A0.125-kg particle undergoes simple harmonic motion along the horizontal between the points =−0.299 m and =0.387 m. the period of oscillation is 0.581 s. find the frequency, the equilibrium position, , the amplitude, the maximum speed, , the maximum magnitude of acceleration, , the force constant, and the total mechanical energy, .

A 0.293-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.285 m and x2 = 0.395 m. The period of oscillation is 0.641 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.

That being said,

f=1/T=1/0.641=1.56 s

xeq=(x1+x2)/2=(-0.285+0.395)/2=0.055 m

A=(x2-x1)/2=(0.395+0.285)/2=0.340 m

x(t)=Acos(wt+ϕ)

v(t)=-wAsen(wt+ϕ)

w=2πf=2*3.14*1.56=9.8 rad/s

vmax=-wA=9.8*0.340=-3.3 m/s

a(t)=-w^2Acos(wt+ϕ)

amax=-w^2A=96.04*0.340=-32.6 m/s^2

w=(k/m)^1/2 => k=mw^2=0.293*96.04=28.1 N/m

Etot=1/2kA^2

Etot=0.5*28.1*0.116=1.63 J

i think the answer would be reserve.