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May 31st, 2015, 10:04 PM   #1
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\$\$\$ entrance exam level \$\$\$

A pendulum has time period T for small oscillations. An obstacle P is situated below the point of suspension O at a distance 3L/4 (from the top). The pendulum is released from rest. Throughout the motion, the moving string makes a small angle with the vertical. Time after which the pendulum returns back to its initial position is what? The length of the pendulum is L.

The correct answer is 3T/4, but why have we assumed half oscillation after the obstacle where the time period does not depend on gravity component?


Explain and I will press the thanks button.

Last edited by skipjack; May 31st, 2015 at 10:14 PM. Reason: no specific reason
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June 1st, 2015, 05:15 PM   #2
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$\displaystyle T = 2\pi \sqrt{\frac{L}{g}}$

released from rest, the time to hit the obstacle is $\displaystyle \frac{T}{4}$

once it hits the obstacle, the new length is $\displaystyle \frac{L}{4}$ which cuts the period $T$ in half ...

after hitting the obstacle, $\displaystyle \frac{T}{8}$ up, $\displaystyle \frac{T}{8}$ down ... back to original length, $\displaystyle \frac{T}{4}$ back to the start.

$\displaystyle \frac{T}{4} + \frac{T}{8} + \frac{T}{8} + \frac{T}{4} = \frac{3T}{4}$
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July 2nd, 2015, 02:40 AM   #3
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will you please explain
how time period gets T/8 up and down after hitting
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