Simple Harmonic Motion
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A 165 g mass attached to a horizontal spring oscillates at a frequency of 2.80 Hz. At t=0 seconds, the mass is at x=5.20 cm and has v_x = -42.0 cm/s. Determine:
A) The period in seconds
B) The angular frequency is rad/s
C) The amplitude in meters
D) The phase constant in rad
E) The maximum speed in m/s
F) The maximum acceleration in m/s^2
G) The total energy in J
H) The position at t=4.80 s in meters
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Solution Summary
The expert examines simple harmonic motion on springs.
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x(t) = Acos(wt+phi)
v(t) = dx(t)/dt = -wAsin(wt+phi)
a(t) = dv(t)/dt = -w2Acos(wt+phi) = -w2x.
f= 2.8 hz
A) The period in seconds
period= T=1/f= 0.3571 seconds
B) The angular frequency is rad/s
angular frequency= w =2pi f= 17.59 rad/s
C) The amplitude in meters
At t=o x= 5.2 cm and v = -42 cm/s
x(t) ...
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