An SHM system is oscillating with angular frequency w = 5.6 rad s-1 and amplitude a = 0.12 m. Calculate the maximum value of the velocity of the oscillator, in m s-1 , to 3 s.f.

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BlackTom AI · Accepted Answer

Consider the standard result for simple harmonic motion: the maximum speed occurs at the equilibrium position and equals v_max = ωA, where ω is the angular frequency and A is the amplitude.

First, identify the given quantities: angular frequency ω = 5.6 rad s−1 and amplitude A = 0.12 m.

Next, apply the formula v_max = ωA by substituting the given values: v_max = 5.6 × 0.12.

Perform the multiplication: 5.6 × 0.12 = 0.672. This is the raw numerical result for the maximum velocity in SI units (m s−1).

Since the problem asks for 3 significant figures, round 0.672 to 3 s.f. The value 0.672 already has three significant figures, so it remains 0.672 m s−1.

Thus, the maximum speed of the oscillator, to three significant figures, is obtained by using the relationship v_max = ωA and results in the value derived from the substitution and multiplication above.

B27FF 1. Periodic Motion and Simple Harmonic Motion

An SHM system is oscillating with angular frequency w = 5.6 rad s-1 and amplitude a = 0.12 m. Calculate the maximum value of the velocity of the oscillator, in m s-1 , to 3 s.f.

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