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PHY 231 1 PHYSICS 231 Lecture 33: Oscillations Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom.

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Presentation on theme: "PHY 231 1 PHYSICS 231 Lecture 33: Oscillations Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom."— Presentation transcript:

1 PHY 231 1 PHYSICS 231 Lecture 33: Oscillations Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom

2 PHY 231 2 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

3 PHY 231 3 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

4 PHY 231 4 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?

5 PHY 231 5 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

6 PHY 231 6 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

7 PHY 231 7 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?

8 PHY 231 8 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)

9 PHY 231 9 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

10 PHY 231 10 time (s) A -A -kA/m kA/m velocity v a x A  (k/m) -A  (k/m) demo: cart on track

11 PHY 231 11 Generally: also add gravitational PE ME = KE + PE spring + PE gravity = ½mv 2 + ½kx 2 + mgh

12 PHY 231 12 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?

13 PHY 231 13 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

14 PHY 231 14 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.

15 PHY 231 15 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)

16 PHY 231 16 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)

17 PHY 231 17 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.


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