A block with mass m = 200 g attached to a massless horizontal spring is oscillat

Question

A block with mass m = 200 g attached to a massless horizontal spring is oscillating with an amplitude of 2.0 cm and a frequency of 2.0 Hz. The spring is maximally stretched to 2 cm past its rest length at t=0. Let the positive x-axis point to the right, and the origin of the x-axis be at the rest length of the spring. What is the angular frequency of the block? Using this angular frequency, write an equation for the position of the block x(t) as a function of time. What is the total mechanical energy of the block-spring system? Draw a diagram of the kinetic energy of the block as a function of position. Be sure to label the x and y axes and indicate the maximum value of the kinetic energy as well as where it goes to zero. What is the maximum velocity of the block? Just as it passes through an equilibrium point moving to the right, a sharp blow directed to the left exerts a 20 N force for 1.0 ms. What is the new velocity of the block? What is the new amplitude of oscillation?

Explanation / Answer

1)

f = w / 2*pie

w = f*2*pie

w = 2 * 2 * 3.14 = 12.6 rad/s

x(t) = A * cos(wt + phy) at t = 0 0.02m / 0.02m = 0.02m * cos(0 + phy) / 0.02 m

x(t) = 0.02m * cos(12.6 rad/s t )

b)

Etot = 0.5 * k * A^2

w = sqrt( k / m)

w^2 = k/m = m*w^2 = k

k = 12.6^2 * 0.2 = 31.75 kg/s

Etot = 0.5 * 31.75 * 0.02^2 = 0.006 N

4)

E = 0.5 * m * Vmax^2 + 0 = 0.006 N

Vmax = sqrt( 2 *E / m)

Vmax = 0.252 m/s

5)

I = F * delta t

I = -20 * 0.001

I = -0.02 Ns

I = Pf - Pi

I = m*vf - m*vi ......... vi = Vmax

vf = (F * delta t + m*vi ) / m

vf = 0.152 m/s

6)

E = U + K

E = 0 + 0.5 * m * vf^2

E = 0.5 * k*A^2 = 0.5 * m * vf^2

A = sqrt( m*vf^2 / k)

A = sqrt( 0.5 * 0.152^2 / 31.75 )

A = 0.038 m = 3.8 cm

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