Question

A 1.33 kg object is attached to a horizontal spring of force constant 2.50 N/cm and is started oscillating by pulling it 6.40 cm from its equilibrium position and releasing it so that it is free to oscillate on a frictionless horizontal air track. You observe that after eight cycles its maximum displacement from equilibrium is only 3.70 cm .

(a) How much energy has this system lost to damping during these eight cycles?
(b) Where did the "lost" energy go? Explain physically how the system could have lost energy.

Answer 1

a.
`\Delta U=3.375\ N.cm=3.375* 10^(-2)\ J`

b. The energy lost during the damping is converted into kinetic energy of the molecules which is heat primarily.

Step-by-step explanation:

Given:

• mass of the attached object,
  `m=1.33\ kg`

• spring constant,
  `k=2.5\ N.cm^(-1)`

• maximum displacement,
  `A=6.4\ cm`

• maximum displacement after damping,
  `x=3.7\ cm`

a)

the energy lost in damping:

`\Delta U=(1)/(2) * k* A^2-(1)/(2) * k* x^2`

`\Delta U=(1)/(2) * 2.5* (6.4-3.7)`

`\Delta U=3.375\ N.cm=3.375* 10^(-2)\ J`

b)

The energy lost during the damping is converted into kinetic energy of the molecules which is heat primarily.

Answer 2

Step-by-step explanation:

a ) Energy of spring = 1/2 k A² where A is amplitude of oscillation and k is force constant .

So initial energy = 1/2 x 2.5 x (6.4 x 10⁻²)²

= 51.2 x 10⁻⁴ J

So final energy = 1/2 x 2.5 x (3.7 x 10⁻²)²

= 17.11 x 10⁻⁴ J

energy lost

= 34.1 J .

This energy is dissipated in the form of heat, sound etc.