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 A particle moves with simple harmonic motion in a straight line, In first   $\tau$   sec. after starting from rest it travels a distance a and in next $\tau$   sec. it travels 2a, in same direction, then 

Answer:

Explanation:

 In   SHM , a particle  starts from rest, we have

i.e,   $x=A\cos \omega t,$     at     t=0,x=A

when $ t=\tau$     then x=A-a   .......(i)

when  $t=2\tau$   , then x=A-3a     .......(ii)

 On comparing Eqs. (i)  and (ii), we get

 $A-a=A\cos \omega \tau$

    $A-3a=A\cos 2\omega \tau$

 As   $\cos2 \omega t=2\cos^{2}\tau-1$

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

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

  $A^{2}-3aA=A^{2}+2a^{2}-4Aa$

$a^{2}=2aA$

 A=2a

  Now,   $A-a=A\cos\omega\tau$

$\Rightarrow$      $\cos\omega\tau=1/2$

$\frac{2\pi}{T}\tau=\frac{\pi}{3}$

   $\Rightarrow $          $ T=6\pi$