A particle performs simple harmonic motion with a time period of 1.4 s and an amplitude of 12 mm - AQA - A-Level Physics - Question 31 - 2022 - Paper 1

Given:

• Amplitude (A) = 12 mm = 0.012 m
• Time period (T) = 1.4 s

Substituting these values into the formula:

$v_{max} = 0.012 imes \frac{2\pi}{1.4}$

Calculating:

$v_{max} = 0.012 \times 4.48799 \\approx 0.05385 \text{ m/s}$

Converting to mm/s:

$v_{max} \approx 54 \text{ mm/s}$

The maximum speed of the particle is 54 mm/s. Thus, the correct answer is:

• C: 54 mm s^{-1}.