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.

Respuesta :

Answer:

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.

Answer:

a. [tex]\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J[/tex]

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

Explanation:

Given:

  • mass of the attached object, [tex]m=1.33\ kg[/tex]
  • spring constant, [tex]k=2.5\ N.cm^{-1}[/tex]
  • maximum displacement, [tex]A=6.4\ cm[/tex]
  • maximum displacement after damping, [tex]x=3.7\ cm[/tex]

a)

the energy lost in damping:

[tex]\Delta U=\frac{1}{2} \times k\times A^2-\frac{1}{2} \times k\times x^2[/tex]

[tex]\Delta U=\frac{1}{2} \times 2.5\times (6.4-3.7)[/tex]

[tex]\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J[/tex]

b)

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