A 5 kg mass connected to a spring oscillates at a frequency of 2.25HZ on a horiz

Question

A 5 kg mass connected to a spring oscillates at a frequency of 2.25HZ on a horizontal frictionless surface. The motion is started bydisplacing the mass .5m and giving it an initian velocity of 10 m/sback twords the equilibrium position. Find the spring constant, theamplitude of the motion, and an expression that describes thedisplacement as a function of time.

I already figured out the spring constant, and it was 999.698 andthat was the correct answer. however i got the wrong answer for theamplitude of motion any help would be appreciated.

Explanation / Answer

the angular frequency of the spring is w = (k/m)^(1/2) or w^2 = (k/m) or k = m * w^2 = m * (2f)^2 = 4^2 * m * f^2 m = 5 kg and f = 2.25 Hz the amplitude of the motion vmax= A * w or A = (vmax/w) -------------(1) the acceleration of the spring m * a = k * x or a = (k * x/m) x = 0.5 m the final speed of the spring v^2 - u^2 = 2a * x or v = (u^2 + 2a * x)^(1/2) u = 10 m/s Here,vmax= v from (1) A = (v/w) the expression for the displacement as a function oftime x = A * cos(wt + )

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