User:Stormx2

$f_{o}=f_{v}\left({\frac {C}{C_{b}}}\right)-f_{c}$ where:

• $f_{o}$ is the output frequency (beat frequency),
• $f_{v}$ is the frequency of the variable oscillator with human input,
• $f_{c}$ is the frequency of the fixed (constant) oscilattor,
• $C_{b}$ is the capacitence of the aerial with no human input (base capacitence),
• $C$ is the capacitence of the aerial at the given point in time.

$f=2^{n/12}\times 440$

$f=2^{1/12}\times 440$

$f\approx 466.16Hz$

$f(x)=\sin(x)$

$f(x)=2(2\pi x\qquad mod\qquad 2)-1$

$f(x)=x-\operatorname {floor} (x)$

${\frac {1}{R_{\mathrm {total} }}}={\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}$.

$Fr=\mu R$

$0=ma$ where:

Proof of Newton's second law will involve a few calculations. Two separate variables need to be calculated, one through the use of the independent variables only. This is the theoretical acceleration, i.e. the acceleration calculated through newton's third law, given by:

$F=ma$ where:

• $F$ is force acting on the object,
• $m$ is mass of the object, and
• $a$ is acceleration of the object.

In experiments 1 and 2, force is changed and mass kept constant. In the third, the inverse is true. This leaves one unknown: acceleration.

Through the use of the dependent variables, the actual acceleration of the truck can be calculated. To do this, the truck is modeled as a particle. Through friction compensation, the forces parallel to the plane are assumed to be in equilibrium, and therefore the acceleration of the system is zero. When a force is added, the system accelerates from it's initial velocity of zero, to the velocity which can be calculated from readings taken from the light gate. Two forumlae will be used:

$v^{2}=u^{2}-2aS_{1}$, and

$v={\frac {S_{2}}{\bar {t}}}$

where:

• $v$ is the velocity of the truck when it is mid-way through the light gate,
• $u$ is the initial velocity of the truck, the moment the force is applied and it begins to move,
• $a$ is the acceleration of the truck, which remains constant,
• $S_{1}$ is the displacement the truck between the two velocity readings,
• $S_{2}$ is the length of the card on the truck,
• ${\bar {t}}$ is the average time taken for the truck to pass through the light gate

These formulae can be combined and rearranged in order to calculate the actual acceleration.

$\left({\frac {S_{2}}{\bar {t}}}\right)^{2}=u^{2}+2aS_{1}$

$\left({\frac {S_{2}}{\bar {t}}}\right)^{2}-u^{2}=2aS_{1}$

${\frac {\left({\frac {S_{2}}{\bar {t}}}\right)^{2}-u^{2}}{2S_{1}}}=a$

Constants can then be inputted to leave only a relationship between the dependent and independent variables:

• $u=0$ because the truck always starts off stationary,
• $S_{1}=0.81$. The truck actually starts before the start line. It must travel $S_{2}/2$ before it is half way over, and therefore 0.71m away from the variable $v$ being measured. $v$ is the average velocity through the light gate, and it therefore the velocity of the truck when it is exactly half way through the gate, and
• $S_{2}=0.2$. This value is given in the initial brief.

$a={\frac {\left({\frac {0.2}{\bar {t}}}\right)^{2}-0^{2}}{2\times 0.81}}$

$a={\frac {\left({\frac {0.04}{{\bar {t}}^{2}}}\right)}{1.62}}$

$a={\frac {0.04}{1.62{\bar {t}}^{2}}}$

$a={\frac {2}{81{\bar {t}}^{2}}}$

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$f={{\sqrt {T \over m/L}} \over 2L}$

$y={{\sqrt {x \over 0.003/0.6}} \over 2\times 0.6}$

$y={{\sqrt {200x}} \over 1.2}$

$y={{\sqrt {100}}{\sqrt {2x}} \over 1.2}$

$y={10{\sqrt {2x}} \over 1.2}$

$y={25{\sqrt {2x}} \over 3}$

JOHNNEYYY!!

${y-y_{1} \over y_{2}-y_{1}}={x-x_{1} \over x_{2}-x_{1}}$

MOAARR!!

$x={{\sqrt {y \over m/0.6}} \over 1.2}$

$x={{\sqrt {0.6y \over m}} \over 1.2}$

$y=3cos(\theta )+3$

$3cos(\theta )+3$

$100$

$x$

$x={\sqrt {100^{2}-(3cos(\theta )+3)^{2}}}$

$x={\sqrt {100^{2}-(3cos(\theta )+3)^{2}}}$

$y=a+bx$

$b={\frac {S_{xy}}{S_{xx}}}$

$a={\bar {y}}-b{\bar {x}}$

$y=0.036+0.93x$

$b={\frac {4\pi ^{2}}{k}}$

$f={\frac {1}{2\pi }}{\sqrt {\frac {k}{m}}}.\,$

$f^{-2}=T^{2}={\frac {4\pi ^{2}}{k}}\times m$

$F=kx\$

$k={\frac {F}{x}}$

$k={\frac {ma}{x}}$

$T^{2}=0.93x$

$T=0.96{\sqrt {x}}$

$f=1.08{\sqrt {\frac {1}{x}}}$