A simple pendulum having bob of mass m and length of string l has time

A simple pendulum having bob of mass m and length of string l has time period of T. If the mass of the bob is doubled and the length of the string is halved, then the time period of this pendulum will be

[amp_mcq option1=”T” option2=”T /√2″ option3=”2T” option4=”√2 T” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2022

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The correct option is B, stating that the new time period will be T/√2.

The time period (T) of a simple pendulum is given by the formula T = 2π√(l/g), where l is the length of the string and g is the acceleration due to gravity. This formula shows that the time period depends only on the length of the pendulum and the acceleration due to gravity, and it is independent of the mass of the bob.
In the given problem, the initial time period is T for a pendulum with mass m and length l. When the mass of the bob is doubled (to 2m) and the length of the string is halved (to l/2), the new time period T’ will be T’ = 2π√((l/2)/g) = 2π√(l/(2g)). We can rewrite this as T’ = 2π√(l/g) * (1/√2). Since T = 2π√(l/g), the new time period T’ is T/√2.

The independence of the time period from the mass of the bob is a key characteristic of a simple pendulum, provided the amplitude of oscillation is small (usually less than 15 degrees). This property allows pendulums to be used as reliable timekeeping devices. Real-world factors like air resistance and the rigidity of the string can affect the period slightly, but for an ideal simple pendulum, mass is irrelevant.

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