Question

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall whle the other end is connected to a movable object. The spring and object are compressed by 0.082 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 11.9 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.041 m relative to its unstrained length?
_______ m/s

Answer 1

given that

   initial Energy =   U  = (1/2) k A²     where A= 0.082 m

the final energy is potential and kinetic, or

    K + U =   (1/2) mv²   + (1/2) kx²    where   x = 0.041 m

The final and initial energies are equal

    (1/2) k A²   = (1/2) m v²   + (1/2) kx²      

eliminate the (1/2) everywhere and divide everything bym

       (k/m)A²   =  v²    + (k/m)x²            

    ω² = k/m       so use this and

      ω²A²   = v² + ω² x²     andsolve for v

    v = [    ω²A²   - ω² x²    ]^1/2   =    ω [  A² - x² ]^1/2 =

         =  11.9 [ 0.082² - 0.041² ]^1/2 =     0.8450 m/s