A 2.0 kg mass, shown in the figure above, is attached to a horizontal Hooke\'s L

Question

A 2.0 kg mass, shown in the figure above, is attached to a horizontal Hooke's Law spring with a spring constant of 600 N/m and displaced 1.0 m to the right from the equilibrium point. The 2.0 kg mass is released and, at a position, 0.50 m to the left of the equilibrium point, it collides with a 4.0 kg mass and sticks. Both masses rest on a frictionless surface.

What is the frequency of the simple harmonic motion after the collision. Assume thast the dimensions of the masses do not need to be considered.

1.6 Hz

1.4 Hz

1.2 Hz

1.0 Hz

E. 0.80Hz

1.6 Hz

1.4 Hz

1.2 Hz

1.0 Hz

E. 0.80Hz

Explanation / Answer

f = (1/2pi) * sqrt( k/m)

here k = spring constant

m = taotal mass = 4 +2 = 6

f = (1/2pi) * sqrt( 600/6)

f = 1.5915 Hz = 1.6 Hz A option

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