Questions
Two simple harmonic motions given by and are super posed on a particle. The amplitude of the resulting motion of the particle is
detailed solution
Correct option is C
y1=10sin100πt=10cos10πt−π2Phase difference ϕ=100πt−π6−100πt−π2=π3∴Resultant amplitude =102+52+2.10.5cosπ3=57Talk to our academic expert!
Similar Questions
A charged particle is deflected by two mutually perpendicular oscillating electric fields such that the displacement of the particle due to each one of them is given by respectively. The trajectory followed by the charged particle is
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
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
