A 5 kg mass is connected to a spring (950 N/m) platform which undergoes simple h

Question

A 5 kg mass is connected to a spring (950 N/m) platform which undergoes simple harmonic motion in the vertical direction. The amplitude of the oscillation is 0.3 m. At t=0 the box is 0.3 m above the equilibrium position.

A)What is the period of oscillation?

B)What is the angular frequency of the box?

C)What is the speed of the box at the equilibrium position

D)What is the total energy of the 5 kg box?

E)What is the acceleration of the box after 5 s (assume the box is at +A m at t = 0 s, and UP is positive)?

Explanation / Answer

a) T = 2*pi root (m/k) = 2*pi* root (5/950) = 0.455 sec

b) angular frequency w = 2*pi /T = 13.80 units

b) let the equation is x =A sin13.8t + c , at t=0 , 0.3 = 0.3 sin 13.8t + c, thus c=0.3

so equation becomes x = 0.3 sin13.8t + 0.3

d)total energy is given by 0.5 kA^2 = 0.5 x 950 x 0.3^2 = 42.75 J

e) putting t= 5sec, x = 0.3 sin 69.01 + 0.3 = 0.58 m, so acceleration of the box after 5 sec = w^2 x = 110.4 m/sec^2

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