A particle performs S.H.M in a straight line with time period greater than 8 seconds. Starting from rest it covers distances a and b in the first and second seconds. Amplitude of its oscillation is

$y_{\text{t}=2\sec }=\text{A}\cos 2\omega$

$y_{\text{t}=0}=\text{A};\text{A}-\text{a}=\text{a}\cos \omega ;\text{a}=\text{A}-\text{A}\cos \omega$

$\cos \omega =\frac{\text{A}-\text{a}}{\text{A}}=1-\frac{\text{a}}{\text{A}}$

$\text{A}-\left(\text{a}+\text{b}\right)=\text{A}\cos 2\omega$

$\text{a}+\text{b}=\text{A}-\text{A}\cos 2\omega =\text{A}-\text{A}\left(2{\cos }^2\omega -1\right)$

$=\text{A}-\text{A}\left[2{\left(1-\frac{\text{a}}{\text{A}}\right)}^2-1\right]$

$=\text{A}-\text{A}\left(2+\frac{{\text{2a}}^{\text{2}}}{{\text{A}}^{\text{2}}}-4\frac{\text{a}}{\text{A}}-1\right)$

$=\text{A}-\text{A}\left(1+\frac{{\text{2a}}^{\text{2}}}{{\text{A}}^{\text{2}}}-4\frac{\text{a}}{\text{A}}\right)$

$=\text{A}-\text{A}-\text{A}\frac{{\text{2a}}^{\text{2}}}{{\text{A}}^{\text{2}}}+\frac{4\text{a}}{\text{A}}\text{A}=-\frac{{\text{2a}}^{\text{2}}}{\text{A}}+4\text{a}$

$\frac{{\text{2a}}^{\text{2}}}{\text{A}}=4\text{a}-\text{a}-\text{b}=3\text{a}-\text{b}$

$\text{A}=\frac{{\text{2a}}^{\text{2}}}{3\text{a}-\text{b}}$