Question

A mass-spring oscillator consists of a 3.40-kg block attached to a spring of spring constant 103 N/m. At time t = 1.40 s, the position and the velocity of the block are x = 0.150 m and v = 3.18 m/s respectively. What is the amplitude of oscillation? What was the position of the block at t = 0? What was the speed of the block at t = 0?

Answer 1

A)
given
m = 3.4 kg
K = 103 N/m
at t = 1.40s, x = 0.15 m, v = 3.18 m/s

let A is the amplitude of motion.

use conservation of energy

(1/2)*k*A^2 = (1/2)*k*x^2 + (1/2)*m*v^2

A^2 = x^2 + (m/k)*v^2

A = sqrt(x^2 + (m/k)*v^2)

= sqrt(0.15^2 + (3.4/103)*3.18^2)

= 0.597 m <<<<<<--------------------Answer

b) angular frequency of the motion, w = sqrt(k/m)

= sqrt(103/3.4)

= 5.50 rad/s

let phi is the phase constant.

x = A*cos(w*t + phi)

at t = 1.40 s,

0.15 = 0.597*cos(5.5*1.4 + phi)

0.15/0.597 = cos(5.5*1.4 + phi)

0.251 = cos(5.5*1.4 + phi)

cos^-1(0.251) = 5.5*1.4 + phi

1.317 = 5.5*1.4 + phi

phi = 1.317 - 5.5*1.4

= -6.383 radians

at t = 0,

x = A*cos(w*t + phi)

= 0.597*cos(0 - 6.383)

= 0.597 m <<<<<<<<------------------Answer

speed of the block at t = 0, v = 0   <<<<<<<<------------------Answer

because it starts from positive extreme position.