A 1.9 kg object oscillates at the end of a vertically hanging light spring once

Question

A 1.9 kg object oscillates at the end of a vertically hanging light spring once every 0.60 s .

Part A

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

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

y(t)=(0.16m)cos(2t0.60s)

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?

Part D

What will be the object's maximum acceleration?

Part E

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

y(t)=(0.16m)cos(0.60st) y(t)=(0.16m)sin(2t0.60s) y(t)=(0.16m)cos(t0.60s)

y(t)=(0.16m)cos(2t0.60s)

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?

Part D

What will be the object's maximum acceleration?

Part E

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

Explanation / Answer

here,

m = 1.9 kg

time period , T = 0.6 s

f = 1/T = 1/0.6 = 1.67 Hz

amplitude , A = 0.16 m

a)

as w = 2*pi*f

y(t) = A*cos(w*t)

the equation of motion is y(t) = (0.16)*( cos(2*pi*t*1.67))

b)

for first equilibrium position , t = T/2

t = 0.6/2

t = 0.3 s

the time taken by it for the first equilibriium poistion is 0.3 s

c)

maximum speed , Vmax = 2*pi*f*A

Vmax = 2*pi*1.67*0.16

Vmax = 1.67 m/s

d)

the maximum accelration , amax = A*( 2*pi*f)^2

amax = 0.16 * ( 2*pi*1.67)^2

amax = 17.6 m/s^2

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