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Simple Harmonic Motion. Spring motion. Let’s assume that a mass is attached to a spring, pulled back, and allowed to move on a frictionless surface…. k. m. k. m. k. m. Simple Harmonic Motion (SHM).
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Spring motion • Let’s assume that a mass is attached to a spring, pulled back, and allowed to move on a frictionless surface…
k m k m k m Simple Harmonic Motion (SHM) • We know that if we stretch a spring with a mass on the end and let it go, the mass will oscillate back and forth (if there is no friction). • This oscillation is called Simple Harmonic Motion, and is actually very easy to understand...
F = -kx a k m x SHM Dynamics • At any given instant we know thatF = mamust be true.But in this case F = -kx and F = ma • So: -kx = ma • F = - kx (Hooke’s Law) • Period is proportional to the square root of mass over spring constant
SHM Dynamics... y = Rcos =Rcos(t) • But wait a minute...what does angularvelocityhave to do with moving back & forth in a straight line ?? y 1 1 1 2 2 3 3 0 x 4 6 -1 4 6 5 5
SHM and Velocity and Acceleration • If you were to plot the position of the mass over time, it would look like a sine wave. • The velocity and acceleration would look a little out of phase
Problem: Vertical Spring • A mass m = 102 g is hung from a vertical spring. The equilibrium position is at y = 0. The mass is then pulled down a distance d = 10 cm from equilibrium and released at t = 0. The measured period of oscillation is T = 0.8 s. • What is the spring constant k? k y 0 -d m t = 0
z L m d mg The Simple Pendulum... • Period is proportional to the square root of the length over gravity • This works when the angle is less than 15o • The period does not depend on the mass of the object
Simple Harmonic Motion • You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T1. • Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T2. Which of the following is true: (a)T1 = T2 (b)T1 > T2 (c) T1 < T2
Solution Standing up raises the CM of the swing, making it shorter! L2 L1 T1 T2 Since L1 > L2 we see that T1 > T2 .
U K E U s -A 0 A Energy in SHM • For both the spring and the pendulum, we can derive the SHM solution by using energy conservation. • The total energy (K + U) of a system undergoing SHM will always be constant! • This is not surprising since there are only conservative forces present, hence K+U energy is conserved.
Simple Harmonic Motion • The End