The variation of velocity of a particle executing SHM with time is shown is fig. The velocity of the particle when a phase change of `(pi)/(6)` takes place from the instant it is at one of the extreme positions will be

Options

A. `3.53(m)/(s)`
B. `2.5(m)/(s)`
C. `4.330(m)/(s)`
D. none of these

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)[-co(s(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)`