A. What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8cm from equilibrium and then released f

Question

A. What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8cm from equilibrium and then released from rest, and whose period is 0.66s? Assume that the displacement at the start of the motion is positive. (express your answer in terms of t using two significant figures.)

B. What will be its displacement after 1.7s? (express your answer to two significant figures and include the appropriate units.)

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Linh Đan 3 days 2021-07-19T14:25:54+00:00 2 Answers 1 views 0

Answers ( )

    0
    2021-07-19T14:27:07+00:00

    Answer:

    a) x = 8.8 cm * cos (9.52 rad/s * t)

    b) x = 8.45 cm

    Explanation:

    This is a Simple Harmonic Motion, and most Simple Harmonic Motion equations start from the equilibrium point. In this question however, we are starting from the max displacement the equations, and thus, it ought to be different.

    From the question, we are given that

    A = 8.8 cm = 0.088 m

    t = 0.66 s

    Now, we need to find the angular speed w, such that

    w = 2π/T

    w = (2 * 3.142) / 0.66

    w = 6.284 / 0.66

    w = 9.52 rad/s

    The displacement equation of Simple Harmonic Motion is usually given as

    x = A*sin(w*t)

    But then, the equation starts from the equilibrium point at 0 sec, i.e x = 0 m

    When you have to start from the max displacement, then the equation would be

    x = A*cos(w*t).

    So when t = 0 the cos(0) = 1, and then x = A which is max displacement.

    Thus, the equation is

    x = 8.8 cm * cos (9.52 rad/s * t)

    At t = 1.7 s,

    x = 8.8 cos (9.52 * 1.7)

    x = 8.8 cos (16.184)

    x = -8.45 cm

    0
    2021-07-19T14:27:08+00:00

    Answer:  

    (a)      $y(t) = 8.8cos(\frac{2\pi }{0.665} t)$

    (b)  and after 1.7s the displacement would be

    $y(1.7s) = 8.8cos(\frac{2\pi }{0.665s}1.7s)$ = 8.456461cm

    Explanation:

    This is periodic motion and the best equation to model this type of motion is sinusoidal equation.

    We will chose cos function because it starts at the max or the amplitude.

    A = 8.8cm, T = 0.66s in following general form gives the answer.

    $y(t)=Acos(\frac{2\pi }{T} t)

    and substituting t = 1.7s in it gives us 8.456461cm, just little bit lower than initial position.

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