Answer:
(i) When the lift goes up [Fig.(a)] with, uniform velocity v, tension in the string, \[T=mg\]. The value of g remains unaffected. The period T remains same as that in stationary lift i.e., \[T=2\pi \sqrt{\frac{l}{g}}\] When the lift goes up with acceleration a [Fig. (b)], the net upward force on the bob is \[T-mg=ma\] \[\therefore \] \[T'=m(g+a)\] The effective value of g is \[(g+a)\] and time period is \[{{T}_{1}}=2\pi \sqrt{\frac{l}{g+a}}\] Clearly, \[{{T}_{1}}<T\]i.e., time period decreases. (iii) When lift comes down with acceleration a [Fig. (c)], the net downward force on the bob is \[mg-T'=ma\] \[\therefore \] \[T'=m(g-a)\] The effective value of \[g\] becomes \[(g-a)\] and time period is \[{{T}_{2}}=2\pi \sqrt{\frac{l}{g-a}}\] Clearly, \[{{T}_{2}}>T\] i.e., time period increases.
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