InterviewSolution
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InterviewSolution
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In the previous problem, the displacement of the particle from the mean position corresponding to the instant mentioned isA. `(5)/(pi)m`B. `(5sqrt3)/(pi)m`C. `(10sqrt3)/(pi)m`D. `(5sqrt3)/(2pi)m` |
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Answer» Correct Answer - B 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)` |
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