Chapter 13 Oscillatory Motion We continue our studies of mechanics but combine the concepts of translational and rotational motion We will revisit the ideal spring In particular we will re examine the restoring force of the spring and its potential energy We will consider the motion of a mass attached to the spring about its equilibrium position This type of motion is applicable to many other kinds of situations pendulum atoms planets Simple Harmonic Motion If we add a mass m to the end of the massless spring stretch it to a displacement x0 and release it The spring mass system will oscillate from x0 to x0 and back k x0 0 x0 Without friction and air m resistance the oscillation would continue indefinitely This is Simple Harmonic Motion SHM SHM has a maximum magnitude of x0 A called the Amplitude One way to understand SHM is to reconsider the circular motion of a particle and rotational kinematics The Reference Circle The particle travels on a circle of radius r A with the line from the center to the particle making an angle with respect to the x axis at some instant in time Now project this 2D motion onto a 1D axis x A cos but t x A x 0 x A x Therefore x A cos t x is the displacement for SHM which includes the motion of a spring A A SHM is also called sinusoidal motion xmax A x0 amplitude of the motion maximum is the angular frequency speed in rad s It remains constant during the motion As we saw previously and the period T are related 2 Define the frequency cycles 1 f sec T 2 2 f T T Units of 1 s Hertz Hz Relates frequency and angular speed frequency As an example the alternating current AC of electricity in the US has a frequency of 60 Hz Now lets consider the velocity for SHM again using the Reference circle v v t sin v t r A v A sin t Amplitude of the velocity is v max A x vt v A Acceleration of SHM a ar cos 2 2 2 vt r ar r r x a ar 2 2 2 ar r A a A cos t The amplitude of the acceleration is amax A Summary of SHM 2 xmax A t 0 2 x A cos t 3 v A sin t v max A t 2 2 2 a A cos t a A 2 t 0 2 max x A 1 t rads v A 1 t rads a A 2 1 Example Problem Given an amplitude of 0 500 m and a frequency of 2 00 Hz for an object undergoing simple harmonic motion determine a the displacement b the velocity and c the acceleration at time 0 0500 s Solution Given A 0 500 m f 2 00 Hz t 0 0500 s 2 f 2 2 00 Hz 4 00 rad s t 4 00 rad s 0 0500 s 0 200 rad 0 628 rad a x A cos t 0 500 m cos 0 628 rad x 0 405 m v A sin t v 0 500 m 4 rad s sin 0 628 rad v 3 69 m s 2 a A cos t 2 a 0 500 m 4 rad s cos 0 628 rad 2 a 63 9 m s Frequency of Vibration Apply Newton s 2nd Law to the spring mass system neglect friction and air resistance k m 0 Fs x A F ma x Consider x direction only Fs kx ma x Substitute x and a for SHM 2 k A cos t m A cos t 2 k m k Angular frequency of vibration for a spring with spring constant k and m attached mass m Spring is assumed x to be massless 1 2 f f 2 1 m T 2 T f k k m Frequency of vibration Period of vibration This last equation can be used to determine k by measuring T and m m 2 m T 4 k 4 2 k T 2 2 Note that f or T does not depend on the amplitude of the motion A Can also arrive at these equations by considering derivatives
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