A block is attached to a spring and rests on a horizontal, frictionless surface. The block is pulled to the right and released from rest whe

Question

A block is attached to a spring and rests on a horizontal, frictionless surface. The block is pulled to the right and released from rest when the clock starts counting and t = 0. The block oscillates back and forth about its equilibrium position, and the motion at any time is described by:x(t) = (3.5 cm)cos[(0.8 rad/s)t].What is the value of the t the second time the block goes through the equilibrium position?

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Dâu 3 years 2021-08-17T09:41:53+00:00 1 Answers 8 views 0

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  1. diemthu
    0
    2021-08-17T09:43:40+00:00
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    Answer:

    Explanation:

    x(t) = (3.5 cm)cos[(0.8 rad/s)t]

    Angular velocity ω = .8

    Time period = 2π / ω

    = 2 X 3.14 / .8

    = 7.85 s

    First time block will pass through equilibrium is T/4 ,

    Second time block will pass through equilibrium

    = 3.T/4

    = 3/4 x 7.85

    = 5.8875 s

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