Question

A 2.8 ?? block attached to a massless spring and is displaced 35 ?? by a force of 927.5 ? across a smooth floor. The box is then let go and undergoes simple harmonic motion. Determine, a. The maximum acceleration of the block. b. The period of oscillation. c. The angular frequency. d. The speed at equilibrium. e. The velocity at ? = 16.2 ?�

Answer 1

here

mass , m = 2.8 kg

amplitude of motion , A = 35 cm

A = 0.35 m

force applied ,F = 927.5 N

let the spring constant be K

F = K * x

927.5 = K * 0.35

K = 2650 N/m

a)

The maximum acceleration of the block , a = F /m

a = 927.5 /2.8 m/s^2 = 331.25 m/s^2

b)

the period of oscillation , T = 2*pi * sqrt(m/K)

T = 2*pi * sqrt(2.8 /2650) s

T = 0.204 s

c)

the angular frequency , w= sqrt(K/m)

w = sqrt(2650 /2.8) = 30.8 rad/s

d)

the speed at equilibrium , vm = A * w

vm = 0.35 * 30.8 m/s = 10.77 m/s

e)

let the velocity be v1 when x1 = 16.2 cm = 0.162 m

using conservation of energy

0.5 * m * vm^2 = 0.5 * m * v1^2 + 0.5 * K * x1^2

2.8 * 10.77^2 = 2.8 * v1^2 + 2650 * 0.162^2

solving for v1

v1 = 9.54 m/s

the velocity is 9.54 m/s