A particle moves with a simple harmonic motion in a straight line. In the first second starting from rest it travels a distance $a$ and in the next second it travels a distance $b$ in the same direction. The amplitude of the motion is

Let the acceleration be f; $f=-{\omega }^{2}x$

Therefore, distance of the particle from the centre at any time t is given by 

$x=r\cos \left(\omega t\right)$ where r is the amplitude

$\text{ when }t=1\mathrm{s},x=r-a$

$\begin{array}{l}\therefore (r-a)=r\cos \omega \\ \cos \omega =\frac{r-a}{r} ........\left(i\right)\end{array}$

$\text{ When }t=2\mathrm{s},x=r-a-b,$

$\text{ therefore }r-a-b=\cos 2\omega$

$\therefore r-a-b=r\left(2{\cos }^2\omega -1\right)$    ………..(ii)

Substituting the value of $\cos \omega$ from Eq. (i) in Eq. (ii),

$\text{ we get }r-a-b=r\left[2\frac{(r-a{)}^2}{r^2}-1\right]=\frac{2(r-a{)}^2}{r}-r$

$\therefore r(3a-b)=2a^2\Rightarrow r=\frac{2a^2}{3a-b}$