A simple pendulum, suspended from the ceiling of a stationary van, has time period T. If the van starts moving with a uniform velocity the period of the pendulum will be [RPMT 2003]
A) Less than T B) Equal to 2T C) Greater than T D) Unchanged
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There is a thumb rule for solving these kinds of problems.
Time period of a simple pendulum in an inertial frame is given by : 2(pi)*sqrt(l/g)
but if the pendulum is oscillating in a non inertial frame accelerating with an acceleration 'a' making an angle alpha with vertical then g in the denominator in sqrt is replaced by geffectivegeffective = sqrt{g^2+a^2+2.g.a.cos(alpha)}
Thus if, the train is accelerating on a horizontal track, then alpha = 90 degress and cos(alpha) = 0.
Thus, geffectivegeffective becomes = sqrt(g^2+a^2)
So, Time period T' = 2*(pi)*sqrt{l/sqrt(g^2+a^2)}
= 2*(pi)*sqrt[l/g.sqrt{1+(a/g)^2}] = T/sqrt[sqrt{1+(a/g)^2}]
Time period of a simple pendulum in an inertial frame is given by : 2(pi)*sqrt(l/g)
but if the pendulum is oscillating in a non inertial frame accelerating with an acceleration 'a' making an angle alpha with vertical then g in the denominator in sqrt is replaced by geffectivegeffective = sqrt{g^2+a^2+2.g.a.cos(alpha)}
Thus if, the train is accelerating on a horizontal track, then alpha = 90 degress and cos(alpha) = 0.
Thus, geffectivegeffective becomes = sqrt(g^2+a^2)
So, Time period T' = 2*(pi)*sqrt{l/sqrt(g^2+a^2)}
= 2*(pi)*sqrt[l/g.sqrt{1+(a/g)^2}] = T/sqrt[sqrt{1+(a/g)^2}]
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Answer:
The 2 nd option is correct
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