The function x = (7.8 m) cos[(5πrad/s)t + π/3 rad] gives the simple harmonic motion of a body. At t = 4.4 s, what are the (a) displacement,

Question

The function x = (7.8 m) cos[(5πrad/s)t + π/3 rad] gives the simple harmonic motion of a body. At t = 4.4 s, what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?

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Mộc Miên 4 years 2021-07-21T17:01:21+00:00 1 Answers 18 views 0

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  1. kimcuc
    0
    2021-07-21T17:03:07+00:00
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    Answer:

    a.3.84m

    b.-106.67m/s

    c.947.3m/s^2

    d.70.17 rad

    e.2.5Hz

    d.0.4secs

    Explanation:

    Given x=(7.8)cos[5πrad/s)t+π/3)]

    a.Displacement at t=4.4

    7.8cos(5π*4.4+π\3)=3.84m

    b.velocity

    V= dx/dr=-5π(7.8)sin(5πrad/s)t+π\3

    at t=4.4

    -5π(7.8)sin(5π*4.4+π\3)=-106.67m/s

    c.acceleration

    a=d^2x/dr^2

    -(5π)^2(7.8) cos (5π*t+π\3)

    at t=4.4

    -(5π)^2(7.8)cos(5π*4.4+π\3)=-947.3m/s^2

    d. Phase =(5πrad/s)t+π\3

    At t=4.4

    5π×4.4+π\3=70.17 rad

    e.frequency

    Given x= 7.8cos(5πt+π\3

    Compare with x=Acos(2πft)

    2πft=5πt

    F=2.5Hz

    f.T=1\f

    T=1/2.5=0.4sec

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