Mechanical vibrations: floating apple

[vxr]

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.

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

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: $a(\frac{1}{2}T)$ = ?

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

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

 

[haruspex]

Problem 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.

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

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: $x(\frac{1}{2}T)$ = ?

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

--

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: $v(\frac{1}{2}T)$ = ?

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

--

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: $a(\frac{1}{2}T)$ = ?

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

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

Looks right.