KVPY 2020-21 SB SX Exam Previous Papers

Question : 24 of 120

Marks: +1, -0

A point particle is acted upon by a restoring force

–kx^3

. The time period of oscillation is T when the amplitude is A. The time period for an amplitude 2A will be :

Solution:

Given

{F}=-{kx}^{3}

-{{d} {U}}/{{dx}}=-{kx}^{3}

\⇒{U}={1}/{4} {kx}^{4}

\∴

Energy of oscillations will be

{E}={1}/{2} {mv}^{2}+{U}={1}/{2} {m}({{d} {x}}/{{dt}})^{2}+{1}/{4} {kx}^{4}

.....(1)
If we pull

{{dx}}/{{dt}}=0

in above equation, we will get amplitude as

{A}=\√^{4}{{{4 E}}/{{k}}}

..... (2)
Also on rearranging equation (1), we get

{dt}=\±{dx} \√{{{m}}/{2 {E}}}(1-{{k}}/{4 {E}} {x}^{4})^{-1 \/ 2}

Now, use

A=\√^{4}{{4 E}/{k}}

, to reduce above equationas

{dt}=\±{dx} \√{{2 {m}}/{{k}}} {A}^{-2}(1-({{x}}/{{A}})^{4})^{-1 \/ 2}

The time period can be found by integrating above equation.

{T}=4 \∫↙{0}↖ {{A}} {dx} \√{{2 {m}}/{{k}}} {A}^{-2}(1-({{x}}/{{A}})^{4})^{-1 \/ 2}

=4 \√{{2 {m}}/{{k}}} {A}^{-2} \∫↙{0} ↖{{A}}(1-({{x}}/{{A}})^{4})^{-1 \/ 2} \.{dx}

Put

{{x}}/{{A}}={u} \⇒{dx}={Adu}

\∴ {T}=4 \√{{2 {m}}/{{k}}} {A}^{-2}({A}) \∫↙{0} ↖{1} {du}(1-{u}^{4})^{-1 \/ 2}

{T}=4 \√{{2 {m}}/{{k}}} {A}^{-1}({I})

Where

{I}=\∫↙{0} ↖{1}(1-{u}^{4})^{-1 \/ 2} {du}

is a numerical value
So from above equation

{T} \∝{A}^{-1}

\∴{{T}_{1}}/{{T}_{2}}={2 {A}}/{{A}} \⇒{T}_{2}={{T}}/{2}

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