A simple pendulum with a period T = 2.50 s is attached to the ceiling of an elev

Question

A simple pendulum with a period T = 2.50 s is attached to the ceiling of an elevator which is initially at rest. Calculate the period of oscillations when the elevator begins to accelerate upwards with acceleration a = 1.50 m/s2.

Calculate the period of oscillations when the elevator accelerates downwards with acceleration a = 2.50 m/s2.

A simple pendulum with a period T = 2.50 s is attached to the ceiling of an elevator which is initially at rest. Calculate the period of oscillations when the elevator begins to accelerate upwards with acceleration a = 1.50 m/s2. Calculate the period of oscillations when the elevator accelerates downwards with acceleration a = 2.50 m/s2.

Explanation / Answer

so.. T = 2*pi * sqrt ( L / g )

so.. 2.5 = 2*pi * sqrt ( L / 9.81 )
so.. length of pendlum = L = 1.553064 m

so.. for 1.5 m/sec2 upwards ..
g_effective = 9.81 + 1.5 = 11.31 m/sec2

so.. Time period = 2*pi*sqrt (L / g_eff ) = 2*pi * sqrt ( 1.553064 / 11.31 ) = 2.3283 secs

so.. for 1.5 m/sec2 downwards ..
g_effective = 9.81 - 1.5 = 8.31 m/sec2

so.. Time period = 2*pi*sqrt (L / g_eff ) = 2*pi * sqrt ( 1.553064 / 8.31 ) = 2.716277 secs

so.. for 2.5 m/sec2 upwards ..
g_effective = 9.81 + 2.5 = 12.31 m/sec2

so.. Time period = 2*pi*sqrt (L / g_eff ) = 2*pi * sqrt ( 1.553064 / 12.31 ) =2.23175 secs

so.. for 2.5 m/sec2 downwards ..
g_effective = 9.81 - 2.5 = 7.31 m/sec2

so.. Time period = 2*pi*sqrt (L / g_eff ) = 2*pi * sqrt ( 1.553064 / 7.31 ) = 2.89612 secs

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