question_answer

A simple pendulum is hung in a stationary lift and its periodic time is T. What will be the effect on its periodic time T if (i)  the lift goes up with uniform velocity \[\upsilon \], (ii)  the lift goes up with uniform acceleration a, and (iii) the lift comes down with uniform  acceleration a?

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.