Answer to Question #95528 in Physics for Sreyas

Question

The velocity of a particle executing SHM is 0.12 ms^-1and 0.04 ms^-1 when its displacements from the mean position are 0.01m and 0.03m respectively. Calculate the amplitude, period and maximum velocity of the particle ?

Answer 1

$v_1=\omega \sqrt{a^2-x_1^2},v_2=\omega \sqrt{a^2-x_2^2}$

$\frac{v_1}{v_2}=\frac{\sqrt{a^2-x_1^2}}{\sqrt{a^2-x_2^2}}$

$\frac{0.12}{0.04}=\frac{\sqrt{a^2-0.01^2}}{\sqrt{a^2-0.03^2}}$

$a=0.032\ m$

$v_1=\omega \sqrt{a^2-x_1^2}\to 0.12=\omega \sqrt{0.032^2-0.01^2}$

$\omega=4\frac{rad}{s}$

$v_{max}=\omega a=(4)(0.032)=0.13\frac{m}{s}$

$T=\frac{2\pi}{\omega}=1.6\ s$