PHY 317K 1st Edition Lecture 14 Outline of Last LectureI. ReviewOutline of Current LectureI. Review ProblemII. Oscillation CharacteristicsIII. Graph and function of displacement oscillationsIV. Velocity OscillationV. Acceleration OscillationVI. Relating Period and Amplitude Current LectureA uniform beam of weight L and mass m = 1.8 kg is at rest on two scales. A uniform block with mass u = 2.7 kg, is at rest on the beam with its center a distance L/4 from the beam’s left end. What do the scales read?A. Scale 1 is larger than scale 2B. Scale 1 is smaller than scale 2C. SameFnet = 0Tnet = 0Oscillations:- Energy is conserved- A is amplitude (amplitude does notaffect period)- Period: T is the period (the time per cycle)- Frequency: F = 1/T and the units are Hertz (Hz)- Omega (angular frequency or angularvelocity): w = 2 pi/T = 2 pi x f Displacement during oscillations:- X(t) = A cos theta- Starting point is A (amplitude) on thegraphVelocity graph of oscillations:- The derivative of the displacement graph- Starts at 0- Then has negative velocity- Then increases velocity- Maximum velocity is –Aw and +Aw- Function is equal to –Aw sin (wt)- Maximum velocity depends on both A and wAcceleration Graph of oscillations:- Derivative of velocity graph- d/dt (-Aw sin wt)- a (t) = -Aw^2 cos wtA mass attached to a spring oscillates back and forth as indicated in the position vs time plot. At point p, the mass has:A. Positive velocity and positive accelerationB. Positive velocity and negative accelerationC. Positive velocity and zero accelerationD. Negative velocity and positive accelerationE. Negative velocity and negative accelerationF. Negative velocity and zero accelerationG. Zero velocity but is still acceleratingH. Zero velocity and zero accelerationRelating Period and Amplitude:- E = ½ k a^2 (maximum energy) is equal to ½ mv^2 + ½ k x^2- Initial potential energy is equal to the kinetic and potential energy at a given time- If you isolate the equation above, v = sqrt ((k/m)(a^2 – x^2))- Vmax is when x = 0 so vmax = sqrt ((k/m)a^2)- Angular velocity is equal to the square root of k/m- T = 2 pi/w so w = 2pi/T- So 2 pi/T = square root k/m- Therefore T = 2 pi square root of
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