A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is connected to a mo

Question

A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is connected to a movable object. The spring and object are compressed by 0.080 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 12.1 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.041 m relative to its unstrained length

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MichaelMet 5 years 2021-07-14T04:41:06+00:00 1 Answers 19 views 0

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  1. haidang
    0
    2021-07-14T04:42:42+00:00
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    Answer:

    -0.831 m/s

    Explanation:

    With amplitude A = 0.08 m and angular frequency ω = 12.1 rad/s, we have the following simple harmonic equations of motion:

    Distance: x(t) = Acos(ωt) = 0.08cos(12.1t)

    Speed: v(t) = -Aωsin(ωt) = -0.08*12.1sin(12.1t) = -0.968sin(12.1t)

    At distance = 0.041, we can solve for 12.1t

    x(t) = 0.041

    0.08cos(12.1t) = 0.041

    cos(12.1t) = 0.5125

    12.1t = 1.033

    t = 0.085s

    Use it to plug into the v(t) equation

    v(0.085) = -0.968sin(12.1*0.085) = -0.968*sin(1.033) = -0.968*0.859 = -0.831 m/s

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