A body of mass 100 g attached to a spring executed SHM of period 2 s and amplitude 10 cm. How long a time is required for it to move from a point 5 cm below its equilibrium position to a point 5 cm above it, when it makes simple harmonic vertical oscillation (take `g=10(m)/(s^2)`)?

Options

A. 0.6 s
B. `(1)/(3)s`
C. `1.5s`
D. `2.2s`

Answer

B

Explanation

Period `=2(pi)/(omega)=2s` or `omega=pi(rad)/(s)`,
amplitude `=10cm`
If the system is released, the equation of motion is
`y=Acosomegat`
`5=10cospit_1` when `y=5cm`
`-5=10cospit_2` when `y=-5cm`
`pit_1=cos^-1((1)/(2))=(pi)/(3)` or `t_1=(1)/(3)s`
`pit_2=cos^-1(-(1)/(2))=(2pi)/(3)` or `t_2=(2)/(3)s`
Time interval`=t_4-t_1=(2)/(3)-(1)/(3)=(1)/(3)s`