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

The correct answer is "0.246".

Step-by-step explanation:

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

`A \rightarrow (A-(A)/(9) )`

`A \rightarrow (8A)/(9)`

As we know,

Energy will be:

⇒
`E_(1)=(1)/(2)KA^2`

and,

⇒
`E_(2)=(1)/(2)K((8A)/(9) )^2`

`=(64KA^2)/(162)`

⇒
`\Delta E=E_1-E_2`

On putting the estimated values, we get

`=(1)/(2)KA^2-(64KA^2)/(162)`

⇒
`(\Delta E)/(E)=((20)/(162)KA^2)/((1)/(2)KA^2)`

`=(40)/(162)`

`=0.246`