question_answer

11. 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 :   JEE Main  Solved  Paper-2014

A) amplitude of motion is 4a

B) time period of oscillations is \[6\tau \]

C) amplitude of motion is 3a

D) time period of oscillations is \[8\tau \]

Correct Answer: B

Solution :

\[\cos (\omega \tau )=\frac{A-a}{A}\]    \[\cos (2\omega \tau )=\frac{A-3a}{A}\] \[\Rightarrow \]\[\frac{A-3a}{A}=2{{\left( \frac{A-a}{A} \right)}^{2}}-1\] \[\Rightarrow \]\[1-\frac{3a}{A}=2+2{{\left( \frac{a}{A} \right)}^{2}}-\frac{4a}{A}-1\] \[\left( \frac{a}{A} \right)={{\left( \frac{a}{A} \right)}^{2}}A=2a\] \[\cos (\omega \tau )=\frac{1}{2}\Rightarrow \omega \tau =\frac{\pi }{3}\Rightarrow \frac{2\pi }{T}.\tau =\frac{\pi }{3}\Rightarrow T=6\tau \]