A 0.245-kg particle undergoes simple harmonic motion along the horizontal x-axis

Question

A 0.245-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.313 m and x2 = 0.437 m. The period of oscillation is 0.503 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.

f= ____________________ Hz

vmax= __________________ m/s

k= __________________ N/m

umax= ________________ J

U= ___________________ J

K= __________________ J

v= _________________ m/s

Explanation / Answer

f=1/T=1/0.503=1.98 s
xeq=(x1+x2)/2=(-0.313+0.437)/2=0.124 m
A=(x2-x1)/2=(0.437+0.313)/2=0.75 m
x(t)=Acos(wt+?)
v(t)=-wAsen(wt+?)
w=2?f=2*3.14*1.98=12.43 rad/s
vmax=-wA=12.43*0.75=-9.33 m/s
a(t)=-w^2Acos(wt+?)
amax=-w^2A=154.5*0.75=-115.87 m/s^2
w=(k/m)^1/2 => k=mw^2=0.245*154.5=37.85 N/m
Etot=1/2kA^2
Etot=0.5*37.85*0.56=10.6 J

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