Rutgers University Department of Physics & Astronomy. 01:750:271 Honors Physics I Fall Lecture 20 JJ II. Home Page. Title Page.

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1 Rutgers University Department of Physics & Astronomy 01:750:271 Honors Physics Fall 2015 Lecture 20 Page 1 of 31

2 1. No quizzes during Thanksgiving week. There will be recitation according to the regular schedule using the Th! T, Fr! Wed switch. 2. No lecture on Wednesday Nov 23rd (Friday classes) Page 2 of 31

3 Midterm Class average: 80.6 Page 3 of 31

4 Simple harmonic motion (SHM) One dimensional motion of a point particle given by x(t) =x m cos(!t + ). x m amplitude =maximum displacement t time! angular frequency phase constant or phase angle Page 4 of 31

5 Period: T = time needed to complete one oscillation Frequency: T = 2! f = number of oscillations per unit time. f = 1 T =! 2. Page 5 of 31

6 Page 6 of 31 back and forth from +x m to x m ~v =0when x(t) = ±x m. v = v max when x(t) =0.

7 same!, x 0 m >x m Page 7 of 31 same x m, T =2T 0,! 0 =2!

8 same x m,! =0, 0 = /4 < 0 ) right shift > 0 ) left shift Page 8 of 31 The value of depends on the displacement and the velocity at time t =0: x m cos = x(0)!x m sin = v x (0)

9 Velocity and acceleration x(t) =x m cos(!t + ) v x (t) = dx dt =!x msin(!t + ) a x (t) = dv x dt =!2 x m cos(!t + ) =! 2 x(t) Page 9 of 31 a x (t) =! 2 x(t)

10 Force law in SHM Newton s 2nd law: F x = ma x + F x = m! 2 x Spring force! Page 10 of 31 SHM is the motion executed by a system subject to a force that is proportional to the displacement of the system but opposite in sign.

11 Note: F x = kx ) k = m! 2 r k! = r m m T =2 k m d2 x dt 2 = kx ) x = x mcos(!t + ) x 0 = x(0) = x m cos v x0 = v x (0) =!x m sin Page 11 of 31

12 3 di erent springs, identical masses which one has the smallest k? Page 12 of 31 A) Graph 1 B) Graph 2 C) Graph 3 D) All have the same k

13 3 di erent springs, identical masses which one has the smallest k? Page 13 of 31 A) Graph 1 B) Graph 2 C) Graph 3 (T =2 p m/k) D) All have the same k

14 Energy in SHM Kinetic Energy K = 1 2 mv2 = m!2 x 2 m sin 2 (!t + ) 2 Potential energy U = 1 2 kx2 = kx2 m 2 cos2 (!t + ) = m!2 x 2 m cos 2 (!t + ) 2 Page 14 of 31

15 K = m!2 x 2 m sin 2 (!t + ) 2 Page 15 of 31 U = m!2 x 2 m cos 2 (!t + ) 2 K + U = m!2 x 2 m 2 = constant

16 An angular SHM: torsion pendulum Resoring torque: from twisting of the suspension wire = apple = angular displacement Equation of motion: d2 dt 2 = apple Page 16 of 31

17 Analogy with linear SHM: d2 dt 2 = $ x $ m apple $ k = m! 2 apple $ md2 x dt 2 = kx Page 17 of 31 = m cos(!t + ),! = r apple T = 2! =2 r apple

18 The simple pendulum Page 18 of 31 Restoring force = tangential component of gravitational force F g sin Not linear in!

19 Small angle approximation: sin ' Good for apple 10 = /18 rad sin(10 )= = rad Page 19 of 31 sin(10 ) /18 sin(10 ) ' < 1%

20 Equation of motion for small : d2 dt 2 = d2 dt 2 = F g = mglsin ' mgl mgl = rotational inertia about an axis? to plane of motion passign through O. Page 20 of 31 Fg = torque of gravitational force about the same axis.

21 Analogy with linear SHM: d2 dt 2 = + mgl = m cos(!t + ) r mgl! = s T =2 mgl Page 21 of 31

22 Pointlike mass, ideal cord: = ml 2 r s g L! = T =2 L g Page 22 of 31

23 Physical Pendulum Same reasoning d2 dt 2 = + mgh = m cos(!t + ) Page 23 of 31 r mgh! = s T =2 mgh

24 Damped oscillations Restoring spring force F s = ky Damping force F d = bv y Page 24 of 31 Assume gravitational force very small compared to first two.

25 Equation of motion along y axis: m d2 y = ky bdy dt2 dt m d2 y dt + bdy 2 dt + ky =0 y(t) =y m e bt/2m cos(! 0 t + ) Page 25 of 31! 0 = r k m b 2 4m 2 ( assume k>b 2 /4m)

26 y(t) =y m e bt/2m cos(! 0 t + )! 0 = r k m b 2 4m 2 cos function with amplitude decreasing in time Mechanical energy also decreasing in time: Page 26 of 31 E(t) ' 1 2 kx2 m e bt/m

27 y(t) =y m e bt/2m cos(! 0 t + ) Page 27 of 31

28 Forced oscillations and resonance Suppose the hand moves with a given frequency! d (driving frequency) Assume gravity is negligible. Equation of motion: m d2 y = ky bdy dt2 dt + F 0 cos(! d t) (note the additional external force) Page 28 of 31

29 m d2 y + ky + bdy dt2 dt = F 0 cos(! d t) y = y m cos(! d t + ) Note: The oscillation frequency equals the frequency of the external force. Page 29 of 31 y m = q(! F 0 /m! 0 = 20! 2d )2 + b 2! 2d /m2 q k/m

30 y = y m cos(! d t + ) y m = F 0 /m q (! 0 2! d 2)2 + b 2! d 2/m2 The amplitude depends on!,! 0 y m is maximal when! d =! 0 Resonance:! d =! 0 ) maximal energy transfer between external force and oscillating system. Page 30 of 31

31 Page 31 of 31

32 Have a Good Thanksgiving! Page 32 of 31