A 3 kg mass is connected to a spring (k = 710 N/m) platform which undergoes simp

Question

A 3 kg mass is connected to a spring (k = 710 N/m) platform which undergoes simple harmonic motion in the vertical direction. The amplitude of the oscillation is 10 cm. At t=0 the box is 10 cm above the equilibrium position. What is the period of oscillation? s What is the angular frequency of the box? rad/s What is the velocity of the box at the equilibrium position? m/s What is the total energy of the 3 kg block? J What is acceleration (magnitude and direction) of the box after 5 s (assume mass is at +A at t = 0s.)? m/s2

Explanation / Answer

T = 2*pi *sqrt(m/k) = 2*pi*sqrt(3/710) = 0.4 s w = 2*pi/f = 15.7 1/2*710*(10/100)^2 = 1/2*m*v^2 v = 1.54 m/s Energy = 1/2*710*(0.1)^2 = 3.55 At 5s = 0.2 s after oscillations. Acceleration = 710*0.1/3 = 23.67 m/s^2

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