_lec34_ppt_ - Lecture 34 Apr 16, 2006 P112 Agenda for today...

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Unformatted text preview: Lecture 34 Apr 16, 2006 P112 Agenda for today New Subject: Simple Harmonic Motion 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... k m k m k m How the spring oscillator works You extend the spring and let it go The spring force and acceleration are to the left, so the speed increases as it approaches the equilibrium position The net force at that point is zero, but it overshoots the equilibrium position and compresses the spring At that point the process start over again. SHM Dynamics At any given instant we know that F = ma must be true. But in this case F = -kx d2x kx = ma = m 2 dt F = -kx k a m SO: x k d2x = x 2 m dt a differential equation for x(t)! SHM Dynamics... d2x k = x 2 dt m define = k m d2x = 2 dt 2 x Where is the angular frequency of motion Try the solution x = A cos( t) dx = dt Asin( t ) d2x = 2 Acos( t ) = 2 x dt 2 This works, so it must be a solution! What does the solution mean? The solution describes the time-dependent position of a body that undergoes simple harmonic motion. x(t) = A cos( t) A: Amplitude (e.g."how far did you compress the spring?") : "angular frequency" (in the case of the spring it depends on spring stiffness and mass of the body) SHM Dynamics... what does angular frequency have to do with moving back & forth in a straight line ?? y = R cos y 1 2 3 = R cos ( t) 1 x 0 1 2 3 2 4 5 6 -1 4 5 6 recap The equation that describes SHM is d2x = 2 dt 2 x The solution (which describes the x-position as a function of time of a body undergoing SHM) is x = A cos( t) For any oscillation you can determine a period T and a frequency f (in Hz) 1 f = T Frequency of SHM The angular frequency is related to the frequency by: 2 =2 f = T For the oscillating ideal spring only: k 2 = =2 f = m T Independent of amplitude A and phase p! Question 1: derive the acceleration and velocity of a body undergoing SHM! Question 1: derive the acceleration of a body undergoing SHM! d2x k ax = 2 = x m dt Why is it negative? SHM Solution x = A cos( t) is not a unique solution. x = A sin( t) is also a solution. The most general solution is a linear combination of these two solutions! x = B sin( t)+ C cos( t) dx = Bcos( t ) dt C sin( t ) d2x = 2 dt 2 Bsin( t ) 2 C cos( t ) = 2 x ok SHM solution x = A cos( t + ) is equivalent to x = B sin( t)+ C cos( t) Will use x = A cos( t + ) as our standard solution. SHM Solution... Drawing of A cos( t ) A = amplitude of oscillation T=2 / A 2 A 2 SHM Solution... Drawing of A cos( t + ) 2 2 SHM So Far The most general solution is x = A cos( t + ) where A = amplitude = angular frequency k = m = phase For a mass on a spring The frequency does not depend on the amplitude!!! We will see that this is true of all simple harmonic motion! The oscillation occurs around the equilibrium point where the net force is zero! Example 1 Draw the position of a block attached to a spring undergoing SHM as a function of time. Keeping all other parameters the same How does it change if A doubles? How does it change if k doubles? How does it change if m doubles? Example 2 A horizontal spring with a spring constant of k=200N/m is attached to a block of mass 0.5kg and stretched out to 0.02m. It starts to oscillate. Assume a frictionless surface. Find the angular frequency, frequency and period of the oscillation. k m Summary Simple Harmonic Motion ...
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This note was uploaded on 03/27/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell University (Engineering School).

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