A 2.5kg object oscillates at the end of a vertically hanging light spring once e

Question

A 2.5kg object oscillates at the end of a vertically hanging light spring once every 0.65s .

Part A)

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

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

y(t)=(0.18m)?cos(t0.65s) y(t)=(0.18m)?cos(2?t0.65s) y(t)=(0.18m)?sin(2?t0.65s) y(t)=(0.18m)?cos(0.65s?t)

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

Part C) What will be its maximum speed?

Part D) What will be the object's maximum acceleration?

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

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

equilibrium point release point

Explanation / Answer

we have amplitude is 0.18 cm

and T= 0.65 s

so T=2pi/w

so w= 2pi/0.65

so equation is

y(t)= Acoswt

y(t)=0.18cos(2pi/0.65 t )

so correct option is B

answer- B

b). to get to equillibrium position for first time it will take T/4 sec

= 0.65/4= 0.1625 s

C). Maximum speed

1/2KA^2= 1/2K*Vmax^2

so Vmax= wA

=2pi/0.65*0.18

=1.739 m/s

d). we have K= w^2*m

=2.58(2pi/0.65)^2

=233.36 m/s

so

F= KA

so max acce= KA/m

= w^2*A

=16.8 m/s^2

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