A particle of mass $m$ moving in a symmetric potential that is well described by $V (x) = A ∣ x ∣^{n}$ has a total energy $\frac{1}{2} m (d x / d t)^{2} + V (x) = E$ . Solving for $d x / d t$ and integrating we find that the period of motion is $τ = 2 2 m \int_{0}^{x_{m a x}} \frac{d x}{( E - A x ^{n} ) ^{1/2}}$ where $x_{m a x}$ is a classical turning point given by $A x_{m a x}^{n} = E$ . Show that $τ = \frac{2}{n} \frac{2 πm}{E} (\frac{E}{A})^{1/ n} \frac{Γ ( 1/ n )}{Γ ( 1/ n + \frac{1}{2} )}$

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A particle of mass

m

moving in a symmetric potential that is well described by

V (x) = A ∣ x ∣^{n}

has a total energy

\frac{1}{2} m (d x / d t)^{2} + V (x) = E

. Solving for

d x / d t

and integrating we find that the period of motion is

τ = 2 2 m \int_{0}^{x_{m a x}} \frac{d x}{( E - A x ^{n} ) ^{1/2}}

where

x_{m a x}

is a classical turning point given by

A x_{m a x}^{n} = E

. Show that

τ = \frac{2}{n} \frac{2 πm}{E} (\frac{E}{A})^{1/ n} \frac{Γ ( 1/ n )}{Γ ( 1/ n + \frac{1}{2} )}

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Question Text	A particle of mass $m$ moving in a symmetric potential that is well described by $V (x) = A ∣ x ∣^{n}$ has a total energy $\frac{1}{2} m (d x / d t)^{2} + V (x) = E$ . Solving for $d x / d t$ and integrating we find that the period of motion is $τ = 2 2 m \int_{0}^{x_{m a x}} \frac{d x}{( E - A x ^{n} ) ^{1/2}}$ where $x_{m a x}$ is a classical turning point given by $A x_{m a x}^{n} = E$ . Show that $τ = \frac{2}{n} \frac{2 πm}{E} (\frac{E}{A})^{1/ n} \frac{Γ ( 1/ n )}{Γ ( 1/ n + \frac{1}{2} )}$
Topic	All topics
Subject	Physics
Class	Class 11

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