Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A small ball is suspended by a thread of length l = 1 m at the point O on the wall, forming a small angle α=2∘ with the vertical (as shown in figure). Then the thread with ball was deviated through a small angle β=4∘ and set free. Assuming the collision of the ball against the wall to be perfectly elastic, the oscillation period of such a pendulum (in seconds) is ___________ (Take g=π2).

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

answer is 1.33.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

The time period of free, oscillation of pendulum T=2πlg=2sTime taken by bob to go from extreme position A to mean position B is =T4Time taken by bob to move from mean position B to position C (where its angular displacement α is half the angular amplitude β) is found from equation α=βsin⁡2πTt.Solving we get t=T12⇒ Total time period of oscillation=2T4+T12=23T=43s
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon