1. PHY 231 1 PHYSICS 231 Lecture 33: Oscillations Remco Zegers Question hours:Monday 9:15-10:15 Helproom

2. PHY 231 2 The (loss of) ability to do work: entropy entropy:  S=Q R /T R refers to a reversible process The equation ONLY holds for a reversible process. example: Carnot engine: Hot reservoir:  S hot =-Q hot /T hot (entropy is decreased) Cold reservoir:  S cold =Q cold /T cold We saw: efficiency for a general engine: e=1-Q cold /Q hot efficiency for a Carnot engine: e=1-T cold /T hot So for a Carnot engine: T cold /T hot =Q cold /Q hot and thus: Q hot /T hot =Q cold /T cold Total change in entropy:  S hot +  S cold =0 For a Carnot engine, there is no change in entropy

3. PHY 231 3 The loss of ability to do work: entropy Now, consider the following irreversible case: T=300 K T=650 K conducting copper wire Q transfer =1200 J entropy:  S=Q R /T This equation only holds for reversible processes. We cut the irreversible process up into 2 reversible processes  S hot +  S cold =Q hot /T hot +Q cold /T cold =-1200/650+1200/300= =+1.6 J/K The entropy has increased!

4. PHY 231 4 quiz (extra credit) V (m 3 ) P (Pa) V min P max P min Consider this cyclic process. The hot reservoir has a temperature of 1000 K and the cold reservoir a temperature of 500 K Which of the following is true? a)This is a Carnot engine and the efficiency=1-T cold /T hot =0.5 b)This is a Carnot engine but the efficiency cannot be calculated from the info given c)This is not a Carnot engine and the efficiency = 0.5 d)This is not a Carnot engine and the efficiency < 0.5 e)This is not a Carnot engine and the efficiency could be any value between 0 and 1. V max

5. PHY 231 5 Hooke’s law F s =-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

6. PHY 231 6 Simple harmonic motion time (s) displacement x a) b) c) A 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

7. PHY 231 7 Simple harmonic motion time (s) displacement x 5cm -5cm 2468 10 a)what is the amplitude of the harmonic oscillation? b)what is the period of the harmonic oscillation? c)what is the frequency of the harmonic oscillation? a)Amplitude: 5cm (0.05 m) b)period: time to complete one full oscillation: 4s c)frequency: number of oscillations per second=1/T=0.25 s

8. PHY 231 8 The spring constant k When the object hanging from the spring is not moving: F spring =-F gravity -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

9. PHY 231 9 acceleration(a) displacement vs acceleration time (s) displacement x A -A Newton’s second law: F=ma  -kx=ma  a=-kx/m -kA/m kA/m

10. PHY 231 10 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? 1 st step: find the spring constant k F spring =-F gravity or -kd=-mg k= mg/d =1*9.8/0.5=19.6 N/m 2 nd step: find the acceleration upon release Newton’s second law: F=ma  -kx=ma  a=-kx/m a=-19.6*0.2/1=-3.92 m/s 2

11. PHY 231 11 energy and velocity E kin (½mv 2 )E pot,spring (½kx 2 ) Sum 0 ½kA 2 ½kA 2 ½mv 2 0 ½mv 2 0 ½k(-A) 2 ½kA 2 A -A conservation of ME: ½m[v(x=0)] 2 =½kA 2 so v(x=0)=±A  (k/m)

12. PHY 231 12 velocity more general Total ME at any displacement x: ½mv 2 +½kx 2 Total ME at max. displacement A: ½kA 2 Conservation of ME: ½kA 2 =½mv 2 +½kx 2 So: v=±  [(A 2 -x 2 )k/m] position Xvelocity Vacceleration a +A0-kA/m 0 ±A  (k/m) 0 -A0kA/m

13. PHY 231 13 time (s) A -A -kA/m kA/m velocity v a x A  (k/m) -A  (k/m) demo: cart on track

14. PHY 231 14 Generally: also add gravitational PE ME = KE + PE spring + PE gravity = ½mv 2 + ½kx 2 + mgh

15. PHY 231 15 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? a)v=  [(A 2 -x 2 )k/m] (speed is always positive!) maximum if x=0:  [A 2 k/m]=0.04  (19.6/0.4)=0.28 m/s b) v=  [(A 2 -x 2 )k/m] at x=-0.015 v=  [((0.04) 2 -(-0.015) 2 )19.6/0.4]=0.26 m/s c) same as b) d)  [(A 2 -x 2 )k/m]=0.28/2=0.14 x=  (A 2 -0.14 2 m/k)=0.035m

16. PHY 231 16 circular motion & simple harmonic motion A x A particle moves in a circular orbit with angular velocity , corresponding to a linear velocity v 0 =  r=  A  The horizontal position as a function of time: x(t)=Acos  =Acos(  t) (  =  t) t=0  v0v0 vxvx The horizontal velocity as a function of time: sin  =-v x /v 0 v x (t)=-v 0 sin  =-  Asin(  t) Time to complete one circle (I.e. one period T): T=2  A/v 0 =2  A/  A=2  /   =2  /T=2  f (f: frequency)  : angular frequency

17. PHY 231 17 Circular motion and simple harmonic motion The simple harmonic motion can be described by the projection of circular motion on the horizontal axis. x harmonic (t)=Acos(  t) v harmonic (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.

18. PHY 231 18 For the case of a spring position Xvelocity Vacceleration a +A0-kA/m 0 ±A  (k/m) 0 -A0kA/m 1) velocity is maximum if v=±A  (k/m) 2) circular motion: v spring (t)=-  Asin  t maximal if v spring =±  A combine 1) & 2)  =  (k/m) Acceleration: a(t)=-(kA/m)cos(  t)=-  2 Acos(  t)

19. PHY 231 19 time (s) A -A -kA/m kA/m velocity v a x A  (k/m) -A  (k/m) x harmonic (t)=Acos(  t) v harmonic (t)=-  Asin(  t) a harmonic (t)=-  2 Acos(  t)  =2  f=2  /T=  (k/m)

20. PHY 231 20 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. a)What is the angular frequency (  ) of the corresponding circular motion? b)What is the period (T) of the harmonic motion? c)What is the frequency (f)? d)What are the functions for x,v and t of the mass as a function of time? Make a sketch of these. a)  =  (k/m)=  =  (100/0.2)=22.4 rad/s b)  =2  /T T= 2  /  =0.28 s c)  =2  f f=  /2  =3.55 Hz (=1/T) d)x harmonic (t)=Acos(  t)=0.1cos(0.28t) v harmonic (t)=-  Asin(  t)=-0.028sin(0.28t) a harmonic (t)=-  2 Acos(  t)=-0.0078cos(0.28t)

21. PHY 231 21 time (s) 0.1 -0.1 -0.0078 0.0078 velocity v a x 0.028 -0.028 0.28 0.56 0.28 0.56

22. PHY 231 22 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? a)In the central position the velocity is maximal b)In the most lhs or rhs position, the magnitude of the acceleration is largest. c)the acceleration is always directed so that it counteracts the velocity d)in the absence of frictional forces, the object will oscillate forever e)the velocity is zero at the most lhs and rhs positions of the object