A particle of mass 100g is dexcribing SHM along a straight lig with a period of 2s and amplitude of 12 cm. What is the knetice energy when it is (i) 3cm (ii) 6cm from its equilibrium position? `(Use pi^(2)=10)`

Explanation

Here, `m=100g,T=2s,A=12cm`
Angular frequency, `omega=(2pi)/(T)=(2pi)/(2)=pirad//s`
Velocity, `v=omegasqrt(A^(2)-Y^(2))`
Case (i) Whwn `y=3cm`
`y=pisqrt((12)^(2)-(3)^(2))=pisqrt(135)cms^(-1)`
`:. ` KE`=(1)/(2)MV^(2)=(1)/(2XX100xxpi^(2)xx135`
`=67500erg`
(ii) When `y=6cm,v=pisqrt(12^(2)-6^(2))`
`=pisqrt(108)cms^(-1)`
`KE=(1)/(2mv^(2))=(1)/(2)xx100xxpi^(2)xx108`
`=54000erg`