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Question

Physics Question on Oscillations

From a fixed support,two small identical spheres are suspended by means of strings of length 1m1\, m each .They are pulled aside as shown and then released BB is the mean position.Then the two spheres collide.

A

at BB after 0.250.25 second

B

at BB after 0.50.5 second

C

on the right side of BB after some time

D

on the right side of BB when the strings are inclined at 1515^\circ with BB

Answer

at BB after 0.50.5 second

Explanation

Solution

Irrespective of amplitude,
both spheres take same time to reach the mean position, i.e., T4s\frac{T}{4} s
T=2πlg=2×3.14×110=2s\therefore T=2 \pi \sqrt{\frac{l}{g}}=2 \times 3.14 \times \sqrt{\frac{1}{10}}=2\, s
So, they will collide at T4=24=0.5s\frac{T}{4}=\frac{2}{4}=0.5\, s