The acceleration of this particle for x x0 is

Question

When a particle of mass $m$ moves on the $x$ -axis in a potential of the form $V (x) = k x^{2}$ it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}},$ as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of $x = 0$ in a way different from $k x^{2}$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass $m$ moving on the $x$ -axis. Its potential energy is $V (x) = α x^{4} (α > 0)$ for $| x |$ near the origin and becomes a constant equal to $V_{0}$ for $| x |$ $\geq X_{0}$ (see figure). [2010]

Q. The acceleration of this particle for $| x |$ $> X_{0}$ is

Smart Achievers · Accepted Answer

(4)

$F = \frac{- d V (x)}{d x}$

$∵$ $As V (x) = constant for x > X_{0}$

$∴$ $F = 0 and hence a = 0 for x > X_{0}$

Solution

1 proportional to $V_{0}$

2 proportional to $\frac{V_{0}}{m X_{0}}$

3 proportional to $\sqrt{\frac{V_{0}}{m X_{0}}}$

4 zero

Ans.
(4)

$F = \frac{- d V (x)}{d x}$

$∵$ $As V (x) = constant for x > X_{0}$

$∴$ $F = 0 and hence a = 0 for x > X_{0}$

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Solution

1 proportional to V0

2 proportional to V0mX0

3 proportional to V0mX0