Mastering Physics Solutions: Exercise 13.12

Mastering Physics Solutions: Exercise 13.12

On December 27, 2011, in Chapter 13: Vibrations and Waves, by Mastering Physics Solutions

Exercise 13.12

Part A = 1.2m/s Click to use the calculator/solver for this part of the problem
Part B = at the equilibrium position
Part C = 1.0m/s Click to use the calculator/solver for this part of the problem

Solutions Below:

A mass-spring system is in SHM in the horizontal direction.

Part A

If the mass is 0.25 kg, the spring constant is 15 N/m, and the amplitude is 15 cm, what is the maximum speed of the mass?

Express your answer using two significant figures.

Spring potential energy is given by:

U = 1/2kA2
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 = 1/2mv2
0.16875 = 1/2(0.25)v2
0.16875 = 0.125v2
1.35 = v2

v = 1.2m/s

Part B

Where does this occur?

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

What is the speed at a half-amplitude position?

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):

Umax = 0.16875J

PE at half amplitude:

U = 1/2kx2:
U = 1/2(15)(0.15/2)2
U = 0.0422J

Subtract from maximum PE:

KE = Umax – U
KE = 0.16875 – 0.0422
KE = 0.12655

Solve for velocity:

KE = 1/2mv2
0.12655 = 1/2(0.25)v2
0.12655 = 0.125v2
1.0124 = v2

v = 1.0m/s (2 significant figures)

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