A 0.291-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.275 m and x2 = 0.493 m. The period of oscillation is 0.659 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.
Respective values are in units
f= Hz xeq=m A=meters vmax= m/s amax= m/s^2 k=N/m Etot=J

Question

A 0.291-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.275 m and x2 = 0.493 m. The period of oscillation is 0.659 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.
Respective values are in units 
f= Hz xeq=m A=meters vmax= m/s amax= m/s^2 k=N/m Etot=J

anonymous · Answer

First let us find the frequency. $$f=\frac{1}{T}$$ frequency is the reciprocal of the period. The period is 0.659 seconds so the frequency is 1/0.659 = 1.52 hertz.

Next the equilibrium position. The equilibrium position is exactly halfway between the the points x1 and x2. This halfway point is (x1+x2)/2 = (-0.275+0.493)/2 = 0.109 meters

The amplitude is the distance between the equilibrium position and the maximum position. That is equal to (x2-0.109) = (0.493-0.109) = 0.384 meters. 

Can you try the rest by yourself?