(1 point) The graph shows the displacement from equilibrium of a mass-spring sys
ID: 2888255 • Letter: #
Question
(1 point) The graph shows the displacement from equilibrium of a mass-spring system as a function of time after the vertically hanging system was set in motion at time t 0 Assume that the units of time are seconds, and the units of displacement are centimeters. The first t-intercept is (0.25, 0) and the first maximum has coordinates (1.25, 3) 3 2 (a) What is the period T of the periodic motion? seconds (b) What is the frequency f in Hertz? What is the angular frequency w in radians/second? = | 1/4 Hertz radians / second (d) Determine the amplitude A and the phase angle (in radians), and express the displacement in the form y(t) = A cos(wt-7), with y in meters. Note from Whitaker: please try both pi, and pi-3.14159, because we are not sure why it sometimes wants one and sometimes the other meters (e) With what initial displacement y(0) and initial velocity y'(0) was the system set into motion? y(0) = meters y'(0) = second meters /Explanation / Answer
(a) The time difference between the starting of t and the maximum point is 1 sec, there would be 4 phases same as this one, therefore , the time period would be 4 sec
T = 4 sec
(b) The frequency of the mass-spring system would be 1/T = 1/4 sec-1
(c) The angular frequency of the mass-spring system would be 2/T = 2/4 = /2 radian
(d) As given that the maximum amplitude of the system is 3 hence, the equation can be written as
y(t) = 3 cos(t/2 - )
Plugging the given condition of (0.25,0) we get
y(0.25) = 3 cos(/8 - ) = 0
=> cos(/8 - ) = 0
=> = 5/8
Therefore, the equation becomes y(t) = 3 cos(t/2 - 5/8)
y'(t) = (3/2)cos(/8 - t/2)
=> y(0) = 3 cos(- 5/8) = -1.14
=> y'(0) = (3/2)cos(/8) = 4.3536
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