The equilibrium condition tells us that all the forces must balance so

$PA-(P+dP)A-\rho gA\,dy=0$

giving us a differential form of the relationship between pressure and depth

$\frac{dP}{dy}=-\rho g$

$\frac{dP}{dy}=-\rho g$

$\int_{P_{1}}^{P_{2}}\,dP=-\int_{y_{1}}^{y_{2}}\rho g\,dy$

$P_{2}-P_{1}=-\rho g(y_{2}-y_{1})$

The depth $h=y_{2}-y_{1}$ and the pressure at the top $P_{2}$ is atmospheric pressure $P_{0}$

We can therefore say that the pressure at depth $h$, $P=P_{1}$ is

$P=P_{0}+\rho gh$

We can see that we need to take in to account the pressure of the atmosphere above the container!

If you measure the pressure in your tires you are actually measuring the pressure difference between the atmospheric pressure. The absolute pressure is therefore the sum of the gauge pressure $P_{G}$, which is what you measure, and the atmospheric pressure $P_{0}$

$P=P_{0}+P_{G}$

In a fluid of uniform density $\rho_{f}$

$F_{B}=F_{2}-F_{a}=\rho_{f}gA(h_{2}-h_{1})=\rho_{f}gA\Delta h=\rho_{f}Vg=m_{f}g$

We can note from this equation that the buoyant force does not depend on the depth of the object or on the density of the object. The buoyant force is determined by the weight of the fluid which is displaced by the object.

In a cylindrical tube of radius $R$ and length $l$ the **volume** rate of flow $Q$ is related to the change in pressure from the beginning of the tube $P_{1}$ to the end of the tube $P_{2}$ by Poiseulle's equation. $\eta$ is the viscosity of the fluid.

$Q=\frac{\pi R^{4}(P_{1}-P_{2})}{8\eta l}$

$m\frac{d^{2}x}{dt^{2}}=-kx$

$x=A\cos(\omega t+\phi)$

$A$ is the amplitude, $\omega=\frac{2\pi}{T}=2\pi f$ and $\phi$ allows us to change the starting point of the motion.

$\frac{dx}{dt}=v=-\omega A\sin(\omega t+\phi)$

$\frac{d^{2}x}{dt^{2}}=a=-\omega^{2}A\cos(\omega t+\phi)$

$-m\omega^{2}A\cos(\omega t+\phi)=-kA\cos(\omega t+\phi)$

which is true if

$\omega^{2}=\frac{k}{m}$

$T=2\pi\sqrt{\frac{m}{k}}$

The restoring force above is $F=-mg\sin\theta$

For small angles $\sin\theta=\theta$ so $F\approx-mg\theta$

and using the relation $s=l\theta$ gives $F\approx-\frac{mg}{L}s$

$f=\frac{\omega}{2\pi}=\frac{1}{2\pi}\sqrt{\frac{g}{l}}$

$T=\frac{1}{f}=2\pi\sqrt{\frac{l}{g}}$

**Any** object attached at a which is not at it's center of gravity can act like a pendulum.

$T=2\pi\sqrt{\frac{I}{mgh}}$

A torsion pendulum has a restoring force provided by a torsion spring.

The torque provided by the spring is

$\tau=-K\theta$

$I\frac{d^{2}\theta}{dt^{2}}=-K\theta$

so a torsion pendulum can execute simple harmonic motion $\theta=\theta_{max}\cos(\omega t+\phi)$ with $\omega=\sqrt{\frac{K}{I}}$

The damping force opposes the motion and can sometimes be approximated as proportional to the speed

$F_{damping}=-bv$

Newton's second law gives us

$ma=-kx-bv$

or

$m\frac{d^{2}x}{dt^{2}}+b\frac{dx}{dt}+kx=0$

The solution to this equation is of the form

$x=Ae^{-\gamma t}\cos(\omega' t)$

where

$\gamma=\frac{b}{2m}$ and $\omega'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}}}$

Depending on the ratio between the damping coefficient and restoring force, a system will display either underdamping, overdamping or critical damping.

If we subject an oscillator to an oscillatory force

$F_{ext}=F_{0}\cos\omega t$

the equation of motion is

$ma=-kx-bv+F_{0}\cos\omega t$

or

$m\frac{d^{2}x}{dt^{2}}+b\frac{dx}{dt}+kx=F_{0}\cos\omega t$

A solution to this equation is

$x=A_{0}\sin(\omega t+\phi)$

where

$A_{0}=\frac{F_{0}}{m\sqrt{(\omega-\omega_{0}^{2})+b^{2}\omega^{2}/m^{2}}}$

$\phi_{0}=\tan^{-1}\frac{\omega_{0}^2-\omega^{2}}{\omega(b/m)}$

If we freeze a wave at a certain time the displacement of the points can often be represented a sinusoidal function, $D(x)=A\sin\frac{2\pi}{\lambda}x$.

If the wave moves so that it takes a time $T$, the period for a wavelength $\lambda$ to pass a point we can say that the velocity of a wave is $v=\frac{\lambda}{T}=f\lambda$.

In order to have a wave which at time $t=0$ has a displacement function $D(x)=\sin\frac{2\pi}{\lambda}x$ propagate with $v$ we can write

$D(x,t)=A\sin\frac{2\pi}{\lambda}(x-vt)$

which using $v=\frac{\lambda}{T}=f\lambda$ can be written

$D(x,t)=A\sin(\frac{2\pi}{\lambda}x-\frac{2\pi t}{T})=A\sin(kx-\omega t)$

$k$ is the wave number, $k=\frac{2\pi}{\lambda}$ and the angular frequency $\omega=2\pi f$

The velocity of the waves propagation, which we call the phase velocity can be expressed in terms of $\omega$ and $k$

$v=f\lambda=\frac{\omega}{2\pi}\frac{2\pi}{k}=\frac{\omega}{k}$

The intensity is the average power per unit area

$I=\frac{\bar{P}}{S}$

As a 3 dimensional wave propagates from a point the area through which the wave passes increases, when the power out put is constant, the intensity decreases as $\frac{1}{r^{2}}$

$\frac{I_{2}}{I_{1}}=\frac{\bar{P}/4\pi r_{2}^{2}}{\bar{P}/4\pi r_{1}^{2}}=\frac{r_{1}^2}{r_{2}^{2}}$

In order for the power output to be constant the amplitude must also decrease as $S_{1}A_{1}^2=S_{2}A_{2}^2$ implying $A\propto\frac{1}{r}$

$\lambda=2l,l,2/3l,l/2,..$ etc.

$f=\frac{v}{\lambda}=\frac{v}{2l},\frac{v}{l},\frac{3v}{2l},2vl,..$ etc.

Our sensitivity to the loudness of sound is logarithmic, a sound that is ten time as intense sounds only twice as loud to us. The sound level $\beta$ is thus measured on a logarithmic scale in decibels is

$\beta=10\log_{10}\frac{I}{I_{0}}$

$I_{0}$ is the weakest sound intensity we can hear $I_{0}=1.0\times 10^{-12}\mathrm{W/m^{2}}$

Beats occur when two waves with frequencies close to one another interfere.

If the two waves are described by

$D_{1}=A\sin2\pi f_{1}t$

and

$D_{2}=A\sin2\pi f_{2}t$

$D=D_{1}+D_{2}$

$D=2A\cos2\pi(\frac{f_{1}-f_{2}}{2})t\sin2\pi(\frac{f_{1}+f_{2}}{2})t$

A general formula for the doppler effect is

$f'=f\frac{(v_{sound}\pm v_{obs})}{(v_{sound}\mp v_{source})}$

Top part of the $\pm$ or $\mp$ sign is for an source or observer moving towards each other, the bottom part is for motion away from each other.