Physics - SHM Question with Solution | TestHub

$\text{x}=-A\; cos\omega \text{t}$

$-\text{A}+\text{a}=-\text{A cos}\omega \tau$...(i)

$-\text{A}+3\text{a}=-\text{A cos}\left(\omega 2\tau \right)$...(ii)

$\therefore -1+\frac{\text{a}}{\text{A}}=-\text{cos}\omega \tau$...(iii)

and$-1+\frac{3\text{a}}{\text{A}}=-\text{cos}\left(2\omega \tau \right)$...(iv)

Adding (iii) and (iv)

$2=3 cos\left(\omega \tau \right)-2{\text{ cos}}^2\left(\omega \tau \right)+1$

$\therefore 2{\text{ cos}}^2\left(\omega \tau \right)-3\text{ cos}\left(\omega \tau \right)+1=0$

$\therefore \text{cos}\omega \tau =\frac{1}{2}$

$\therefore \omega \tau =\frac{\pi }{3}$

$\therefore \text{T}=6\tau$