A body of mass 2kg is executing simple harmonic motion. Its displacement y (in cm) at t seconds is given by y=6sin(200t+π6). Find its maximum kinetic energy.
A
144kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
48kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
144J
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D144J Given,
mass of body (m)=2kg
Displacement y=6sin(200t+π6)
Comparing with, y=Asin(ωt+ϕ) A=6cm,ω=200rad/s,ϕ=π6
Maximum kinetic energy (KEmax)=12mω2A2 =12×2×(200)2×(6×10−2)2 =144J