SOLUTION: a particle undergoes simple harmonic motion with an amplitude of 5cm and an angular velocity of 10pi rad/s^-1 calculate the maximum velocity, the velocity when it is 2cm from the e

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Question 1013649: a particle undergoes simple harmonic motion with an amplitude of 5cm and an angular velocity of 10pi rad/s^-1 calculate the maximum velocity, the velocity when it is 2cm from the equilibrium position, the maximum acceleration of the particle and the period of oscillation
Answer by fractalier(6550) (Show Source):
You can put this solution on YOUR website!
In one-dimensional simple harmonic motion, the position of the particle can be found by

Here A = .05 m and w = 10pi, so that
x(t) = .05*sin (10(pi)t)
The velocity is the 1st derivative, or
v(t) = 5(pi)*cos(10(pi)t)
All you need is the time when it is 2 cm from equilibrium...you can find that from the x(t) equation...
To find the maximum velocity, take a 2nd derivative and set it equal to zero...
The period T = 1/f and w = 2(pi)f...you can plug in and find that...