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madchemist

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"A particle moves along the x-axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration -0.320 m/s^2. Assume it moves with simple harmonic motion for 4.50 s and x=0 is its equilibrium position. Find its position and velocity at the end of this time interval."

x=Amplitude * cos2(pi)ft

Found f using T= radical (x/a) * 2pi = 5.769 so f= 0.1734 Hz

However, even with solving for f I'm still left with two unknowns, i.e. A and t. Please help...

x=Amplitude * cos2(pi)ft

Found f using T= radical (x/a) * 2pi = 5.769 so f= 0.1734 Hz

However, even with solving for f I'm still left with two unknowns, i.e. A and t. Please help...

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