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1/3/2012 Hooke’s Law F=-kx Simple Harmonic Motion Physics 123: Eyres Hooke’s Law Springs and Kinematics F=-kx • • • • • F is the Force on ________ caused by ________ • X is really the ____________ • K is ____________ • The negative sign means_____________ F is the Force on Object caused by Spring X is really the Spring Extension/Compression K is Spring Constant The negative sign means that extension/compression are in the opposite direction to F. • Let’s Review the motion of the object at different locations on an oscillating spring. • Find the sign of the – Position – Velocity – Acceleration http://en.wikipedia.org/wiki/File:Muelle.gif Motion on a Spring Review the Quantities Situation Position (+ to Right) velocity Pulled to right and released X=+A Component Symbol Relation to Rotational Quantity It comes from this: X=0 Tangential Displacement s θr Circumference = 2πr Tangential Velocity vt ωr 2πr ∆θ = r Τ ∆t Tangential Acceleration at αr look for the pattern Radial (Centripetal) Acceleration ar or ac rω ω2 Moving left Moving Right X=0 Completed cycle acceleration v 2 = r (r ω )2 r 1 1/3/2012 SHM: Hanging Spring SHM: Hanging Spring • The restoring force must be proportional to the displacement, or approximately so. • Note: When the spring is stretched with a mass in equilibrium it is stretched by d. Any system that undergoes simple harmonic motion: • When the system is displaced from equilibrium there must exist a restoring force that tends to restore it to equilibrium. • The restoring force must be proportional to the displacement, or approximately so. • • • • • ΣFon mass =Fspring-mg ΣFon mass =-k(Δl)-mg ΣFon mass =k(x+d)-mg ΣFon mass =kx+kd-mg ΣF=k(x) http://en.wikipedia.org/wiki/Simple_harmonic_motion Springs in SHM • Tangential velocity • Maximum energy • A little algebra 2πA T 2 2 1 1 2 kA = 2 mvmax vmax = 2πA kA = m T 2 m T = 2π k Equations of Motion • x = xmax cos θ • x = A cos θ • v = -vmax sin θ • v = -Aω sin θ • a = -amax cos θ • a = -A ω2 cos θ Pendulum in SHM 2 • Ft = - m g sin θ • Ft = - m g θ=-mgs Ft = - m g k= s = − ks L mg L T = 2π T = 2π m mg L L g T = 2π m k Equations of Motion • What are the assumptions for which these equations can be used? • What if you have a different situation? x=A cos (2πƒt) = A cos ωt v = -2πƒA sin (2πƒt) = -A ω sin ωt a = -4π2ƒ2A cos (2πƒt) = -Aω2 cos ωt 2 1/3/2012 A Problem The motion of an object is described by the equation Find • (a) the position at t = 0 and t = 0.60 s, • (b) the amplitude • (c) the frequency • (d) the period (a) 0.30 m, 0.24 m (b) A Problem πt x = (0.30m) cos 3 x = A cos θ • A 50.0-g object is attached to a horizontal spring with a force constant of 10.0 N/m and released from rest with an amplitude of 25.0 cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless? Things that are constant: Things that are not constant: ω= Position θ Mass x Mass velocity Mass Accel α= vt= 0 t= m k T = 2π .05kg = 2π (.0707 s) = 0.44s 10 Nm f = .71Hz A = .25m 0.30 m (c) 1 6 Hz (d) 6.0 s Solve for everything Times T = 2π ar= 12.5 cm f= Reading SHM Graphs • SHM_graphs.xlsx • Look at the graphs and find: T, f, A, ω, α, vt, ar • At a given time find for the shadow: T, θ, x, v, a • If k = 5 N/m, what is the mass on the spring? T= t=0.44 sec A= Problem #16 Solve for everything Motion on a Spring Situation Position (+ to Right) velocity acceleration Pulled to right and X=+A released (speed up) 0 Max to left (-) Moving left speed up Left (-) (-) 0<x<+A ω=14.13 rad/sec Position Times Move L, at center X=0 Left (-) Max 0 Move L (slows) -A<x<0 (-) (+) Far to Left change direction -A 0 Max (+) Moving Right speed -A<x<0 up (+) (+) Move R at center X=0 (+) Max 0 Move R slows 0<x<+A (+) (-) At far right X=+A Things that are constant: Things that are not constant: t=0 θ Mass x 0 25 cm Mass velocity 0 Mass Accel -50 m/s2 α=0 vt=3.53m/s ar=50 m/s2 (-) Max t=0.074 sec π/3 60° t=0.44 sec 2π 360° 12.5 cm -3.06 m/s -25 m/s2 F=2.25Hz T=0.44 s 25 cm 0 -50 m/s2 A=25 cm Completed cycle 3