Question

A mass-spring oscillator consists of a 3.45-kg block attached to a spring of spring constant 72.0 N/m. At time t 3.00 s, the position and the velocity of the block are x-0.160 m and v -4.05 m/s What is the amplitude of oscillation? 0.901 m You are correct reviousies r receipt no. is 157-144 What was the position of the block at t-0 Submit An Tries 6/10 Previous Tries swer What was the speed of the block at t 0?

Answer 1

given
m = 3.45 kg
K = 72 N/m
at t = 3s, x = 0.16 m, v = 4.05 m/s

a) let A is the amplitude of motion.

Apply 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(0.16^2 + (3.45/72)*4.05^2)

= 0.901 m

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

= sqrt(72/3.45)

= 4.57 rad/s

let phi is the initial phase,

at t = 3s,

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

0.16 = 0.901*sin(4.57*3 + phi)

0.16/0.901 = sin(4.57*3 + phi)

0.178 = sin(4.57*3 + phi)

sin^-1(0.178) = 4.57*3 + phi

0.1789 = 4.57*3 + phi

==> phi = 0.1789 - 4.57*3

= -13.5311 rad

at t = 0 s,

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

= 0.901*sin(4.57*0 - 13.5311)

= -0.740 m <<<<<<<<<<<-------------Answer

c) at t = 0s,

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

= 0.901*4.57*cos(4.57*0 - 13.5311)

= 2.34 m/s <<<<<<<<<<<-------------Answer