Questions
Two pendulums of lengths 121cm and 100cm start vibrating. At some instant the two are in the mean position in the same phase. After how many vibrations of the shorter pendulum the two will be in phase in the mean position?
detailed solution
Correct option is B
T1=2π121g,T2=2π100gLet shorter pendulum makes n vibrations; then the longer will make one less than n to come in phase again.nT2=n−1T1ORn.2π100g=n−12π121gSolving, we get; n=11Talk to our academic expert!
Similar Questions
Two springs of spring constants are joined in series. The effective spring constant of the-combination is given by:
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests