Standing Waves

If you have a solid understanding of what Travelling waves are (see previous sub-section if you need a refresher) then when you add up a sine wave moving to the right with a wave moving to the left, you get a standing wave.

y(x,t) = \sin(kx-\omega t) + \sin(kx + \omega t)

Using \sin(a) + \sin(b) = 2 \sin((a+b)/2) \cos((a-b)/2) and simplifying the above equation we get :

y(x,t) = 2\sin(kx)\cos(\omega t)

A plot of this looks like the following:

Red circles denote the nodes where the amplitude of vibration is 0. Anti nodes on the other hand are the regions of maximum amplitude oscillations.

For the most part when one is referring to standing waves, it is customary to just talk about the resultant wave that you see above.

But one should understand that the way you form this is by taking a right moving wave and adding it up with a left-moving wave:

Blue – Traveling wave moving to the right. Red- Traveling wave moving to the left

Beats + Doppler effect : How do radar guns work?

Concepts involved:

Doppler Effect and Beats

Radar stands for RAdio Detection And Ranging. The way radar guns work to find out the speed of an object is that a high frequency radio wave is transmitted from the gun onto the object whose speed you would like to measure.

This wave bounces off the moving object and then returns back to the radar gun. A circuit in the gun amplifies the signals and adds them together.

Case – I

Let’s consider a stationary object. Any wave that hits the object reflects back and get back to you in the same frequency. If you add these signals up you would get a wave with the following amplitude:

y(t)  =\sin(2 \pi f t) + \sin(2 \pi f t + \phi)

Case-II

Now if that object is moving with some velocity, your reflected wave would be doppler shifted. This means the frequency of the received wave is different than the transmitted wave. And when you add those two signals up you get:

y = sin(2 \pi f t) + sin(2 \pi f_1 t) \rightarrow \text{Beats Phenomenon}

f_{beats} = | f - f_1 |

Since each velocity corresponds to a particular beat frequency, radar guns use the beat frequency as a measure to find out how fast an object is moving.

If you would like to learn more about the circuit underlying radar guns, a useful resource would be this video.

Can you swim in a sea of Sulphur Hexafluoride?

In this demonstration, Sulphur Hexafluoride which is denser than air is poured onto a fish tank. Since it is denser than air, it accumulates in the bottom of the container and a buoyant force acts to keep the aluminum boat afloat in this sea of Sulphur Hexafluoride.

You can use the principles that you have learned in the last few sections to tackle this problem as well.

The only difference is that \rho_f \rightarrow \text{Density of Sulphur Hexafluoride} . And if \rho_{avg} > \rho_f , the boat would sink

Questions to ponder:

  • People float on the dead sea because the density of sea water is greater than the average human density. If you consider a sea of Sulphur Hexafluoride, what are the conditions under which a human could possibly float on it?

Source for the Sulphur Hexafluoride demonstration – https://www.youtube.com/watch?v=1PJTq2xQiQ0

The travelling wave: An intuition

The aim of this section is to understand the traveling wave solution.

We all know about our favorite function – ‘The sinusoid’.

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y becomes 0 whenever sin(x) = 0 i.e x = n \pi  (n=1,2,3 ...)

Now the form of the traveling sine wave is as follows:

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When does the value for y become 0 ? Well, it is when

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As you can see this value of x is dependent on the value of time t ,which means that the points where y=0 moves to the right as time progresses.

\text{(t=0)} x = n \pi

\text{(t=1)} x = n \pi  + \omega

\text{(t=2)} x = n \pi + 2\omega

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Here is a slowly moving forward sine wave for reference.

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You can now try to convince yourself that a sine wave moving to the left is given by :

y= A sin(x + \omega t )

** The general form of a traveling wave is y = sin(kx - \omega t) . For the sake of simplicity we have considered the case where k=1 but understand that you can apply the same argument to any value of k as well.

Vibrating string fixed at both ends

Consider a string that is clamped at x = 0 and x= L (i.e y(0) = 0; y(L) = 0 ) undergoing traverse vibrations. And you would like to know the motion of the string.

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Maybe you know a priori that the solutions are sinusoids but you have no information on its wave number.

So you start trying out every single possibility of the wave number.

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Few more solutions that satisfy the boundary condition

The important thing to understand here is that If there weren’t any boundary conditions that was imposed on the string then all possible sinusoidal wave would be a solution to the problem.

But the existence of a boundary condition acts as a constraint and restricts the total number of possibilities.

Fun Sidenote:

Perhaps you have stumbled upon the word ‘quantization’, ‘quantized’, ‘discrete energy states’ when people talk about atoms. If not, you will most probably hear it somewhere when you are taking an undergraduate class in physics.

If you have an electron in a hydrogen atom, there are only specific energy levels it can be observed to occupy when its energy is measured because the electron is trapped in the atom which is a type of Boundary condition -> restricts the total number of states possible -> identical to the string scenario -> leads to quantization

But if the electron is unbound or there are no boundary conditions, the electron can in theory take any energy state it wants. You cannot have quantized states if you do not have boundary conditions.

Simple Harmonic Oscillator: An intuition

You have a mass suspended on a spring. We want to know where the mass will be at any instant of time.

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The physical solution

Now before we get on to the math, let us first visualize the motion by attaching a spray paint bottle as the mass.

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Oh, wait that seems like a function that we are familiar with – The sinusoid.

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Without even having to write down a single equation, we have found out the solution to our problem. The motion that is traced  by the mass is a sinusoid.

But what do I mean by a sinusoid ?

If you took the plotted paper and tried to create that function with the help of sum of polynomials i.e x, x2, x3 … Now you this what it would like :

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By taking an infinite of these polynomial sums you get the function Since this series of polynomial occurs a lot, its given the name – sine.

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A note on the cosine

Since cosine is merely the sine function pushed to the side, the analogy works the same way.

Therefore we can conclude that the motion described a mass attached to a spring is a sine or a cosine function:

y(t) = A \sin( \omega t + \phi) \ \text{or} \  A \cos(\omega t + \phi)

Hopefully this gives you an intuition on why the solution to a simple harmonic oscillator is given by the sine or the cosine function. Following runs through how you would go about getting it from solving a differential equation.

\text{solution} \ x = A \cos(\omega t + \phi) \  \text{or} A \sin(\omega t + \phi)

Note on average density and why ships do not sink

Floating on the dead sea

Let’s ask a very generic question: I hand you an object and ask you to predict whether the object would float or sink. How would you go about doing that ? Well, you can measure the mass of the object and the volume of the object and can derive this quantity called Average Density (\rho_{avg} )

\rho_{avg} = m_{object}/V_{object}

It is the average density of the entire object as a whole. If this object is submerged in a fluid of density \rho_f , then we can draw the following force diagram:

If \rho_{avg} > \rho_{f} , we note that this generic object would sink and if \rho_{avg} < \rho_{f} it would float!. Therefore in order to make any object float in water, you need to ensure its average density is less than the density of the fluid its submerged in!

Why does a ship stay afloat in sea?

A ship is full of air! Although it is made from iron which sinks in water but with all the air that it is full of, it’s average density (m_{ship}/V_{ship} ) drops down such that \rho_{avg-ship} < \rho_{sea-water} .

Fun Experiment:

If you drop some raisins in soda, you will notice that they raise up and fall down like so (Try it out!):

Source

This is because air bubbles that form on the top of the raisin decrease its average density to the point that its able to make the raisin raise all the way from the bottom to the top. BUT once it reaches the top all the air bubbles escape into the atmosphere (its average density increases) and the raisin now falls down.

Questions to ponder:

  • Why do people not sink in the dead sea ?
  • How are submarines/divers able to move up and down the ocean ? How would you extend the average density argument in this case.
  • Why do air bubbles in soda always want to raise up ?
  • If the total load that needs to on a ship is 25 tons. What should be the total volume of the ship in order to remain afloat if the density of sea water is 1029 kg/m3,

How do hot air balloons work?

Source

One of the most surprising things about air that may not be intuitive is that it is a fluid and like any other fluids exerts a pressure on objects.

Standing on earth with layers of air above us, we are constantly being ‘weighed down’ by a pressure of ~1atm at all times.

All objects in air are also assisted by a buoyant force that is caused from the pressure difference between the top and bottom surfaces.

Let’s now consider the hot-air balloon in particular. A force diagram is probably the best way to start:

Envelope is the the actual fabric which holds the air inside the balloon

Therefore we see that in order for the hot air balloon to float, we need to have the buoyant force compensate for the net weight from the balloon, the load on it and the air inside it:

F_B > w_{envelope} + w_{load} + w_{hot-air}


Say you built a modest hot-air balloon that just barely managed to get off the ground. How would you make it go higher ?

You don’t want to play around with w_{load} because you don’t want to throw out any of your passengers or your supplies. You can’t really play around with w_{envelope} without re-making the balloon again.

BUT you can reduce the weight of the hot air inside the balloon ( w_{hot-air} = m_{hot-air} g = (\rho_{hot-air} V_{hot-air}) g ) ! How ?

Recall the ideal gas law P = \rho R T . Assuming that the pressure and the volume of the hot-air balloon does not change we note that:

\rho_{hot-air} \propto 1/T_{hot-air}

meaning that if we want to reduce the density of air, we just need to crank up the temperature.

And with a spectacular burner onboard our hot air balloon, we can easily increase the temperature of the air inside the balloon!

That’s pretty much how hot-air balloons work! You increase the temperature of the air inside the balloon to go up, decrease the temperature to go down and skillfully adjust the temperature to hover.

Questions to ponder:

  • When you increase the temperature of air, its density decreases. What do you think happens to all the molecules that were previously inside ? Do they exit the balloon ?
  • Different shaped hot-air balloons is a common sight. Do you think that the Buoyant force changes for each shape? (Review the formula for Buoyant force discussed in the previous post)
  • How do you think Helium balloons work ?

Buoyancy

Check out the video description on YouTube on the details of how buoyancy is related to each clip in the video

Useful Preliminaries :

\rho = \frac{m}{V}

w = mg = \rho Vg

p = \frac{F}{A}

p = p_0 + \rho g h

Pressure depends on depth:

In a tank filled with a fluid, the bottom most part of the tank experiences a higher pressure than the top most part of the tank.

This is simply because if you are at the bottom, there are more layers of the fluid above your head “weighing you down” than compared to the top where you have fewer layers.

Fact: Marina Trench is 1000 times the pressure at the sea level.

As a result of the pressure depending on depth, any object placed in a fluid experiences a pressure difference between its top and bottom surfaces.

The top is at a lower pressure and the bottom at a higher pressure. Therefore, there is an upward force called the ‘Buoyant Force’ that acts on the object when you submerge it in a fluid.

Anything that you submerge underwater will feel lighter than it actually is because the Buoyant force acts upward on the objects an help you counteract the effect of gravity.

F_{net} = mg - F_B

(i) If mg>F_B , object sinks

(ii) If mg<F_B , object floats

(i) If mg=F_B , object remains stationary. (It is neutrally buoyant)strived(i) If mg=F_B , object remains stationary. (It is neutrally buoyant)

Note on the Buoyant force:

  • Air is also a fluid and also offers a Buoyant force on any object.
  • The Buoyant force is a property of the fluid and has nothing to do with the nature of the object that you submerge. A 1 m^3 of Iron, Styrofoam, Lead, etc all would feel the same Buoyant force.

An equation to represent Buoyant force:

So far, all of this has been qualitative. Let’s try to obtain an expression for the Buoyant force that we can work with.

Consider an object submerged in a fluid with density \rho_f as follows:

F_B = (P_{high} - P_{low}) A

F_B = \left[ p_0 + \rho_f g (h+d) - (p_0 + \rho_f g h ) \right] A

where \rho_f -> Density of the fluid.

F_B = \rho_f  g (d A )

F_B = \rho_f g V_{object}

Notice that the Buoyant force only depends on the volume occupied by the object and not it’s density and as a result all objects with the same volume irrespective of its density experience the same Buoyant force!

We can make this even better by realizing that V_{object} = V_{fluid- displaced} because when an object submerges in water it pushes away all the fluid that was already there to occupy it for itself.

How much fluid does it need to displace ? For submerged objects it’s exactly how much volume it needs to accommodate the object in the fluid. Therefore we can rewrite the above formula like so:

F_B = (\rho_f  V_{fluid-displaced}) g

F_B = m_f g (since \rho = m/V )

This is the Archimedes principle. It reads that the Buoyant force experienced by any object is equal to the weight of the fluid displaced.

Eureka!