Question

A small mass executes linear SHM about a point o with amplitude r and period t It's displacement from o at time t by 8 after passing through O is

Answer 1

Answer:

$Displacement=\frac{r}{\sqrt{2} }$

Explanation:

Displacement of a particle executing S.H.M at time t is given by :

x=A·sin(ωt)

where A : amplitude; ω: angular frequency (2π/T)

Hence displacement here is given by:

$x= r . sin(\frac{2\pi }{T}t)$

$x= r . sin(\frac{2\pi }{T}.\frac{T}{8} )$

$\boxed{x= \frac{r}{\sqrt{2} }}$

Hope this answer helped you :)