A) \[{{x}^{2}}\]
B) \[{{e}^{x}}\]
C) \[x\]
D) \[{{\log }_{e}}x\]
Correct Answer: A
Solution :
[a] In this problem acceleration [a] is given in terms of displacement\[(x)\]to determine the velocity with respect to position or displacement we have to apply integration method. |
From given information\[a=-\text{ }kx,\]where a is acceleration, \[x\] is displacement and \[k\] is a proportionality constant. |
\[\frac{v\,dv}{dx}=-kx\] \[\left[ \because \frac{dv}{dt}=\frac{dv}{dx}\left( \frac{dx}{dt} \right)=\frac{v\,dv}{dx} \right]\] |
\[\Rightarrow \] \[v\,dv=-k\,xdx\] |
Let for any displacement from 0 to\[x,\]the velocity changes from\[{{v}_{0}}\]to\[v\]. |
\[\Rightarrow \]\[\int_{{{v}_{0}}}^{v}{v\,dv}=-\int_{0}^{x}{k\,x\,dx}\]\[\Rightarrow \]\[\frac{{{v}^{2}}-v_{0}^{2}}{2}=-\frac{k{{x}^{2}}}{2}\] |
\[\Rightarrow \]\[m\left( \frac{{{v}^{2}}-v_{0}^{2}}{2} \right)=-\frac{mk\,{{x}^{2}}}{2}\] |
\[\Rightarrow \]\[\Delta K\propto {{x}^{2}}\] (\[\Delta K\]is loss in\[KE\]) |
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