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 May 31st, 2015, 10:04 PM #1 Newbie   Joined: May 2015 From: INDIA Posts: 28 Thanks: 1 \$\$\$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
 June 1st, 2015, 05:15 PM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,034 Thanks: 1621 $\displaystyle T = 2\pi \sqrt{\frac{L}{g}}$ 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. $\displaystyle \frac{T}{4} + \frac{T}{8} + \frac{T}{8} + \frac{T}{4} = \frac{3T}{4}$ Thanks from rexdex
 July 2nd, 2015, 02:40 AM #3 Newbie   Joined: May 2015 From: INDIA Posts: 28 Thanks: 1 will you please explain how time period gets T/8 up and down after hitting

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### a pendulum has time period T for small oscillation An obstacle P is situated below the point of suspension O at a distance 3l/4 The pendulum is released from rest throughout the motion the moving string makes small angle with vertical.Time after which the

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