a particle oscilates between the points x = 40 mm and x = 160 mm with an acceler

Question

a particle oscilates between the points x = 40 mm and x = 160 mm with an acceleration a=k(100-x), where a and x are expressed in mm/s2   and mm, respectively, and k is a constant.  the velocity of the particle is 18mm/s when x=100 mm and is zero at both x=40 mm and x=160 mm. Determine A) the value of k, B) the velocity when x= 120mm

Explanation / Answer

a = k(100-x) mm/s2 at x = 100mm , v = 18mm/s v =0 at x = 40mm and 160mm dv/dt = k(100-x) (dv/dx)(dx/dt) = k(100-x) =>(dv/dx)v = k(100-x) (since dx/dt =v) =>vdv = k(100-x)dx integrating both sides we get (v^2)/2 = kx(100-x/2) + c now at x = 100mm , v = 18mm/s => 324/2 = 100k(100-50) +c =>c = 162 - 5000k => (v^2)/2 = kx(100-x/2) + 162 - 5000k v = 0 at x = 40 =>0 = 40k(100-20) + 162-5000k =>1800k = 162 => k = 0.09 B) velocity at x = 120mm v = 0.09*120(100-60) + 162 -5000*0.09 = 144in/s

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