A particle moves with simple harmonic motion in a straight line. In first s, after starting from rest it travels a distance a, and in next s it travels 2 a, in same direction, then: [Main 2014]

Correct Answer is : time period of oscillators is 6τ

time period of oscillators is 6τ

time period of oscillation is 8τ

Correct Answer is : time period of oscillators is 6τ

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JEE Main 2014 Paper

Section: Physics

Question : 11 of 90

Marks: +1, -0

\tau s

a

\tau s

2 a

Solution:

In simple harmonic motion, starting from rest,

t=0, x=A

x=A \cos \omega t \; \; . . .(i)

When

t=\tau, x=A-a

When

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

From equation

(i)

A-a=A \cos \omega \tau \; \; . . .(ii)

A-3a=A \cos 2 \omega \tau \; \; . . .(iii)

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

From equation

(ii),(iii)

and

(iv)

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

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

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

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

\Rightarrow \; \frac{a}{A}= \; \frac{1}{2}

Now,

A-a=A \cos \omega \tau

\Rightarrow \cos \omega \tau= \; \frac{A-a}{A} \Rightarrow \cos \omega \tau= \; \frac{1}{2}

\; \text{ or, } \; \frac{2\pi}{T} \tau = \; \frac{\pi}{3} \Rightarrow T-6\tau

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