Suppose that a particle of mass 0.24 kg acted upon by a spring undergoes simple

Question

Suppose that a particle of mass 0.24 kg acted upon by a spring undergoes simple harmonic motion. The particle moves back and forth along the x axis between points x= -0.20 m and x= 0.20m. The period of the motion is 1.2s, frequency is .833Hz and angular frequency is 5.235, amplitude is 0.20m, and phase constant is 1.571 radians.

What is the total energy of this motion?
At what time is the kinetic energy zero? At what time is the potential energy zero?
At what time is the kinetic energy equal to potential energy?

Explanation / Answer

a)total energy=0.5m2 A2 =0.5*0.24*5.235*5.235*0.2*0.2=0.1315J

b)

time period=1.2s

at t=0s ,0.6s,1.2s,1.8s,2.4s............. PE is zero

at t=0.3s,0.9s,1.5s,2.1s............. KE is zero

c)KE=PE

0.5mv2 =0.5m2x2

0.5m2 A2 cos2(t-kx)=0.5m2 A2 sin2(t-kx)

cos(t-kx)=sin(t-kx)

t-kx=/2-t-kx

solve for t

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