Asked by Sandhya
A mass m = 1.50 kg oscillates on an ideal,massless horizontal spring with a constant k = 73.5 N/m.The amplitude of oscillation is 10.0 cm.The system is frictionless. The positive x-direction is to the right. What is the mass's natural period of oscillation? At t = 0, the mass is located at x = 0 and is moving to the left.What is the mass's position and velocity as functions of time? If at t = 0, the mass is located at x = -5.00 cm and is moving to the right. What is the mass's position as a function of time?
Answers
Answered by
drwls
The period of oscillation is
P = 2 pi sqrt(m/k)= 0.898 s
For your other questions, choose a value of the phase angle phi such that
x = 10 cm * sin (2 pi t/P + phi)
and the direction of motion (the sign of dx/dt) is correct.
Fot the x=0 @ t=0 case, phi = pi
P = 2 pi sqrt(m/k)= 0.898 s
For your other questions, choose a value of the phase angle phi such that
x = 10 cm * sin (2 pi t/P + phi)
and the direction of motion (the sign of dx/dt) is correct.
Fot the x=0 @ t=0 case, phi = pi
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