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

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 2

Answer:

a. $\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J$

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, $m=1.33\ kg$

• 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:

$\Delta U=\frac{1}{2} \times k\times A^2-\frac{1}{2} \times k\times x^2$

$\Delta U=\frac{1}{2} \times 2.5\times (6.4-3.7)$

$\Delta U=3.375\ N.cm=3.375\times 10^{-2}\ J$

b)

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