Exercise 13.12

Part A = 1.2m/s

Part B = at the equilibrium position

Part C = 1.0m/s

Solutions Below:

Part A

Express your answer using two significant figures.

Spring potential energy is given by:

U = 1/2kA^{2}

U = 1/2(15)(0.15)^{2}

U = 0.16875J

At maximum speed, all of the potential energy will be converted to kinetic energy:

KE = 1/2mv^{2}

0.16875 = 1/2(0.25)v^{2}

0.16875 = 0.125v^{2}

1.35 = v^{2}

v = 1.2m/s

Part B

A. at a half-amplitude position

B. at a third-amplitude position

C. at a quarter-amplitude position

D. at the equilibrium position

D. at the equilibrium position

Part C

Express your answer using two significant figures.

Find potential energy at half-amplitude, subtract from maximum potential energy (to find kinetic energy) and then determine the velocity using the formula for kinetic energy:

Maximum PE (from Part A):

U_{max} = 0.16875J

PE at half amplitude:

U = 1/2kx^{2}:

U = 1/2(15)(0.15/2)^{2}

U = 0.0422J

Subtract from maximum PE:

KE = U_{max} – U

KE = 0.16875 – 0.0422

KE = 0.12655

Solve for velocity:

KE = 1/2mv^{2}

0.12655 = 1/2(0.25)v^{2}

0.12655 = 0.125v^{2}

1.0124 = v^{2}

v = 1.0m/s (2 significant figures)