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(II) A 1.60 -kg object oscillates from a vertically hanging light spring once every 0.55 s. (a) Write down the equation giving its position $y(+$ upward) as a function of time $t,$ assuming it started by being compressed 16 $\mathrm{cm}$ from the equilibrium position (where $y=0 )$ , and released. $(b)$ Howlong will it take to get to the equilibrium position for the first time? (c) What will be its maximum speed? (d) What will be its maximum acceleration, and where will it first be attained?

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a. $$(0.16 m) \sin (11.4) t$$b. $$0.137 \mathrm{sec}$$c. $$1.82 \mathrm{m} / \mathrm{s}$$d. $$20.8 \mathrm{m} / \mathrm{s}^{2}$$

Physics 101 Mechanics

Chapter 14

Oscillators

Motion Along a Straight Line

Motion in 2d or 3d

Periodic Motion

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Hope College

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Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

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In physics, an oscillation is the repetitive variation, typically in time, of some measure about a central value or between two or more different states. The oscillation may be periodic or aperiodic.

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(II) A 1.60 -kg object osc…

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so for party we want right when we want to. Ah, right. The equation. You know that the amplitude is going to be equal 2.16 meters and we know that omega the angular velocity will be equal to two pi divided by t the period. And this will be equal to two pi divided by 0.55 seconds. And this is equaling 11.4 radiance per second. Therefore, why the function for this placement in terms of time will be equal. Why of tea will be equal to 0.16 meters and a Times co sign of 11.4 t and so this will be your answer for a party. Part B is asking us to find the time it takes to reach equilibrium. This will be equal to the period divided by four. So this would be equal to 0.55 seconds divided by four and this legal 40.1375 seconds and this will be your final answer for part B. This will be the amount of time it takes to reach equilibrium and then for part C, we want to find the maximum velocity. This will be equal to the angular velocity times the amplitude and this is equaling 11.4 radiations per second and then times the amplitude of 0.16 meters. And this is giving us 11.4 0.16 1.8 to 4 meters per second. And lastly for party, we want to find ah, the maximum acceleration at the release point. So this will be equal to omega squared times the amplitude. So this will be 11.4 radiance for a second quantity squared times the amplitude of 0.16 meters. And we find that the maximum acceleration at the release point is gonna be equal. Took 20 uh, point seven 94 meters per second squared. So this would be our final answer for party. That is the end of the solution. Thank you for watching.

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