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A mass is attached to a spring of spring constant 60 N/m along a horizontal, frictionless surface. The spring is initially stretched by a force of 5.0 N on the mass and let go. It takes the mass 0.50 s to go back to its equilibrium position when it is oscillating. What is the frequency of oscillation?

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MechEngineer

Answer:

0.5 Hz

Explanation:

If it takes 0.5s to go back to the equilibrium position, that means it would take 1 s to reach maximum compression, and 2s to go back to the initial stretching position, aka complete 1 cycle of oscillation.

So the period is T = 2s. Then the frequency would be

[tex]f = \frac{1}{T} = \frac{1}{2} = 0.5 Hz[/tex]

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