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 ?
v 1 = ω a 2 − x 1 2 , v 2 = ω a 2 − x 2 2 v_1=\omega \sqrt{a^2-x_1^2},v_2=\omega \sqrt{a^2-x_2^2} v 1 = ω a 2 − x 1 2 , v 2 = ω a 2 − x 2 2
v 1 v 2 = a 2 − x 1 2 a 2 − x 2 2 \frac{v_1}{v_2}=\frac{\sqrt{a^2-x_1^2}}{\sqrt{a^2-x_2^2}} v 2 v 1 = a 2 − x 2 2 a 2 − x 1 2
0.12 0.04 = a 2 − 0.0 1 2 a 2 − 0.0 3 2 \frac{0.12}{0.04}=\frac{\sqrt{a^2-0.01^2}}{\sqrt{a^2-0.03^2}} 0.04 0.12 = a 2 − 0.0 3 2 a 2 − 0.0 1 2
a = 0.032 m a=0.032\ m a = 0.032 m
v 1 = ω a 2 − x 1 2 → 0.12 = ω 0.03 2 2 − 0.0 1 2 v_1=\omega \sqrt{a^2-x_1^2}\to 0.12=\omega \sqrt{0.032^2-0.01^2} v 1 = ω a 2 − x 1 2 → 0.12 = ω 0.03 2 2 − 0.0 1 2
ω = 4 r a d s \omega=4\frac{rad}{s} ω = 4 s r a d
v m a x = ω a = ( 4 ) ( 0.032 ) = 0.13 m s v_{max}=\omega a=(4)(0.032)=0.13\frac{m}{s} v ma x = ωa = ( 4 ) ( 0.032 ) = 0.13 s m
T = 2 π ω = 1.6 s T=\frac{2\pi}{\omega}=1.6\ s T = ω 2 π = 1.6 s
