A 1.8kg object oscillates at the end of a vertically hanging light spring once e

Question

A 1.8kg object oscillates at the end of a vertically hanging light spring once every 0.50s .

Part A

Write down the equation giving its position y (+ upward) as a function of time t . Assume the object started by being compressed 16cm from the equilibrium position (where y = 0), and released.

Part B

How long will it take to get to the equilibrium position for the first time?

Express your answer to two significant figures and include the appropriate units.

Part C

What will be its maximum speed?

Express your answer to two significant figures and include the appropriate units.

Part D

What will be the object's maximum acceleration?

Express your answer to two significant figures and include the appropriate units.

Part E

Where will the object's maximum acceleration first be attained?

Explanation / Answer

PART A:

Since it started at a compressed position, we use cosine,

y = Acos([2pi/T]t)

where

A = amplitude
T = period
t = time

Thus,

y = 16cos(2pi t/0.50 s) cm

y = 16cos(12.57 t s^-1) cm or y = 0.16cos(12.57 t s^-1) m

PART B:

It will be at equilibium in 1/4 of a cycle.

A cycle is 0.50 s.

Thus, it will next be in equilibrium after 0.050/4 s = 0.125 s

PART C:

Note that the maximum speed is

v_max = 2piA/T

Thus,

v_max = 2(3.1416)(16 cm)/(0.50 s)

v_max = 201 cm/s or 2.01 m/s

PART D:

The maximum acceleration, meanwhile, is

a_max = A(2pi/T)^2

Thus, plugging in,

a_max = 2527 cm/s^2 or 25.27 m/s^2

PART E:

Note that for a simple harmonic oscillator, maximum acceleration is attained at the highest compression.

Thus, it is at the RELEASE POINT where it attains maximum acceleration.

DONE! It was nice working with you!

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