To find the angular frequency, amplitude, and velocity of the particle as it passes through the equilibrium position, we can use the given information about the particle's motion.
- Angular Frequency (ω): The angular frequency (ω) of a particle undergoing simple harmonic motion is given by the formula:
ω = 2πf
where f is the frequency of oscillation. In this case, it is mentioned that the particle makes 2 oscillations in 1 second, so the frequency is 2 Hz. Therefore, the angular frequency is:
ω = 2π * 2 = 4π rad/s
- Amplitude (A): The amplitude (A) is the maximum displacement from the equilibrium position. We can find it using the given information that the particle has a velocity of 20 cm/s when it is 3 cm from the equilibrium position. The velocity of a particle undergoing simple harmonic motion is given by:
v = ωA
Given v = 20 cm/s and A = 3 cm, we can rearrange the equation to solve for A:
A = v / ω = 20 cm/s / (4π rad/s) ≈ 1.59 cm (rounded to two decimal places)
Therefore, the amplitude of the motion is approximately 1.59 cm.
- Velocity at the Equilibrium Position: At the equilibrium position, the displacement is zero, so the velocity of the particle is at its maximum. The maximum velocity in simple harmonic motion is equal to the angular frequency multiplied by the amplitude:
v_max = ωA
Substituting the values we have:
v_max = (4π rad/s) * 1.59 cm ≈ 19.88 cm/s (rounded to two decimal places)
Therefore, the velocity of the particle as it passes through the equilibrium position is approximately 19.88 cm/s.