Amplitude of a simple harmonic motion equation

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The discussion focuses on deriving the amplitude of a simple harmonic motion from the equation x(t) = Acos(ωt + ϕ). Given a 200 g block with a spring constant of 10 N/m, the initial position is 20 cm below equilibrium, and it has an upward velocity of 100 cm/s at t = 0s. The formula A = sqrt((x0)^2 + (v0/ω)^2) is used to calculate the amplitude, where x0 is the initial displacement and v0 is the initial velocity. The derivation involves squaring the velocity equation v = dx/dt = -Aωsin(ωt + φ) and simplifying it to relate amplitude, position, and velocity. Understanding this derivation clarifies how the amplitude is determined in simple harmonic motion.
nehcrow
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if you have the equation: x(t) = Acos(ωt + ϕ)

and you have the following information: A 200 g block hangs from a spring with spring constant 10 N/m. At t = 0s the block is 20 cm below the equilibrium position and moving upward with a speed of 100 cm/s.

The answers give a short method of calculating the amplitude but I have no idea how the reached it, basically what they did was: A = sqrt((x0)^2 + (v0/ω)^2)
how did they reach that conclusion??
 
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velocity v = dx/dt = -Aωsin(ωt + φ)
Square both the sides and simplify.
 
Thanks!
 

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