For periodic motion of small amplitude A, the time period T of this particle is proportional to [2010]

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. For periodic motion of small amplitude A, the time period T of this particle is proportional to

Smart Achievers · Accepted Answer

(2)

Potential energy, $x$ given

$∴$ $α = \frac{Potential energy}{x^{4}} = \frac{M L^{2} T^{- 2}}{L^{4}} = [M L^{- 2} T^{- 2}]$

Now, $\frac{1}{A} \sqrt{\frac{m}{α}} = \frac{1}{L} \sqrt{\frac{M}{M L^{- 2} T^{- 2}}} = T$

Solution

1 $A \sqrt{\frac{m}{α}}$

2 $\frac{1}{A} \sqrt{\frac{m}{α}}$

3 $A \sqrt{\frac{α}{m}}$

4 $\frac{1}{A} \sqrt{\frac{α}{m}}$

Ans.
(2)

Potential energy, $V \propto x^{4}$ given

$∴$ $α = \frac{Potential energy}{x^{4}} = \frac{M L^{2} T^{- 2}}{L^{4}} = [M L^{- 2} T^{- 2}]$

Now, $\frac{1}{A} \sqrt{\frac{m}{α}} = \frac{1}{L} \sqrt{\frac{M}{M L^{- 2} T^{- 2}}} = T$

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Solution

1 Amα

2 1Amα

3 Aαm

4 1Aαm