You are watching an object that is moving in SHM. When the object is displaced 0

Question

You are watching an object that is moving in SHM. When the object is displaced 0.590 m to the right of its equilibrium position, it has a velocity of 2.40 m/s to the right and an acceleration of 8.60 m/s2 to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

Explanation / Answer

in SHM, f = -kx => ma = -kx => k = -ma/x applying conservation of mechanical energy 1/2mv^2 - 1/2kx^2 = - 1/2kX^2 where, m = mass of the object v = velocity of the object k = spring constant x = initial position of the shm X = amplitude of shm => 1/2mv^2 - (1/2)(-ma/x)x^2 = - (1/2)(-ma/x)X^2 => 1/2mv^2 + 1/2max = 1/2maX^2/x => v^2 + ax = aX^2/x => 2.4^2 + 8.6*0.59 = 8.6X^2/0.59 => 5.76 + 5.074 = 14.576X^2 => X^2 = 10.834/14.576 => X^2 = 0.743 => X = 0.862 m distance travelled by object before it stops = 0.862 - 0.590 = 0.272 m

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