A 9.00 kg mass is connected to a spring with a spring constant of 4.00 N/m. The velocity is given by the expression v(t) = (12.8 cm/s) cos(ω t + π/4). What is the amplitude of the oscillations?
A) 13.6 cm
B) 19.2 cm
C) 35.3 cm
D) 40.4 cm
E) 45.5 cm

Question

Quizplus · Accepted Answer

The amplitude of oscillation can be found using the formula for the velocity of a mass-spring system in simple harmonic motion, $v(t) = -A\omega \sin(\omega t + \phi)$, where $A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant. Given $v(t) = (12.8 \, \text{cm/s}) \cos(\omega t + \pi/4)$, we can compare this to the standard form by recognizing that the cosine function can be shifted to a sine function with a different phase, but the amplitude and angular frequency remain the same. The amplitude of the velocity function does not directly give the amplitude of the oscillation, but it's related to the maximum velocity.The angular frequency $\omega$ can be found from the spring constant $k$ and the mass $m$ using the formula $\omega = \sqrt{\frac{k}{m}}$. Substituting $k = 4.00 \, \text{N/m}$ and $m = 9.00 \, \text{kg}$, we get:$$\omega = \sqrt{\frac{4.00}{9.00}} = \frac{2}{3} \, \text{s}^{-1}$$The maximum velocity $V_{\text{max}}$ is given as $12.8 \, \text{cm/s}$, which is equal to $A\omega$. Therefore, to find the amplitude $A$, we rearrange the equation to $A = \frac{V_{\text{max}}}{\omega}$:$$A = \frac{12.8 \, \text{cm/s}}{\frac{2}{3} \, \text{s}^{-1}} = 19.2 \, \text{cm}$$Thus, the amplitude of the oscillations is 19.2 cm.

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