Questions
The total mechanical energy of a harmonic oscillator of amplitude 1 m and force constant 200 N/m is 100 J then
Remember concepts with our Masterclasses.
80k Users 60 mins Expert Faculty Ask Questions
a
The minimum PE is zero
b
the maximum PE is 100 J
c
the minimum PE is 50 J
d
the maximum PE is 50 J
Ready to Test Your Skills?
Check Your Performance Today with our Free Mock Tests used by Toppers!detailed solution
Correct option is A
given force constant k=200N/mtotal energy TE=100 Jmaximum kinetic energy is KE =12mw2A2 maximum kinetic energy KE=12kA2 maximum kinetic energy KE=12×200×12 ∴KEmax=100 JTotal energy TE=KEmax+PEmin100=100+PEminPEmin=0 JSimilar Questions
When the displacement is half the amplitude, the ratio of potential energy to the total energy 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
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
Popular Books
HC Verma Part 1HC Verma Part 2RS AgarwalRD SharmaLakhmir SinghDK GoelSandeep GargTS GrewalNCERT solutions
Class 12 NCERT SolutionsClass 11 NCERT SolutionsClass 10 NCERT SolutionsClass 9 NCERT SolutionsClass 8 NCERT SolutionsClass 7 NCERT SolutionsClass 6 NCERT Solutions
