The potential energy of a particle of mass 0.1 kg, moving along the x-axis, is where x is in meters. It can be concluded that (a) the particle is acted upon by a variable force -) the minimum potential energy during motion is-20 j (c) the speed of the particle is maximum at x 2m (d) x 2 is position of unstable equilibrium.
Answers
Answer:
Explanation:
=> Speed of the particle is maximum in mean position. [ F = 0 ]. i.e.
F = − 10 x + 20 = 0
⇒ x = 2 m
=> Speed of the particle is maximum at x = 2 m .
F = -dU/dx
= -10 (x - 2) N
=> Therefore, motion is simple harmonic with mean position at x = 2 m.
ω = √ k / m
= √ 10 / 0.1
= 10 rad/s
T = 2π / ω
= π / 5 s
Thus, the period of oscillation of the particle is π/5 s.
Answer:Speed of the particle is maximum in mean position. [ F = 0 ]. i.e.
F = − 10 x + 20 = 0
⇒ x = 2 m
= Speed of the particle is maximum at x = 2 m .
F = -dU/dx
= -10 (x - 2) N
= Therefore, motion is simple harmonic with mean position at x = 2 m.
ω = √ k / m
= √ 10 / 0.1
= 10 rad/s
T = 2π / ω
= π / 5 s
Hence , the period of oscillation of the particle is π/5 s.
Explanation: