For the spring pendulum shown in the given figure, the value of spring constant is 3 x 104 N/m and amplitude of oscillation is 0.1 m. The total mechanical energy of oscillating system is 200 J. Mark out the correct option(s).
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
Minimum PE of the oscillating system is 50 J
b
Maximum PE of the oscillating system is 200 J.
c
Maximum KE of the oscillating system is 200 J.
d
Minimum KE of the oscillating system is 150 J.
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Total mechanical energy of the oscillating system is, E=Kmax+Umin = Kmin+Umax ,Kmin=0 at extreme position.So, Umax=E=200J Kmax=mvmax22=m×A2ω22=KA22=150JSo, Umin=50J