A simple pendulum has time period T when there is no air resistance. If a small air resistance now acts on the oscillating bob, then:

the period is initially greater than T and decreases with time
the period is initially less than T and increases with time
the period is less than T and stays constant
the period is greater than T and stays constant

Correct answer: the period is greater than T and stays constant

Solution

With light damping, the angular frequency becomes omega' = sqrt(omega0² - (b/2m)²), which is slightly less than omega0, so the period T' = 2*pi/omega' is slightly greater than T. Since the damping coefficient is constant, T' does not change with time (only amplitude decays). Hence the period is greater than T and remains constant.

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