Question

A 0.250 kg mass is attached to a spring with k=18.9 N/m. At the equilibrium position, it moves 2.89 m/s. What is the amplitude of the oscillation? (Unit=m)

Answer 1

Answer: y = R*sin(2.75*t - δ)

Here δ is just the time offset and for our purposes is pretty irrelevant.  You can in fact set it to zero since we can say we begin timing when the mass crosses equilibrium.  So

y = R*sin(2.75*t)

We want to find a way to use the information "At the equilibrium position, it moves 2.89 m/s."  I am going to use some calculus here since it makes things so much easier.  If you haven't taken calculus yet, most likely your course has given you a formula to use instead.

We know y=0 when t=0, so y is at equilibrium when t=0.  To say it moves 2.89 m/s is then to say that

y'(0) = 2.89.

From here we can differentiate the displacement function, set t=0 and solve for R.  Using the chain rule:

y'(t) = 2.75*R*cos(2.75*t)

y'(0) = 2.75*R

2.75*R = 2.89

R = 1.051

Explanation: Since this is harmonic motion we can assume there is no damping force.  The frequency of the oscillation is given by ω=√(k/m)=√(18.9/2.5)=2.75.  Keep in mind this is angular frequency, i.e. radians per second, not wavelengths per second.