At what time t1 does the block come back to its original equilibrium position (x=0) for the first time?
Express your answer in terms of some or all of the variables: A, k, and m.
I tried working out the problem myself but I came up with the wrong answer! My answer: (arccos 0^2 x m) / k
2 Answers

The differential equation for a harmonic oscillator is
m d²x(t)/dt² = kx
with the general solution
x(t) = A*cos(ωt +φ)
ω = √(k/m)
φ can be derived from initial condition. I think for this problem the oscillator started its motion at x=A, hence φ=0
x(t1) = A*cos(ωt1) = 0
or
cos(ωt1) = 0
The cosine is zero for
ωt1 = 1/2 π , 3/2 π, 5/2 π;7/2 π,…
hence
t1 = π / (2ω) = π/2 * √(m/k)

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Sorry, you seem to have missed out the equation, and probably some earlier parts of the question. try again?