A body executing linear simple harmonic motion has a velocity of 3 m/s when its displacement is 4 cm and a velocity of 4 m/s when its displacement is 3 cm. What is the amplitude of oscillation?

Veloctity in SHM is given by $\mathrm{v}=\omega \sqrt{a^2-y^2}$

At y=4cm=0.04m, v= 3m/s     $\therefore 3=\omega \sqrt{a^2-(0.04{)}^2}.........\left(1\right)$

At y=3 cm=0.03m, v=4m          $\therefore 4=\omega \sqrt{a^2-(0.03{)}^2}.......\left(2\right)$

Dividing (2) by (1) ,We get $\sqrt{\frac{a^2-(0.03{)}^2}{a^2-(0.04{)}^2}}=\frac{4}{3}\text{ or }\frac{a^2-(0.03{)}^2}{a^2-(0.04{)}^2}=\frac{16}{9}\text{ or }$

$9a^2-9\times (0.03{)}^2=16a^2-16\times (0.04{)}^2\text{ or }7a^2=16\times (0.04{)}^2-9\times (0.03{)}^2$

$=0.0256-0.0081=0.0175\mathrm{a}=0.05\mathrm{m}=5\mathrm{cm}$