Physics 2101 S c e t c i cti n o 3 n 3 March 31st Announcements: Quiz today about Ch. 14 Class Website:

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1 Physics 2101 Section 3 March 31 st Announcements: Quiz today about Ch. 14 Class Website: 3/

2 Simple Harmonic Motion (SHM) Kinematics x(t) = x max cos( ωt + ϕ) Position v(t) = dx dt = x maxω sin ωt + ϕ a(t) = dv dt = x maxω 2 cos ωt + ϕ ( ) Velocity [ = x max ω ] v max [ 2 ] ( ) Acceleration a max = x max ω 2 a(t) = ω 2 x(t) In SHM, the acceleration is proportional to the displacement but opposite sign. The proportionality is the square of the angular frequency.

3 SHO and Hooke s Law (again ) Dynamics F = kx ma = kx a = k m x In SHM, the acceleration is proportional to the displacement but opposite sign. The proportionality is the square of the angular frequency. a(t) = ω 2 x(t) Block spring is linear SHO: ω 2 = k m ω = k m NOTE: Independent of amplitude f = ω f = 1 k 2π 2π m T = 1 f T = 2π m k Larger mass > longer period f k g g p

4 SHO: relationships x(t) = x max cos( ωt + ϕ) v(t) = x max ω sin( ωt + ϕ) a(t) = x max ω 2 cos( ωt + ϕ) shifted by π 2 shifted by π 2 vt () = 0 whenxt () =± x max vt () =± v when xt () = 0 max How to determine x max and φ : INITIAL CONDITIONS v(0) = x ω sin ω(0) + ϕ Knowing x(0) and v(0), max max ( ) ( ω ϕ ) x(0) = x cos (0) + x(0) = x max cos( ϕ) v(0) = tan( ϕ) ωx(0) x(0) x max = v(0) cos tan 1 ωx(0)

5 Problem 15 12: What is the phase constant for the harmonic oscillator with the velocity function v(t) shown in the figure, with x(t) x(t) = x m cos(ωt + φ) Answer: v(t) = x m ω sin(ωt + φ) v m = x m ω v = v(0) = v sin( φ) v s s 4 sin( φ) = = ; φ 53 v 5 v m m

(a) What are the angular frequency, frequency, and period of the resulting motion? ω = f = ω = 1.6 Hz 2π k m = 65 N/m 0.68 kg = 9.8 rad/s T = 1 f = 0.

6 Example A block whose mass m is 680 g is fastened to a spring whose constant k is 65 N/m. The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0. (a) What are the angular frequency, frequency, and period of the resulting motion? ω = f = ω = 1.6 Hz 2π k m = 65 N/m 0.68 kg = 9.8 rad/s T = 1 f = 0.64 s (b) What is the phase angle and amplitude of the oscillation? ( ) ( ) v 0 = x max ω sin ω(0) + ϕ x 0 = x max cos ω(0) + ϕ v(0) = tan( ϕ) ωx(0) 0 m/s ϕ = tan 1 = 0 (9.8 rad/s)( 0.11 m/s) x(0) = x max cos() ϕ x max = x(0) cos tan 1 0 (c) What is the maximum speed and acceleration? ( ()) ( )( ) = 11 cm vmax = xmaxω = 11 cm rad/s = m/s ( )( ) 2 amax = xmaxω = 11 cm 9.8 rad/s = 11 m/s 2 2

7 Energy swapping in SHO The elastic property of the oscillating system (spring) stores potential energy The inertia property (mass) stores kinetic energy: Total mechanical energy is constant Knowing: U(t) = 1 kx 2 = 1 kx max cos 2 ( ωt + ϕ) ( ) KE(t) = 1 mv2 = 1 mω 2 x 2 sin 2 ωt + ϕ 2 2 max = 1 m k 2 m x 2 max sin 2 = kx max sin ωt + ϕ ( ωt + ϕ) ( ) E mech = U(t) + KE(t) = 1 kx 2 2 max( cos 2 ( ωt + ϕ)+ sin 2 ( ωt + ϕ) ) = 1 kx 2 2 max = 1 mv 2 2 max Exchange between KE and U periodically during oscillation

8 Clicker Question Question 10 3 A mass of is attached to one end of a spring and is free to slide on a frictionless horizontal surface. The mass is pulled back from its equilibrium position and released. What is the speed and kinetic energy of the mass when it passes through the equilibrium position (x=0)? 1. Maximum 2. Zero 3. Need more info

9 Problem A block of mass 1 kg (m) is on a horizontal surface (a shake table) that is moving back and forth horizontally with simple harmonic motion of frequency 2.0 Hz. (f) The coefficient of static friction between block and surface is 0.5 (μ s ). m = 1 kg f = 2 Hz μ s = 0.5 How great can the amplitude of the SHM be if the block is not to slip along the surface? 1) Angular frequency is: ω = 2πf = 2π (2 Hz) = 12.6 rad /s 2) Motion of shake table is: x(t) = x max cos( ωt) a(t) = x max ω 2 cos( ωt) a max = x max ω 2 3) When the block just slips, it means the static friction is equal to the normal force: 4) Putting together: x ˆ : f s = ma SHM f s,max = μ s N yˆ : N mg = 0 = μ s mg ma SHM,max μ s mg = f s,max = ma SHM,max = μ s mg μ s (0.5)(9.8 m / s 2 x 031 a SHM,max = μ s g = x max ω 2 max = sg = ω rad / s = m ( ) 2

10 Problem15 13: In the figure two springs of spring constant k l and k r are attached to a block of mass m. Find the frequency and period. Let the positive x be to the right then the force on the block at x is: k l k r ( xk + xk )= ma l r ( a = xk + xk ) l r = ω 2 a m f = 2πω T = 1 f

11 Spring oscillations x(t) () curves for the experiments involving a particular spring box systems oscillating in SHM (equal k s). Rank (greatest to least) the curves according to: a) the system s angular frequency b) the spring s potential energy at t=0 c) the box s velocity at t=0 d) the box s kinetic i energy at t=0 e) the box s maximum kinetic energy t = 0

12 Problem 15 35: A block of Mass M at rest is attached to a spring of constant k. A bullet of mass mand velocity vstrikes and is embedded in the block. (a) find the speed of the block immediately after the collision: (b) The equation for the displacement of the SHM. (a) () Use conservation of momentum. (M + m)v f = mv mv v f = M + m x (b) Find the equation of SHM. x(t) = x m cos(ωt + φ) v(t) = x m ω sin(ωt + φ) π xt () = xm cos( ωt ) 2 π vt () = vf sin( ωt ) 2 And At t = 0, x = 0 and v = v f x = v f m ω

PHYSICS - CLUTCH CH 15: PERIODIC MOTION (NEW)

PHYSICS - CLUTCH CH 15: PERIODIC MOTION (NEW) !! www.clutchprep.com CONCEPT: Hooke s Law & Springs When you push/pull against a spring (FA), spring pushes back in the direction. (Action-Reaction!) Fs = FA = Ex. 1: You push on a spring with a force