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 of25.0 cm from equilibrium is at t = 0.200 s.

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

(b) What is the the value of the spring constant?

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

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

Explanation / Answer

m = 0.6 kg

A = 50 cm = 0.5 m


a))
we know, x = A*cos(w*t)

at t = 0.2 s, x = 0.25 m

0.25 = 0.5 *cos(w*0.2)

1/2 = cos(w*0.2)

cos(pi/3) = cos(w*0.2)


==>w = pi/0.6

==>2*pi/T = pi/0.6

==> T = 2*0.6

==> T = 1.2 s

b) w = sqrt(k/m)

pi/0.6 = sqrt(k/m)


==> k = m*(pi/0.6)^2

==> k = 0.6*(pi/0.6)^2

==> k = 16.43

c)

v = dx/dt

v = -A*w*sin(w*t)

= -0.5*(pi/0.6)*sin(pi*0.2/0.6)

= -2.26 m/s

d) F = -k*x

m*a = -k*x

a = -k*x/m

= -16.43*0.25/0.6

= -6.85 m/s^2

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