Question

A simple harmonic oscillator consists of a block of mass 3.90 kg attached to a spring of spring constant 120 N/m. When t = 1.00 s, the position and velocity of the block are x = +0.129 m and v = +3.415 m/s.

(a) What is the amplitude of the oscillations?
----- m

(b) What was the position of the mass at t = 0 s?
x = ----- m

(c) What was the velocity of the mass at t = 0 s?
v = ----- m/s

Answer 1

given

m = 3.9 kg

k = 120 N/m

angular frequency, w = sqrt(k/m) = sqrt(120/3.9) = 5.547 rad/s

a) let A is the amplitude of motion.

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

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

= sqrt(0.129^2 + 3.9*3.415^2/120)

= 0.629 m

b) use, x = A*cos(w*t + phi) (here phi is phase constant)

at t = 1s, x = 0.129 m

0.129 = 0.629*cos(5.547*1 + phi)

0.129/0.629 = cos(5.547 + phi)

0.205 = cos(5.547 + phi)

cos^-1(0.205) = 5.547 + phi

1.364 = 5.547 + phi

==> phi = (2*pi + 1.364) - 5.547

= 2.1 rad

so, at t = 0s,

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

= 0.629*cos( 0 + 2.1)

= 0.318 m

c) v = -A*w*sin(w*t + phi)

at t = 0s,

v = -0.629*5.547*sin(2.1)

= 3.01 m/s