SHMHard
Question
The potential energy of a particle of mass 0.1 kg, moving along the x-axis, is given by U = 5x (x − 4) J, where x is in meters. It can be concluded that
Options
A.the particle is acted upon by a constant force
B.the speed of the particle is maximum at x = 2 m
C.the particle executes SHM
D.the period of oscillation of the particle is (π/5) sec
Solution
F = 
= − [10x − 20] = 20 − 10x
acceleration a =
= 100(2−x) = −ω2(x − 2)
a = 0 at x = 2 So V is maximum at x = 2
This is equation of S.H.M so particle executes S.H.M
also ω2 = 100 ⇒ ω = 10
T =
= π/5 sec.
= − [10x − 20] = 20 − 10xacceleration a =
= 100(2−x) = −ω2(x − 2) a = 0 at x = 2 So V is maximum at x = 2
This is equation of S.H.M so particle executes S.H.M
also ω2 = 100 ⇒ ω = 10
T =
= π/5 sec.Create a free account to view solution
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