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13 October, 00:43

An object is undergoing shm with period 0.900 s and amplitude 0.320 m. at t = 0 the object is at x = 0.320 m and is instantaneously at rest. calculate the time it takes the object to go (a) from x = 0.320 m to x = 0.160 m and (b) from x = 0.160 m to x = 0.

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  1. 13 October, 02:18
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    For an object at Simple Harmonic Motion (SMH), we can write the equation of motion as:

    x (t) = A cos[ (2 pi/P) t]

    where

    P is the period = 0.9

    t is time

    A is the amplitude = 0.32

    Therefore for this object we have the equation:

    x (t) = 0.32 cos[ (2 pi/0.9) t]

    A. Calculate t from x = 0.320 m to x = 0.160 m

    We know that at t = 0 the value of x would be: x=0.32 since cos 0 = 1. So we only have to find the value of t when x = 0.16m. Using the equation:

    0.16 = 0.32 cos[ (2 pi/0.9) t]

    0.5 = cos[ (2pi/0.9) t]

    pi/3 = (2 pi/0.9) t

    t = 0.15 s

    Therefore it takes 0.15 s to go from 0.32 to 0.16m

    B. Calculate t from x = 0.160 m to x = 0

    The time at which x = 0 is:

    0 = 0.32 cos[2 pi/0.90) t]

    pi/2 = (2 pi/0.90) t

    t = 0.225s

    Therefore it takes 0.225s-0.150s = 0.075s to go from 0.16 to 0 m.
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