- #1

vxr

- 25

- 2

- Homework Statement
- An apple floats in a barrel of water. If you lift the apple above its floating level by ##h = 0.02m## and release it, it bobs up and down with a period of ##T = \frac{3}{4}s##. Assuming that the motion is simple harmonic, find the position ##x##, velocity ##v## and acceleration ##a## of the apple at the times corresponding to ##1/4## and ##1/2## of period ##T##.

- Relevant Equations
- ##x = Acos(\omega t)##

Can someone double check my calculations? I will skip ##\theta## shift angle in all calculations for simplicity.

##x(t) = Acos(\omega t) \quad \land \quad \omega = \frac{2\pi}{T} \quad \land \quad A = \frac{1}{50}m = h##

##x(\frac{1}{4}T) = Acos(\frac{2\pi}{T} \frac{T}{4}) = Acos(\frac{\pi}{2}) = 0m##

##x(\frac{1}{2}T) = Acos(\frac{2\pi}{T} \frac{T}{2}) = Acos(\pi) = \frac{1}{50}cos(\pi) = -\frac{1}{50}m##

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##v = \frac{dx}{dt} = \frac{d}{dt} \Bigg( Acos(\omega t) \Bigg) = -A \omega sin (\omega t)##

##v(\frac{1}{4}T) = \frac{-2A \pi}{T} sin \Big( \frac{2\pi}{t} \frac{T}{4} \Big) = \frac{-2A \pi}{\frac{3}{4}} sin \Big( \frac{\pi}{2} \Big) = \frac{-8A \pi}{3} = -0.168 \frac{m}{s}##

##v(\frac{1}{2}T) = \frac{-2A \pi}{\frac{3}{4}} sin \Big( \frac{2\pi}{T} \frac{T}{2} \Big) = \frac{-8A \pi}{3} sin (\pi) = 0 \frac{m}{s}##

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##a = \frac{dv}{dt} = \frac{d^2 x}{dt^2} = \frac{d}{dt} \Bigg( -A\omega sin(\omega t)\Bigg) = -A\omega^2 cos(\omega t) = -\omega^2 x##

##a(t) = -\omega^2 x(t)##

##\Longrightarrow a(\frac{1}{4}T) = 0 \frac{m}{s^2}##

##a(\frac{1}{2}T) = -\omega^2 * (-\frac{1}{50}m) = \omega^2 \frac{1}{50}##

##\begin{cases} a(\frac{1}{2}T) = \omega^2 \frac{1}{50} \\ \omega = \frac{2\pi}{T}\end{cases} \Longrightarrow a(\frac{1}{2}T) = \frac{1}{50} \frac{4\pi^2}{T^2} = \frac{4\pi^2}{50} \frac{1}{(\frac{3}{4})^2} = \frac{4\pi^2}{50} \frac{16}{9} = \frac{64 \pi^2}{450} \frac{m}{s^2}##

##x(t) = Acos(\omega t) \quad \land \quad \omega = \frac{2\pi}{T} \quad \land \quad A = \frac{1}{50}m = h##

**1.1:**##x(\frac{1}{4}T)## = ?##x(\frac{1}{4}T) = Acos(\frac{2\pi}{T} \frac{T}{4}) = Acos(\frac{\pi}{2}) = 0m##

**1.2:**##x(\frac{1}{2}T)## = ?##x(\frac{1}{2}T) = Acos(\frac{2\pi}{T} \frac{T}{2}) = Acos(\pi) = \frac{1}{50}cos(\pi) = -\frac{1}{50}m##

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**2.1:**##v(\frac{1}{4}T)## = ?##v = \frac{dx}{dt} = \frac{d}{dt} \Bigg( Acos(\omega t) \Bigg) = -A \omega sin (\omega t)##

##v(\frac{1}{4}T) = \frac{-2A \pi}{T} sin \Big( \frac{2\pi}{t} \frac{T}{4} \Big) = \frac{-2A \pi}{\frac{3}{4}} sin \Big( \frac{\pi}{2} \Big) = \frac{-8A \pi}{3} = -0.168 \frac{m}{s}##

**2.2:**##v(\frac{1}{2}T)## = ?##v(\frac{1}{2}T) = \frac{-2A \pi}{\frac{3}{4}} sin \Big( \frac{2\pi}{T} \frac{T}{2} \Big) = \frac{-8A \pi}{3} sin (\pi) = 0 \frac{m}{s}##

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**3.1:**##a(\frac{1}{4}T)## = ?##a = \frac{dv}{dt} = \frac{d^2 x}{dt^2} = \frac{d}{dt} \Bigg( -A\omega sin(\omega t)\Bigg) = -A\omega^2 cos(\omega t) = -\omega^2 x##

##a(t) = -\omega^2 x(t)##

##\Longrightarrow a(\frac{1}{4}T) = 0 \frac{m}{s^2}##

**3.2:**##a(\frac{1}{2}T)## = ?##a(\frac{1}{2}T) = -\omega^2 * (-\frac{1}{50}m) = \omega^2 \frac{1}{50}##

##\begin{cases} a(\frac{1}{2}T) = \omega^2 \frac{1}{50} \\ \omega = \frac{2\pi}{T}\end{cases} \Longrightarrow a(\frac{1}{2}T) = \frac{1}{50} \frac{4\pi^2}{T^2} = \frac{4\pi^2}{50} \frac{1}{(\frac{3}{4})^2} = \frac{4\pi^2}{50} \frac{16}{9} = \frac{64 \pi^2}{450} \frac{m}{s^2}##