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What is the equation for the position, at any point in time, of a 2 kg mass on a spring with force constant 200 n/m, if the maximum displacement from equilibrium is 2 cm?

User Windsooon
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Final answer:

The equation for the position, at any point in time, of a 2 kg mass on a spring with a force constant of 200 N/m is x(t) = (0.02) * cos √(200/2) * t + ϕ).

Step-by-step explanation:

The equation for the position, at any point in time, of a 2 kg mass on a spring with a force constant of 200 N/m can be derived using Hooke's law and the principles of simple harmonic motion (SHM). According to Hooke's law, a spring's force is proportionate to how far it deviates from its equilibrium position. The equation for the position as a function of time can be written as:
x(t) = A * cos(ωt + ϕ) where:
x(t) is the position of the mass at time t
A is the amplitude or maximum displacement from the equilibrium
ω is the angular frequency given by ω = sqrt(k/m) where k is the force constant and m is the mass
ϕ is the phase constant

In this case, the maximum displacement from equilibrium is 2 cm, which is equivalent to 0.02 m. Substituting the given values into the equation, we have: x(t) = (0.02) * cos(sqrt(200/2) * t + ϕ)

User Vjk
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