Question

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

Answer 1

$U_0=\frac{1}{2}K{A}^2=U_{max}$
$4+\frac{1}{2}k{A}^2=20$
or $\frac{1}{2}k{A}^2=16$ J
(1) $U=U_0+\frac{1}{2}k{X}^2$
$=U_0+\frac{1}{2}k{\left(\frac{A}{2}\right)}^2$
$=4+\frac{16}{4}=8$ J
(2) $KE=\frac{1}{2}k(A^2-\frac{A^2}{16})$
$=\frac{15}{16}\left(\frac{1}{2}k{A}^2\right)=\frac{15}{16}\times 16=15J$
(3) $KE=\frac{1}{2}k(A^2-0)$
$=\frac{1}{2}k{A}^2=16J$
(4) $KE=\frac{1}{2}k(A^2-\frac{A^2}{4})$
$=\frac{3}{4}\left(\frac{1}{2}k{A}^2\right)=\frac{3}{4}\left(16J\right)=12$ J