Question

A block of mass

m = 6.14 kg

is attached to a spring with spring constant

k = 1682 N/m

and rests on a frictionless surface. The block is pulled, stretching the spring a distance of 0.135 m, and is held still. The block is then released and moves in simple harmonic motion about the equilibrium position. (Assume that the block is stretched in the positive direction.)

(b) Where is the block located 3.24 s after it is released? (Give the displacement from the equilibrium. Include the sign of the value in your answer.)
m

(c) What is the velocity of the mass at that time? (Indicate the direction with the sign of your answer.)

Answer 1

here,

mass ,m = 6.14 kg

spring constant , K = 1682 N/m

amplitude ,A = 0.135 m

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

w = sqrt(1682/6.14) = 16.55 rad/s

the displacement , x(t) = A * cos(w*t)

b)

at t = 3.24 s

x(3.24) = 0.135 * cos(16.55 * 3.24)

x(3.24) = - 0.13 m

c)

let the velocity of mass at this time be v

using conservation of energy

0.5 *K * A^2 + 0.5 * K * x^2 + 0.5 * m * v^2

1682 * ( 0.135^2 - 0.13^2) = 6.14 * v^2

solving for v

v= 0.54 m/s

the velocity of mass is 0.54 m/s