A)
B)
C)
D)
Correct Answer: D
Solution :
\[d{{U}_{(x)}}=-Fdx\] |
\[\therefore \,\,{{U}_{x}}=-\int_{0}^{x}{Fdx}=\frac{k{{x}^{2}}}{2}-\frac{a{{x}^{4}}}{4}\] |
\[U=0\]at \[x=0\] and at \[x=\sqrt{\frac{2k}{a}};\,\Rightarrow \] we have potential energy zero twice (out of which one is at origin). |
Also, when we put \[x=0\]in the given function, we get \[F=0\]. But \[F=-\frac{dU}{dx}\] |
\[\Rightarrow \] At \[x=0;\] \[\frac{dU}{dx}=0\] i.e. the slope of the graph should be zero. These characteristics are represented by [d]. |
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