In the previous problem, the displacement of the particle from the mean position corresponding to the instant mentioned is

Options

A. `(5)/(pi)m`
B. `(5sqrt3)/(pi)m`
C. `(10sqrt3)/(pi)m`
D. `(5sqrt3)/(2pi)m`

Answer

B

Explanation

From the graph `T=(5-1)=4s`
(distance between the two adjacent shown in the figure)
and `v_(max)=5(m)/(s)`,`omegaA=5(m)/(s)`
`((2pi)/(T))A=5impliesA=(5T)/(2pi)=(5xx4)/(2pi)=(10)/(pi)` m
Also `omega=(2pi)/(4)=(pi)/(2)(rad)/(s)`
The equation of velocity can be written as
`v=5sin((pi)/(2))t(m)/(s)`
At extreme position, `v=0`, `sin((pi)/(2))t=0` or `t=2s`
Phase of the particle velocity at that insstant corresponding to the above equation `=pi`
Therefore, when a phase change of `(pi)/(5)` takes place, the resulting phase`=pi+(pi)/(6)`
`v=5sin((pi+(pi)/(6)))=-5sin(pi)/(6)=-5((1)/(2))`
`=2.5(m)/(s)` (numerically)
`(dy)/(dt)=vimpliesdy=dt`
`dy=int5sin((pit)/(2))dt=(10)/(pi)[-cos(((pit)/(2)))]+C`
Since at `t=0`, the particle is at the extreme position, there fore at `r=0`,`y=-(10)/(pi)`
`-(10)/(pi)=-(10)/(pi)costheta+CimpliesC=0`
`y=-(10)/(pi)cos(((pit)/(2)))`
Clearly a phase change of `(pi)/(6)` corresponds to a time difference of `(T)/(2pi)((pi)/(6))=(T)/(12)=(4)/(12)=(1)/(3)s`
`y=-(10)/(pi)cos((pi)/(6))=-(10)/(pi)((sqrt3)/(2))`
`=(5sqrt3)/(pi)m` (numerically)
Acceleration `a=(dv)/(dt)=(d)/(dt)(5sin((pit)/(2)))=(5pi)/(2)(cospit)/(2)`
`a` at `t=(1)/(3)s=(5pi)/(2)cos((pi)/(6))=(5pisqrt3)/(4)(m)/(s^2)`
Maximum displacement `x_(max)=A=(10)/(pi)m`
and maximum accceleration, `a_(max)=omega^2A`
`=((pi)/(2))^2xx(10)/(pi)=(5pi)/(2)(m)/(s^2)`