# velocityA 50.0 g object is attached to a horizontal spring with a force constant of 10.0 N/m and released from rest with an amplitude of 25.0 cm.  What is the velocity of the object when it is...

velocity

A 50.0 g object is attached to a horizontal spring with a force constant of 10.0 N/m and released from rest with an amplitude of 25.0 cm.  What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?

neela | High School Teacher | (Level 3) Valedictorian

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F = -10x, where 10/m is the spring constant, when the object is x distance from the equilibrium point, (or the middle point of the 25 cm amplitude).

Therefore, the acceleration d2x/dt^2 = (F/mass) = --10/(50gm) = 10x/(0.05) = -200x.

Therefore, d/dt (v) = -200x, where v =dx/dt is the velocity of the object.

d/dt(v) = (dv/dx)(dx/dt) = vd/dx= -200x. Or. vdv =-200x*dx

( v^2)/2 =( -200x^2)/2 + C  on integration.

But when  v=12.5cm= 0.125m, v =0. Therefore,

v^2 = 200*(0.125^2-x^2 ). When x is the position of the particle from the mean position

At the half way of equilibrium position(or mean position), x=0. So  v = [200(0.125^2)]^(1/2) =  1.7678 m/s