A 0.400 -kg object attached to a spring with a force constant of 8.00 N/m vibrat

Question

A 0.400-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.2 cm. The maximum speed is 54.6 cm/s. CALCULATE THE MAXIMUM VALUE OF ITS ACCELERATION

A 1.50-kg object attached to a spring moves without friction (b = 0) and is driven by an external force given by the expression F = 5.60sin(2?t), where F is in newtons and t is in seconds. The force constant of the spring is 29.0 N/m. The resonance angular frequency of the system is 4.4 s^-1. The angular frequency of the driven system is 6.28 s^-1. FIND THE AMPLITUDE OF THE MOTION

Explanation / Answer

Solving for maximum acceleration:
a = - (k/m)Xo
a = - (8/0.400)(0.122)
a = - 2.44 m/sec^2

Amplitude

A=(Fo/m)/ (? (?^2??o^2)^2+(b?/m)^2)

b=0

wo^2 = k/m = 29/1.5 = 19.33
Angular frequency of the driving force =w=2*pi.

A=(5.6/1.5kg)/ (? ((2?)^2?(19.33)^2)+(0)^2 )

A=3.73/20.15=.185m = 18.5 cm

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