A pendulum has time period T for small oscillations.An obstacle is placed directly beneath the pivot,so that only the lowest one quarter of the string can follow the pendulum bob when it swings in the left of its resting position.the pendulum is released from the rest at a certain point A.The time taken by it to return to that point is? Plz explain it at the earliest.
it is 3T/4. time period will change to t/2 when string length for oscillation is restricted to be l/4. in this case it will complete half the oscillation and again length will become l. so time for restricted part will be t/2/2 ==t/4 and for next part it will take t/2 so net time = t/2+t/4=> 3t/4.
it is 3T/4. time period will change to t/2 when string length for oscillation is restricted to be l/4. in this case it will complete half the oscillation and again length will become l. so time for restricted part will be t/2/2 ==t/4 and for next part it will take t/2 so net time = t/2+t/4=> 3t/4.
The answer has to be 3T/4. When the string oscillates towards the right from the rest point the time to come back to the rest position will remain same which is T/2. Now when it swings to the left, since the length of the string is l/2, the time taken will be π
Hence the total time taken to come back to the original position is 3T/4
Post Comments
vineetgupta_1991 said – Sun, 14 Dec 2008 13:07:24 -0000 ( Flag Edit Link )
DEAR KATHIKIIT, PLZ EXPLAIN HOW U HAVE DONE IT & THE ANSWER “T/4”U HAV GIVEN IT JUST NOW IS WRONG.