Questions
An object of mass 0.2 kg executes simple harmonic motion along the x-axis with a frequency of 25/π Hz. At the position x = 0.04 m, the object has a kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation is
detailed solution
Correct option is B
E=12mω2A2=12m(2πf)2A2 A=12πf2Em Putting E = K+U, we get A=12π(25/π)2×(0.5+0.4)0.2=0.06mTalk to our academic expert!
Similar Questions
One end of an ideal spring is connected with a smooth block with the other end with rear wall of driving cabin of a truck as shown in figure. Initially, the system is at rest. If truck starts to accelerate with a constant acceleration ‘a’ , then the block (relative to truck ):
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