Question
In: Physics
A mass of 1.62 kg stretches a vertical spring 0.318 m. If the spring is stretched...
A mass of 1.62 kg stretches a vertical spring 0.318 m. If the spring is stretched an additional 0.137 m and released, how long does it take to reach the (new) equilibrium position again? ______s
Solutions
Expert Solution
The spring constant of the spring is,
k = mg/y
= (1.62 kg)(9.8 m/s2) / (0.318 m)
= 49.92 N/m
The angular frequency is,
w = [k / m]1/2
= [(49.92 N/m) / 1.62 kg]1/2
= 5.55 rad/s
The period is,
T = 2(pi)/ w
= 2(pi) / w
= 2(pi) / 5.55 rad/s
= 1.13 s
The required, time period is T/4 = 0.283 s
genius_generous answered 3 months agoRelated Solutions
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