Simple Harmonic Motion + derivatives?

Question

I need help with this physics problem.
Consider the function x(t) = Asin(ωt+θ)
a. find the first derivative of x(t)
b. find the second derivative of x(t)
c. by direct substitution show that x(t) = Asin(ωt+θ) satisfies the following equation: d^2x/dt^2 + [k/m]x=0

I have no idea where i should start with this problem.

Answer 1

a. First derivative:

x'(t) = d/dt (x(t)) = A*ω*cos(ωt+θ)

b. Second derivative:

x''(t) = d/dt (x'(t)) = - { A*(ω^2)*sin(ωt+θ) }

c. d^2x/dt^2 + [k/m]x=0

Start with LHS and prove that is equals RHS

x''(t) + (k/m)*x(t) = 0

- { A*(ω^2)*sin(ωt+θ) } + (k/m)*Asin(ωt+θ) = 0

cancelling out A & sin(ωt+θ) throughout the LHS, as both of them are not always equal to zero, the eqn smlifies to:

(k/m) - (ω^2) = 0

ω = sqrt(k/m)

Hence, the eqn is satisfied only under the condition that ω = sqrt(k/m)

Answer 2

Using the chain rule
x´(t) =A*w cos(wt+@) I don´t have mat signs

x´´(t)= -Aw^2sin(wt+@)

if k/m =w^2 -Aw^2sin(wt+@)+Aw^2sin(wt+@) is 0

Normally in harmonic motion you call k the resort´s constant and m the mass