Question

A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does the energy of the mass/spring system decrease over that time?

Answer 1

Answer:

The correct answer is "0.246".

Explanation:

Given that the amplitude is decreased by a factor of 9, then

$A \rightarrow (A-\frac{A}{9} )$

$A \rightarrow \frac{8A}{9}$

As we know,

Energy will be:

⇒  $E_{1}=\frac{1}{2}KA^2$

and,

⇒  $E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2$

          $=\frac{64KA^2}{162}$

⇒  $\Delta E=E_1-E_2$

On putting the estimated values, we get

           $=\frac{1}{2}KA^2-\frac{64KA^2}{162}$

⇒  $\frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}$

          $=\frac{40}{162}$

          $=0.246$