A 4.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The velocity is given by the expression v(t) = 8.00 cm/s sin($\omega$t)+ 10.0 cm/s cos($\omega$ t).What is the maximum acceleration of the simple harmonic motion? A) 10.0 cm/s² B) 16.2 cm/s² C) 19.2 cm/s² D) 20.2 cm/s² E) 24.5 cm/s²

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Quizplus · Accepted Answer

The maximum acceleration in simple harmonic motion is given by $ a_{max} = \omega^2 A $, where $ A $ is the amplitude of the velocity. The angular frequency $ \omega $ is calculated using $ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{9.00}{4.00}} = 1.5 \, \text{rad/s} $. The amplitude $ A $ of the velocity is $ \sqrt{(8.00)^2 + (10.0)^2} = \sqrt{164} \approx 12.8 \, \text{cm/s} $. Thus, $ a_{max} = (1.5)^2 \times 12.8 \approx 19.2 \, \text{cm/s}^2 $.

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