A 0.750 kg air-track glider is attached to each end of the track by two coil spr
ID: 1415077 • Letter: A
Question
A 0.750 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.500 N to displace the glider to a new equilibrium position, x= 0.190 m.
A)Find the effective spring constant of the system. 2.63 N/m
B)The glider is now released from rest at x= 0.190 m. Find the maximum x-acceleration of the glider.
6.67×10-1 m/s^2
C) Find the x-coordinate of the glider at time t= 0.730T, where T is the period of the oscillation.
D)Find the kinetic energy of the glider at x=0.00 m.
I Figured out A and B but i cant figure out C and D HELP!!!
Explanation / Answer
C.
x(t) = A*cos (wt + phi)
A = Xmax = 0.19 m
w^2 = k/m
to find the phi set t = 0
0.19 = 0.19*cos (0 + phi)
cos phi = 1
phi = 0 deg
x(t) = A*cos wt
T = 2*pi*sqrt(m/k)
0.730*T = 0.730*2*pi*sqrt(m/k)
t = 0.730*2*3.14*sqrt(0.75/2.63)
t = 2.448 sec
wt = [sqrt(2.63/0.75)]*2.448 = 4.584
x(t) = 0.19*cos wt
x(t) = 0.19*cos 4.584 = -0.0243 m
D.
using Energy conservation
max KE = max PE
the velocity is greatest at equilibrium so KE will be max at x = 0.00 m
max KE = 0.5*kA^2
max KE = 0.5*2.63*0.19^2
max KE = 0.0474 J
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