Answer:
The correct answer is "0.246".
Explanation:
Given that the amplitude is decreased by a factor of 9, then
[tex]A \rightarrow (A-\frac{A}{9} )[/tex]
[tex]A \rightarrow \frac{8A}{9}[/tex]
As we know,
Energy will be:
⇒ [tex]E_{1}=\frac{1}{2}KA^2[/tex]
and,
⇒ [tex]E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2[/tex]
[tex]=\frac{64KA^2}{162}[/tex]
⇒ [tex]\Delta E=E_1-E_2[/tex]
On putting the estimated values, we get
[tex]=\frac{1}{2}KA^2-\frac{64KA^2}{162}[/tex]
⇒ [tex]\frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}[/tex]
[tex]=\frac{40}{162}[/tex]
[tex]=0.246[/tex]
