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)

For the best answers, search on this site https://shorturl.im/zwRmd

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