A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $$x$$  is proportional to-

Solution :
$$\eqalign{ & {\text{Given : retardation}} \propto {\text{displacement}} \cr & {\text{i}}{\text{.e}}{\text{., }}a = - kx \cr & {\text{But }}a = v\frac{{dv}}{{dx}} \cr & \therefore \frac{{vdv}}{{dx}} = - kx \Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x x \,dx \cr & \left( {v_2^2 - v_1^2} \right) = - k\frac{{{x^2}}}{2} \cr & \Rightarrow \frac{1}{2}m\left( {v_2^2 - v_1^2} \right) = \frac{1}{2}mk\left( {\frac{{ - {x^2}}}{2}} \right) \cr & \therefore {\text{Loss in kinetic energy,}}\,\,\therefore \Delta K \propto {x^2} \cr} $$