particle of mass 250 g250 g executes a simple harmonic motion under a periodic force F=(–25x) N𝐹=(–25𝑥) N. The particle attains a maximum speed of 4 m s−14 m s-1 during its oscillation. The amplitude of the motion is ______cmcm
Question
particle of mass 250 g250 g executes a simple harmonic motion under a periodic force F=(–25x) N𝐹=(–25𝑥) N. The particle attains a maximum speed of 4 m s−14 m s-1 during its oscillation. The amplitude of the motion is ______cmcm
Solution
The maximum speed in simple harmonic motion is given by the formula:
v_max = ωA
where ω is the angular frequency and A is the amplitude of the motion.
The force in simple harmonic motion is given by the formula:
F = -mω²x
where m is the mass of the particle and x is the displacement from the equilibrium position.
From the given force equation F = -25x, we can equate this to -mω²x and solve for ω:
25 = mω²
Substituting the given mass m = 250g = 0.25kg, we get:
25 = 0.25ω²
Solving for ω, we get:
ω = sqrt(25/0.25) = 10 rad/s
Substituting ω = 10 rad/s and v_max = 4 m/s into the equation for maximum speed, we get:
4 = 10A
Solving for A, we get:
A = 4/10 = 0.4 m = 40 cm
So, the amplitude of the motion is 40 cm.
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