A simple pendulum with a period T 500 s is attached to the c

A simple pendulum with a period T= 5.00 s is attached to the ceiling of an elevator which is initaially at rest.

a.) Calculate the period of oscillations when the elevator begins to accelerate upwards with acceleration a=0.50 m/s².

b.) Calculate the period of oscillations when the elevator accelerates downwards with the acceleration a=1.00 m/s².

Solution

Time period= 2*pi*sqrt(L/g)

A) if acclerated upwards then g\' = g+ 0.5 = 9.8+0.5 = 10.3 m/s^2

Now T*sqrt(10.3) = 5*sqrt(9.8) => T = 5*sqrt(9.8)/sqrt(10.3) = 4.877 sec Answer

B) g\' = 9.8-1 = 8.8

T = 5*sqrt(9.8)/sqrt(8.8) = 5.276 sec Answer