Simple Harmonic Motion Amplitude

[Teclis]

Homework Statement

This is problem 21 b) from Section 8.1 of Marsden and Weinstein Calculus II

A mass of 1 kg is hanging from a spring. If x = 0 is the equilibrium position, and x = 1 and x' = 1 when t = 0. The wave is observed to oscillate with a frequency of twice a second. What is the amplitude of the wave?

Relevant Equations

x = Acos(wt) + Bsin(wt)

Using A = x₀, B = v₀/ω

I get

ω = 4π, A = 1, B = 1/4π

then converting to phase/magnitude form

\sqrt{A^{2} + B^{^{2}}} = \alpha

\sqrt{1^{2} + \left ( \frac{1}{4\pi }\right )^{^{2}}} = \alpha = \frac{1}{4\pi }\sqrt{16\pi^{2} +1}

However the answer in the back of the book has

α = 1

Is the answer in the back of the book incorrect? If the books answer is correct, could someone please point out my mistake?

 

[PeroK]

Homework Statement:: This is problem 21 b) from Section 8.1 of Marsden and Weinstein Calculus II

A mass of 1 kg is hanging from a spring. If x = 0 is the equilibrium position, and x = 1 and x' = 1 when t = 0. The wave is observed to oscillate with a frequency of twice a second. What is the amplitude of the wave?
Homework Equations:: x = Acos(wt) + Bsin(wt)

Using A = x₀, B = v₀/ω

I get

ω = 4π, A = 1, B = 1/4π

then converting to phase/magnitude form

\sqrt{A^{2} + B^{^{2}}} = \alpha

\sqrt{1^{2} + \left ( \frac{1}{4\pi }\right )^{^{2}}} = \alpha = \frac{1}{4\pi }\sqrt{16\pi^{2} +1}

However the answer in the back of the book has

α = 1

Is the answer in the back of the book incorrect? If the books answer is correct, could someone please point out my mistake?

If $\alpha = 1$ means that the amplitude is $1$ unit, then at $t=0$ the spring would be at its maximum displacement ($x(0) = 1$ unit). That contradicts $x'(0) \ne 0$.

I get what you get.