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An object is undergoing simple harmonic motion with period 1.210 s and amplitude 0.610 m. At t equals 0 the object is at x equals 0. How far is the object from the equilibrium position at time 0.485 s?
Wiki User
∙ 14y ago
I got -0.495 m. I can't promise you this is correct, but here's my method:
the position as a function of time is x(t)=A*cos(sqrt(k/m)*t)
you already have A and t values, and you can solve for sqrt(k/m) by using the period they gave you.....
T=2pi/(sqrt(k/m))
sqrt(k/m)=2pi/T
Plug and chug. Bada bing.
Wiki User
∙ 14y ago
At time t = 0.485 s, the object is 0.610 m away from the equilibrium position, because the displacement at any time is given by x(t) = A * sin(2πt/T), where A is the amplitude and T is the period. Substituting t = 0.485 s, A = 0.610 m, and T = 1.210 s into the equation will give you the position of the object at that time.
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The height of a wave is twice the measure of its amplitude. Amplitude is the measure of a wave's maximum displacement from its equilibrium position.
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It measures the maximum displacement of a particle in the wave from the equilibrium position. The equilibrium position is also known as the resting position. When displacement of the particle equal to the amplitude, we say that it is at the amplitude, where it is furthest from its equilibrium position.
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The amplitude of the oscillation, which is the maximum displacement from the equilibrium position, is determined by the mass of the hanging object and the spring constant. The maximum displacement occurs when the object is released from its initial position and the system is in simple harmonic motion.
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