Potential energy of a particle at mean position is 4 J and at extreme position is 20 J. Given that amplitude of oscillation is A. Match the following two columns

Answer

A-R:B-P;C-Q;D-S

Explanation

T.E.=`U_(extreme)=20j`
`K_(mean)=(1)/(2)mA^(2)omega^(2)=20-4 =16j`
`K(x=(A)/(2))=(1)/(2)momega^(2)(A^(2)-(A^(2))/(4))=(3)/(4)(16)=12j`
`U(x=(A)/(2))=20-12=8J`
`K(x=(A)/(4))=(1)/(2)momega^(2)(A^(2)-(A^(2))/(16))=(15)/(16)(16)=15J`