A mass moving horizontally on a spring executes simple harmonic motion. The 0.15

Question

A mass moving horizontally on a spring executes simple harmonic motion. The 0.151-kg mass moves along the x-axis between the points x-0.283 m and x2-0.415 m, with a period of oscillation equal to 0.567 s. For this motion, determine the frequency, f, the equilibrium position, Xeg, the amplitude, A, the maximum speed, Vmax, the maximum magnitude of acceleration, amax, the spring constant, k, and the total mechanical energy, Eto Number Hz Number eq Number A= Im Number (Scroll down for more answer blanks.) m/s

Explanation / Answer

mass=m=0.151 kg

amplitude=(x2-x1)/2=(0.415-(-0.283))/2=0.349 m

time period=0.567 seconds

part 1:

frequency=1/time period=1/0.567=1.7637 Hz

part 2:

equilibrium position=x1+amplitude=-0.283+0.349=0.066 m

part 3:

amplitude=0.349 m

part 4:maximum speed=angular frequency*amplitude

=2*pi*f*A

=2*pi*1.7637*0.349=3.8675 m/s

part 5:

maximum acceleration=angular frequency^2*amplitude

=(2*pi*f)^2*A

=(2*pi*1.7637)^2*0.349

=42.858 m/s^2

part 6:

angular frequency=sqrt(k/m)

==>2*pi*1.7637=sqrt(k/0.151)

==>k=0.151*(2*pi*1.7637)^2

==>k=18.543 N/m

part 7:

total energy=0.5*mass*maximum speed^2

=0.5*0.151*3.8675^2

=1.1293 J

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