SHMHard
Question
The potential energy of a 1 kg particle free move along the x-axis is given by
V(x) =
J
The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is
V(x) =
J The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is
Options
A.2
B.3/√2
C.√2
D.1/√2
Solution
kEmax = ET - Umin
Umin(±1) = - 1/4J
KEmax = 9/4 J ⇒ U =
J
Umin(±1) = - 1/4J
KEmax = 9/4 J ⇒ U =
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