Question

A simple harmonic oscillator consists of a block of mass 3.70 kg attached to a spring of spring constant 260 N/m. When t = 1.60 s, the position and velocity of the block are x = 0.199 m and v = 3.920 m/s. (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t = 0 s?

Answer 1

Spring constant, k = 260 N/m
Mass of the block, m = 3.70 kg
Angular velocity, = SQRT[k/m]
= SQRT[260 / 3.70]
= 8.38 rad/s

Consider the position of the block as
x = A sin( * t + C) ...(1)
Where A is the amplitude, t is the time, and C is a phase constant.   
At t = 1.60 s, x = 0.199 m
0.199 = A * sin(8.38 * 1.60 + C) ...(2)

Differentiating (1),
Velocity, dx/dt = A * cos(t + C)
At t = 1.60 s, v = 3.92 m/s
3.92 = 8.38 * A * cos(t + C)
0.468 = A * cos(t + C) ...(3)

a)
(2)^2 + (3)^2 gives
0.199^2 + 0.468^2 = A^2 * cos^2(t + C) + A^2 * sin^2(t + C)
0.258 = A^2 * [cos^2(t + C) + sin^2(t + C)
0.258 = A^2 * 1
A = SQRT[0.258]
= 0.508 m

b)
Substituting A in (2),
0.199 = 0.508 * sin(8.38 * 1.60 + C)
0.392 = sin(13.41 + C)
13.41 + C = sin^-1(0.392)
13.41 + C = 0.402
C = - 13.01 rad

At t = 0,
x(0) = A * sin( * 0 + C)
= 0.508 * sin(-13.01) [Angle is in radians]
= - 0.218 m

c)
At t = 0,
v(0) = A * cos( * 0 + C)
= 0.508 * 8.38 * cos(-13.01)
= 3.85 m/s