A particle executes SHM with a time period of 2 s and amplitude 5 cm. What will be the displacement and velocity of that particle at t = 1/3 ?

Options

A. 3.5 cm and 4.32 cm`s^(-1)`
B. 4.33 cm and 7.85 cm`s^(-1)`
C. 5.12 cm and 6.22 cm`s^(-1)`
D. 7.85 cm and 4.33 cm`s^(-1)`

Answer

B

Explanation

Here, T = 2 s, A = 5 cm and `t = (1)/(3)s`
i. For the particle starting from mean position, (i.e. `phi` = 0) displacement,
`y = A sin omega t = A "sin"(2 pi)/(T)t`
`=5 sin ((2pi)/(2)xx(1)/(3))=5"sin"(pi)/(3)=5 xx (sqrt(3))/(2)=4.33`cm
ii. Velocity, `v = (dy)/(dt)=(d(Asin omegat))/(dt)=A omega cos omega t`
`= (2piA)/(T)"cos"(2pi)/(T)t=(2pi xx 5)/(2)"cos"(pi)/(3)`
`= 5 xx 3.14 xx 0.5 = 7.85 cm s^(-1)" "[because "cos"(pi)/(3)=0.5]`