A block of mass m is attached to the lower end of a vertical spring of force constant k . The block is held such that the spring is in its unstretched position, and then released from rest. Let y be the downward displacement of the block. Which of the following correctly describes the graph of the block's velocity v versus displacement y during its first downward motion?

Step-by-step solution

By the principle of conservation of mechanical energy between the release point ( y=0 ) and a downward displacement y : K + U = 0 ( 1 2 mv^2 - 0 ) + ( 1 2 ky^2 - mgy ) = 0 1 2 mv^2 = mgy - 1 2 ky^2 v^2 = 2gy - k m y^2 Rearranging the terms by completing the square: v^2 = - k m (y^2 - 2mg k y ) v^2 = - k m [ (y - mg k )^2 - ( mg k )^2 ] v^2 + k m (y - mg k )^2 = mg^2 k Dividing by mg^2 k : v^2 ( mg^2 k ) + (y - mg k )^2 ( m^2g^2 k^2 ) = 1 This is the equation of an ellipse. Since we are considering only the downward