A .650-kg mass is attached to a spring of spring constant 13.0N/m along a horizontal frictionless surface. The object oscillates in simple harmonic motion and has an amplitude of 13.0 cm.

Question

a. What is the maximum speed of the mass?

b. What is the total energy of the system?

c. What is the speed of the object at a displacement of 6.5cm?

Answer 1

Let

m = 0.65 kg be the mass of object,

k = 13 N/m be the spring constant,

A = 13 cm = 0.13 m be the amplitude.

b.

When the object is at a displacement x and has velocity v, the total energy

of the system

E = E_s + E_k

where

E_s = kx²/2 is the energy of the spring, (1)

E_k = mv²/2 is the kinetic energy of the object. (2)

So

E = kx²/2 + mv²/2 (3)

At maximum displacement x=A and the speed v=0; substituted into (3)

E = kA²/2 (4)

Substituting actual numbers

E = 13×0.13²/2 = 0.10985J

a.

Maximum speed occurs when displacement x=0, and all the energy is in the kinetic energy E_k.

Then

E = E_k

Substituting from (2) and (4)

kA²/2 = mv²/2

From that

v = A√(k/m)

Substituting actual numbers

v = 0.13×√(13/0.65) = 0.58 m/s

c.

Given x = 6.5 cm = 0.065 m

use (3) and (4)

kA²/2 = kx²/2 + mv²/2

From that

v = √(k(A²-x²)/m)

Substituting actual numbers

v = √(13×(0.13²-0.065²)/0.65) = 0.5 m/s