Lecture26-2A_000

# Lecture26-2A_000 - Chapter 15 Chapter 15 Oscillatory Motion...

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Unformatted text preview: Chapter 15 Chapter 15 Oscillatory Motion III Oscillatory Motion III Chapter 9 Chapter 9 Gravitation Gravitation S H O S H O -- Velocity and Acceleration Velocity and Acceleration It is useful to consider the velocity and acceleration as it relates to the displacement. For this we will use the solution that includes the phase. The velocity is the first derivative: From this we see that the velocity is out of phase with the displacement. When the displacement is maximum, the velocity is zero. Similarly when the velocity is maximum the displacement is zero. The acceleration is the second derivative: The acceleration always has the opposite sign of the displacement, i.e. the object is under the influence of a restoring force! x t A cos t v t dx dt A sin t a t d 2 x dt 2 A 2 cos t Uniform Circular Motion and SHO Uniform Circular Motion and SHO We can think of SHO as the x component of an object undergoing circular motion with a uniform angular velocity w . In the figure q = w (t) and x = A cos q . The tangential velocity is w A and the x component of this velocity is proportional to sin q . Also from the figure we see that the x component of the velocity is pointing in negative direction when sin q is positive. Hence, v = - w A sin q . This should help you to understand why we used w t as the argument for the solution to the displacement of an object under a linear restoring force. Energy in Simple Harmonic Oscillations Energy in Simple Harmonic Oscillations For the mass spring system the potential energy is U(x) = k x 2 , where x is the displacement from equilibrium. The kinetic energy is K = m v 2 . Assuming that x = A cos( w t) we find: U 1 2 kA 2 cos 2 t K 1 2 m 2 A 2 sin 2 t owever w 2 = k / m . Hence the total energy is: E U K 1 2 kA 2 cos 2 t 1 2 kA 2 sin 2 t 1 2 kA 2 The kinetic and potential energy are out of phase so that when one is a minimum the other is a maximum and vice versa. Their total is a constant! Example: Vertical Mass Spring Example: Vertical Mass Spring There are now two forces acting on the mass m . The force of gravity and that due to the spring. The resulting differential equation from Newtons 2 nd is: mg kx m d 2 x dt 2 d 2 x dt 2 k m x g The solution x h = Acos( w t+ f ) satisfies: Simply adding x 1 = mg/k to the homogeneous solution yields the full solution: x x h x 1 A cos t mg / k The mass continues to oscillate at the same frequency as before. It simply oscillates about a new equilibrium position, x 1 = mg/k ....
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## This note was uploaded on 04/16/2008 for the course PHYS PHYS2A taught by Professor Hicks during the Spring '08 term at UCSD.

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Lecture26-2A_000 - Chapter 15 Chapter 15 Oscillatory Motion...

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