Waves Tutorial 4 - Standing Waves

Sometimes waves appear to be standing still, i.e. the crests and the troughs appear to stay in the same place.  We can see them in water, especially water surrounded by walls.  We call them standing waves or stationary waves.  Any kind of wave can form a standing wave.

 

Musical instruments depend on standing waves:

Stationary waves are formed when two progressive waves are superposed:

If we send an incident wave down a string, which is fixed at the end, the wave is reflected at the fixed end and undergoes a phase change of p radians or 180o.  There is no phase change at the free end.

 

 

Question 1

What are the conditions needed for a standing wave?

Answer

 

Formation of Standing Waves

If we send a continuous stream of waves down the string, they are reflected and a standing wave gets set up.  The frequency will be the same, the amplitude very nearly the same and the speed will be the same.  The directions are opposite.  The phase change of p radians causes cancellation at the fixed end.  This region of zero displacement is called a node.

 

In a progressive wave, points X and Y would be in antiphase, p radians out of phase.  However, because the wave is reflected, the phase is changed by p radians.  So they are now 2p radians out of phase, which means that they are in phase.  Superposition is constructive.   The amplitude is now at a maximum, and this is called an antinode.

 

Notice:

Question 2

What is meant by a node and an antinode?  A standing wave loop in a string is 66 cm long.  What is the wavelength in metres?

Answer

 

We can set up a standing wave using 3 cm wave apparatus.

We move the probe between the transmitter and the reflector and we detect maximum readings and minimum readings with the probe, which is connected to a microammeter.  The maximum readings coincide with the antinodes; the minimum readings with the nodes.

 

A similar experiment can be done using sound waves.

 

 

Comparing Standing Waves with Progressive Waves

For all standing wave patterns, these two points are true:

 

  1. The amplitude varies according to position from zero at a node, to maximum at an antinode.  The amplitude of a given point is always the same.

  1. The phase difference between two particles is zero if the points are between adjacent noted.  It is 180o if they are either side of a node.

 

If the points are a separated by an even number of nodes, they are in phase.

 

 

Therefore P and Q are in phase with each other (as are R and S, and T and U).  P and S are in antiphase, but P is in phase with U.

 

Property

Stationary Waves

Progressive Waves

Frequency

All particles vibrate at the same frequency, except at the nodes where there is no vibration

All particles vibrate at the same frequency throughout the wave.

Amplitude

Amplitude varies between zero at the nodes and maximum at the antinodes.  S will vibrate at a bigger amplitude to R.

The amplitude is the same for all particles.

Phase difference between two points.

Phase is np rad, where n is the number of nodes between the two points.

 

Standing Waves in a String

Consider a taut string that is between two fixed points:

 

There are some rules for standing waves in strings:

 

Frequency of a Stationary Wave

We can investigate standing wave in a string using equipment like this:

 

It is called Melde's Apparatus.  If we start the frequency of the vibration at a low level, increasing it slowly, we see little of significance until at a certain value, a single large vibration loop is seen. This is due to resonance and is called the fundamental frequency or the first harmonic. The second harmonic has two vibration loops.  It is twice the fundamental frequency. We will study resonance in Physics Unit 4.

 

In the AQA syllabus, the fundamental frequency is called the first harmonic, and the term fundamental frequency will not be used.  The term overtone will not be used either.  However you will see these terms used in text books and other syllabuses.

 

The frequency at which resonance happens depends on:



 

The Speed of a Wave in a String

This is on the EDEXCEL Syllabus.

The speed of a wave in a string depends on:

The equation is:

Where:

The symbol m is 'mu', a Greek lower case letter 'm'.

 

We can show this graphically.  The speed varies with direct proportionality with the square root of the tension.

 

The speed is inversely proportional to the square root of the square root of the mass per unit length.

 

Measuring the speed of a wave on a string is not at all easy to do directly.  However we can use the idea of the fundamental frequency in a stationary wave to decide what the wave speed is.

 

Fundamental Frequency (First Harmonic) of a String

Consider a stationary wave that is formed by a wave travelling a v m s-1 along a taut string between two fixed points l m apart.  The tension in the string is T N.  The string has a mass per unit length of m kg m-1.

 

 

We know the wave equation:


 

Since this is a stationary wave at the first harmonic, we can say that the wavelength is 2l.  Therefore:

 

We also know that the speed is related to the tension by the equation:

 

 

It doesn't take a genius to see that:

 

So we can rearrange to make the fundamental frequency, f0 the subject of the formula:

 


The Physics Codes are:

 

The symbol m is 'mu', a Greek lower case letter 'm'.

 

 


If you look inside a piano, the bass strings are not much longer than the middle note strings, but they are much thicker.

 

Picture: Frost Nova, Wikimedia Commons.

 

Worked Example

A string is 2.3 m long between two fixed point.  A length of 2.50 m has a mass 53.0 g.  It is under a tension of 25.6 N.  Calculate the first harmonic, and give your answer to an appropriate number of significant figures.

Answer

Use:

 

Find the mass per unit length first:

m = 0.0530 kg ÷ 2.50 m = 0.0212 kg m-1

 

Put the numbers in:

f0 = (1 ÷ (2 × 2.3 m) × (25.6 N ÷ 0.0212 kg m-1)0.5 = 7.55 Hz = 7.6 Hz (2 s.f.)

When rearranging the equation, it is easiest to square the equation:

 

The mass per unit length is worked out from the total length of the string which may well be different from the length between the two fixed points.

 Wire masses tend to be given in grams.  Make sure you convert the mass in grams to kilograms.  (You do know how to do this, don't you?)

 

 

 

Energy and Power in a Wave on a String (A-level Extension)

Consider a wave of amplitude A m and wave length l m travelling on a string of mass per unit length m kg m-1 at a speed of v m s-1 at a frequency of f Hz.

 

The frequency is represented by the relationship:

 

The energy associated with one wavelength is given by:

 

This is consistent with the relationship we met in Tutorial 1:

 

The constant is:

 

We can check the consistency of this by looking at the units.  In tutorial 1, we saw that the units for k were J m-2 In base units:

 

1 J = 1 kg m2 s-2

The units for k are therefore:

J m-2 = kg m2 s-2 × m-2 = kg s-2

 

The units for k in the equation above are:

kg m-1 × (rad2) s-2 × m

 

Since rad is a dimensionless unit, we can say that k has units of:

kg s-2

 

Therefore the units of k are consistent in both cases.

 

 

We know that power is related to energy by:

power (W) = energy (J) ÷ time (s)

 

In waves, we know that each wave takes a time period T.  Each wave has an energy per wave of El.  So we can write an equation for the power of the wave:

 

 

So we can write:

Now we know that:

wave speed (m s-1) = wavelength (m) ÷ Time period (s)

 

So we can write:

 

We can go one step further since we know that:

So we can substitute to give:

It tidies up to give:

 

This is not on the syllabus, but you might be asked to derive it using equations that you have been given.

 

Question 3

 

 

A length of wire has a mass of 1.25 g for a length of 1.60 m. 

 

In an experiment to measure mains frequency that is shown above, it is required to fit between two fixed points 1.15 m apart and is placed between the poles of a magnet.  It is found to resonate at a first harmonic with a frequency of 50.0 Hz. 

 

(a) Calculate the tension needed.

(b) Calculate the speed of the waves in this wire.

Answer

Question 4

(Extension)  In the experiment in Question 4, the amplitude was found to be 20 mm.

(a)  Using your answers to Question 4, work out the power required.

(b)  The apparatus is connected to power supply that gives an AC potential difference of 2.5 V RMS.  What is the current?

Give your answers to an appropriate number of significant figures.

Answer

 

 

Required Practical

We can carry out an experiment using a vibration generator and a string under load:

 

We can do the experiment in two different ways:

The graphs are like this:

This graph shows that frequency2 is directly proportional to the tension.

 

This graph shows the the frequency is inversely proportional to the length of the string.

 

If you can, use a frequency meter to get the value for the frequency.  Most signal generators in school or college labs are not that well calibrated.  Alternatively you should check the calibration of the signal generator using a CRO.  If you haven't done the CRO, your tutor will show you how to do this.