Conslants Part A A21 kg object oscillates at the end of a vertically hanging lig

Question

Conslants Part A A21 kg object oscillates at the end of a vertically hanging light spring once every 0.55 s. Write down the equation giving its position y (+upward) as a function of time t. Assume the object started by being compressed 18 ? f om the equilibrium position where V·0) and released Note: the eq i b i position is defined here as that location of the mass at rest when it is freely hung from the spring, not the unstretched position of the spring with no mass. Oft)-(0.18 m), cos( O y(t)-(0.18 m), sin( O s(t) (0.18 m).cos(0.55 s). t) Submit Part B How long will it take to get to the equilibrium position for the first time? Express your answer to two siqgnificant fiqures and include the appropriate units

Explanation / Answer

Part A -

The equation for the simple harmonic motion is -

y(t) = yo*cos(?t) , ? = 2?/T = 2?/0.55 rad/s, yo = 0.18, w = 11.42 rad/s

=> y(t) = 0.18*cos(2?t/0.55s)

First option is the correct answer.

Part B -

y(t) = 0.18*cos(2?t/0.55) . = 0.18*cos(11.42t) = 0

cos(11.42t) = 0

=>11.42t = ?/2

=> t = 0.137 s

Part C -

v(t) = y'(t) = - ?yosin(?t)

=> v(t) = - 11.42*0.18sin(11.42t)

=> v(t) = - 2.06*sin(11.42t)

This expression gets it's maximum value when sin(11.42t) = ±1

Vmax = ± 2.06 m/s

Part D -

a(t) = V'(t) = - ?²yocos(?t)

a(t) = - (11.42)²*0.18cos(11.42t)

=> a(t) = - 23.47cos(11.42t)

Now, the acceleration is maximum when cos(11.42t) = 1 [the maximum value for the cosine function is 1] .

So -

a(max) = - 23.47 m/s²

Part E -

The maximum acceleration shall be firstly attained when t=0s and y = yo = 0.18 m .

Means at the release point.

So, first option is the correct answer.

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