QJEEM14Ph92ec: JEE Main 2014 Physics - Class 11 JEEM14Ph92ec Solution: A particle moves with simple harmonic motion in a straight…

Question

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A particle moves with simple harmonic motion in a straight line. In first

\tau s,

after starting from rest it travels a distance

a,

and in next

\tau s

it travels

2a,

in same direction, then:

Accepted Answer

In simple harmonic motion, starting from rest,

At

t=0,

x=A

x = A\cos \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

When

t = \tau ,\,\,x = A - a

When

t = 2\,\tau ,\,x = A - 3a

From equation

(i)

A - a = A\cos \omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

A - 3a = A\cos 2\omega \,\tau \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)

As

\cos 2\omega \,\tau = 2{\cos ^2}\omega \tau - 1...\left( {iv} \right)

From equation

(ii),

(iii)

and

(iv)

{{A - 3A} \over A} = 2{\left( {{{A - a} \over A}} \right)^2} - 1

\Rightarrow {{A - 3a} \over A} = {{2{A^2} + 2{a^2} - 4Aa - {A^2}} \over {{A^2}}}

\Rightarrow {A^2} - 3aA = {A^2} + 2{a^2} - 4Aa

\Rightarrow 2{a^2} = aA \Rightarrow \,\,\,\,\,\,\,A = 2a

\Rightarrow {a \over A} = {1 \over 2}

Now,

A-a=A

\cos \omega \tau

\Rightarrow \cos \omega \tau = {{A - a} \over A} \Rightarrow \,\,\cos \omega \tau = {1 \over 2}

or,

{{2\pi } \over T}\tau = {\pi \over 3} \Rightarrow \,\,\,T - 6\,\tau

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