hello sir,
Asked by
| 24th Jan, 2009,
04:05: PM
Expert Answer:
Use T = 2 pi (square root of L/g)
T1/T2 = (square root of L1)/square root of L2 = 11/10
So they will be in the same phase in 110 seconds which is the hcf of time period of the two pendulums.
Now you can find how many vibrations the shorter pendulum makes in 110s.
Answered by
| 28th Jan, 2009,
02:40: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change