$\omega_1=\sqrt{\frac{k}{m}}$
$\omega_2=\sqrt{\frac{k}{m+2}}$
$\frac{\omega_1}{\omega_2}=\sqrt{\frac{m+2}{m}}$
Since $\omega=\frac{2 \pi}{T}$
$\frac{T_2}{T_1}=\sqrt{\frac{m+2}{m}}$
$\left(\frac{5}{4}\right)^2=\frac{m+2}{m}$
$\frac{25}{16}=\frac{m+2}{m}$
$25 \,m=16 \,m+32$
$9 \,m=32$
$m=3 \frac{5}{9} \,kg$
$m=3.5 \,kg$
If the position and velocity of the particle at $t=0\, {s}$ are $2\, {cm}$ and $2\, \omega \,{cm} \,{s}^{-1}$ respectively, then its amplitude is $x \sqrt{2} \,{cm}$ where the value of $x$ is ..... .

$\mathrm{y}=\mathrm{A}_{0}+\mathrm{A} \sin \omega \mathrm{t}+\mathrm{B} \cos \omega \mathrm{t}$
Then the amplitude of its oscillation is given by