A partical is performing SHM and at t = (3/4)T , is at position (√3/2)A and moving towards the origin (where A is amplitude and T is the time period of the body in SHM). Equilibrium position of the partical is at x=0. Then what was the position and direction of the partical at t=0? While answering please show me the working that u did to get the answer. Thx a lot!

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, xcec, the amplitude, A, the maximum speed, Vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.

Explanation:

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

false br

higher potential

explanation:

because it is in a high speed so it must move into a of high potential