SHMHard
Question
The potential energy of a particle of mass 100g. moving along the x - axis, is given by U = 5x(x - 4) J, where x is in meter. Select alternative (s).
Options
A.The particle execute SHM with mean position at x = 4 m
B.The particle execute SHM with mean position at x = 2 m
C.The particle execute SHM with time period π/5 second
D.The particle execute SHM with time period π/10 second
Solution
U = 5x2 - 20x ⇒ F = - 
= - 10x + 20 = - (x - 2)
Acceleration a =
= - 100(x - 2)so ω2 = 100 ⇒ ω2 = 100 ⇒ ω = 10
Time period T =
sec
= - 10x + 20 = - (x - 2)Acceleration a =
= - 100(x - 2)so ω2 = 100 ⇒ ω2 = 100 ⇒ ω = 10Time period T =
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