m) period: time to complete one full oscillation: 4s frequency: number of oscillations per second=1/T=0.25 s PHY 231

The spring constant k When the object hanging from the spring is not moving: Fspring =-Fgravity -kd =-mg k = mg/d k is a constant, so if we hang twice the amount of mass from the spring, d becomes twice larger: k=(2m)g/(2d)=mg/d PHY 231

displacement vs acceleration displacement x A time (s) -A Newton’s second law: F=ma  -kx=ma  a=-kx/m acceleration(a) kA/m -kA/m PHY 231

example A mass of 1 kg is hung from a spring. The spring stretches by 0.5 m. Next, the spring is placed horizontally and fixed on one side to the wall. The same mass is attached and the spring stretched by 0.2 m and then released. What is the acceleration upon release? PHY 231

energy and velocity Ekin(½mv2) Epot,spring(½kx2) Sum 0 ½kA2 ½kA2 ½mv2 0 ½mv2 0 ½k(-A)2 ½kA2 A -A conservation of ME: ½m[v(x=0)]2=½kA2 so v(x=0)=±A(k/m) PHY 231

velocity more general Total ME at any displacement x: ½mv2+½kx2 Total ME at max. displacement A: ½kA2 Conservation of ME: ½kA2=½mv2+½kx2 So: v=±[(A2-x2)k/m] position X velocity V acceleration a +A -kA/m ±A(k/m) -A kA/m PHY 231

A x time (s) -A demo: cart on track velocity v kA/m a -kA/m A(k/m) PHY 231

Generally: also add gravitational PE ME = KE + PEspring + PEgravity = ½mv2 + ½kx2 + mgh PHY 231

An example A 0.4 kg object, connected to a light spring with a spring constant of 19.6 N/m oscillates on a frictionless horizontal surface. If the spring is compressed by 0.04 and then released determine: a) the maximum speed of the object b) the speed of the object when the spring is compressed by 0.015 m c) when it is stretched by 0.015m d) for what value of x does the speed equal one half of the maximum speed? PHY 231

circular motion & simple harmonic motion A particle moves in a circular orbit with angular velocity , corresponding to a linear velocity v0=r=A  The horizontal position as a function of time: x(t)=Acos=Acos(t) (=t)  v0 vx The horizontal velocity as a function of time: sin=-vx/v0 vx(t)=-v0sin=-Asin(t) Time to complete one circle (I.e. one period T): T=2A/v0=2A/A=2/ =2/T=2f (f: frequency) : angular frequency A x t=0 PHY 231

Circular motion and simple harmonic motion The simple harmonic motion can be described by the projection of circular motion on the horizontal axis. xharmonic(t)=Acos(t) vharmonic(t)=-Asin(t) where A is the amplitude of the oscillation, and =2/T=2f, where T is the period of the harmonic motion and f=1/T the frequency. PHY 231

For the case of a spring ±A(k/m) 1) velocity is maximum if v=±A(k/m) position X velocity V acceleration a +A -kA/m ±A(k/m) -A kA/m 1) velocity is maximum if v=±A(k/m) 2) circular motion: vspring(t)=-Asint maximal if vspring=±A combine 1) & 2) =(k/m) Acceleration: a(t)=-(kA/m)cos(t)=-2Acos(t) PHY 231

xharmonic(t)=Acos(t) time (s) -A =2f=2/T=(k/m) velocity v A(k/m) vharmonic(t)=-Asin(t) -A(k/m) kA/m a aharmonic(t)=-2Acos(t) -kA/m PHY 231

Example A mass of 0.2 kg is attached to a spring with k=100 N/m. The spring is stretched over 0.1 m and released. What is the angular frequency () of the corresponding circular motion? What is the period (T) of the harmonic motion? What is the frequency (f)? What are the functions for x,v and t of the mass as a function of time? Make a sketch of these. PHY 231

question An object is attached on the lhs and rhs by a spring with the same spring constants and oscillating harmonically. Which of the following is NOT true? In the central position the velocity is maximal In the most lhs or rhs position, the magnitude of the acceleration is largest. the acceleration is always directed so that it counteracts the velocity in the absence of frictional forces, the object will oscillate forever the velocity is zero at the most lhs and rhs positions of the object PHY 231