Question

A simple pendulum has time period 't'. Its time period in a lift which is moving upwards with acceleration 3 ms⁻² is(a) $t\sqrt{\frac{9.8}{12.8}}$(b) $t\sqrt{\frac{12.8}{9.8}}$(c) $t\sqrt{\frac{9.8}{6.8}}$(d) $t\sqrt{\frac{6.8}{9.8}}$

Answer 1

For calculation of new time period in an accelerated frame we have to replace 'g' vector by 'g effective' vector, where g effective is the resultant vector of 'g' and '-a' where a is the acceleration. Here lift is moving upwards hence -a is downwards hence g effective is= 9.8 + 3 = 12.8

Now use this g effective in place of 'g' for calculating the new time period.

Hope this helps. Have a nice day :-)