A block with a mass of 0.600 kg is connected to a spring, displaced in the posit

Question

A block with a mass of 0.600 kg is connected to a spring, displaced in the positive direction a distance of 50.0 cm from equilibrium, and released from rest at t = 0. The block then oscillates without friction on a horizontal surface. After being released, the first time the block is a distance of 35.0 cm from equilibrium is at t = 0.200 s.

(a) What is the block's period of oscillation?
s

(b) What is the the value of the spring constant?
N/m

(c) What is the block's velocity at t = 0.200 s? (Indicate the direction with the sign of your answer.)
m/s

(d) What is the block's acceleration at t = 0.200 s? (Indicate the direction with the sign of your answer.)
m/s2

Explanation / Answer

Given that

The mass block (m) =0.600kg

The block displaced to a distance of (A) =50cm =0.50m

a)

We know that

x =Acoswt

At t =0.200s it displaced to a distance of x =35cm =0.35m

0.35 =0.50cos(w*0.200)

45.572*0.0174533rad=w*0.200

w=0.7953/0.200=3.976rad/s

We know that

w =2pi/T then the time period T =2pi/w =6.28/3.976rad/s=1.579s

b)

The value of spring constant is w2 =k/m

k =w2*m =(3.976rad/s)2*(0.600kg)=9.485

c)

The block's velocity is given by

v =dx/dt =d(Acoswt)/dt =-Awsinwt=-(0.50)(3.976rad/s)sin(3.976rad/s*0.200)

=0.0275m/s

d)

We know that

F =-kx

The accelration of the block is given by

ma=-kx

a =-(k/m)x=(9.485/0.600)(0.500) =7.906m/s2

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