A particle starts oscillating SHM from its equilibrium position than, the ratio of kinetic energy and potential energy of the particle at time T/12 isa.2:1b.3:1c.4:1d.1:4
Kavan , 7 Years ago
Grade 11
3 Answers
Gaurav Gupta
Last Activity: 7 Years ago
Khimraj
Last Activity: 7 Years ago
Total energy is given as TE = (½)mw2A2
Kinetic energy is given as KE =( ½ )mv2 = (½)mw2A2cos2(wt).
Potential energy is given as PE = TE – KE = (½)mw2A2sin2(wt).
Value of wt = (360/T)*(T/12) = 30
So KE/PE = cot2(wt) = cot2(30) = 3.
Hope it clears.
ankit singh
Last Activity: 4 Years ago
Let the displacement be x
KE=21K(A2−x2)
PE=21Kx2
Given:
KE=PE
21K(A2−x2)=21Kx2
x=2ALet the displacement be x
KE=21K(A2−x2)
PE=21Kx2
Given:
KE=PE
21K(A2−x2)=21Kx2
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