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- Homework Statement
- An object oscillates in simple harmonic motion, reaching a maximum velocity of ##1.2m/s## whenever it passes through the central position, which occurs every ##3.6s##.

Find the amplitude and maximum acceleration of the harmonic motion.

- Relevant Equations
- ##v_{max}=\omega A##, ##a_{max}=\omega^2 A##, ##\omega=\frac{2\pi}{T}##

Since it passes through the origin every ##3.6s## the period is ##T=3.6s## hence ##\omega=\frac{2\pi}{\omega}=\frac{2\pi}{3.6}\frac{rad}{s}## thus ##A=\frac{v_{max}}{\omega}=\frac{1.2}{\frac{2\pi}{3.6}}m\simeq 0.69m## and ##a_{max}=\omega^2 A=(\frac{2\pi}{T})^2 A=(\frac{2\pi}{3.6})^2 \cdot 0.69\simeq 2.1\frac{m}{s^2}##.

What I have done makes sense to me so I don't understand why the solutions to this problem state that ##A=2.8m## and ##a_{max}=0.52\frac{m}{s^2}##. Comment are welcome, thanks.

What I have done makes sense to me so I don't understand why the solutions to this problem state that ##A=2.8m## and ##a_{max}=0.52\frac{m}{s^2}##. Comment are welcome, thanks.