Fig17-8(a), both potential and kinetic energies oscillate with time t and vary between zero and maximum value of-kx Both U and K vary with twice the frequency of the displacement and velocity 3. The total mechanical energy E is E=K+U=-kx 2 (17-14) E k K(x=E-U(x) K(x) Fig17-8(b)
• Fig17-8(a), both potential and kinetic energies oscillate with time t and vary between zero and maximum value of . • Both U and K vary with twice the frequency of the displacement and velocity. 3. The total mechanical energy E is 2 2 1 m kx 2 2 1 m E = K +U = k x (17-14) m x m − x K(x) U(x) x Fig 17-8 (b) 2 2 1 U(x) = kx K(x) = E −U(x) E
At the maximum displacement K=0,U=k At the equilibrium position U=0,K=k Eq(17-14)can be written quite generally as K+u=imy 2+-kx2=ikx. 2 (17-15) then v (17-16)
At the maximum displacement , . At the equilibrium position , . Eq(17-14) can be written quite generally as 2 2 1 m K = kx 2 2 1 m K = 0 U = kx U = 0 2 2 2 2 1 2 1 2 1 x m K +U = mv + k x = k x ( ) 2 2 2 x x m k vx = m − ( ) 2 2 x x m k vx = m − (17-16) (17-15) then or
2: Sample problem 17-2 Fg17-5 000000 X In Fig 17-5, m-2.43kg, k-22IN/m, the block is stretched in the positive x direction a distance of 11.6 cm from equilibrium and released. Take time t=0 when the block is released. the horizontal surface is frictionless (a)What is the total energy? (b) What is the maximum speed of the block? (c)What is the maximum acceleration? (d)What is the position, velocity, and acceleration att=0.215s2
Sample problem 17-2 In Fig 17-5, m=2.43kg, k=221N/m, the block is stretched in the positive x direction a distance of 11.6 cm from equilibrium and released. Take time t=0 when the block is released, the horizontal surface is frictionless. (a) What is the total energy? (b) What is the maximum speed of the block? (c) What is the maximum acceleration? (d) What is the position, velocity, and acceleration at t=0.215s? m x o • Fig 17-5