STANDING WAVES

STANDING WAVES

The blue wave is moving to the right and the green wave is moving to the left. As is the case in any situation in which two waves meet while moving along the same medium, interference occurs. The blue wave and the green wave interfere to form a new wave pattern known as the resultant. The resultant in the animation below is shown in black. The resultant is merely the result of the two individual waves - the blue wave and the green wave - added together in accordance with the principle of superposition.The result of the interference of the two waves above is a new wave pattern known as a standing wave pattern.

Standing waves are produced whenever two waves of identical frequency interfere with one another while traveling opposite directions along the same medium.

Standing wave patterns are characterized by certain fixed points along the medium which undergo no displacement. These points of no displacement are callednodes.(Nodes can be remembered as points ofno displacement). The nodal positions are labeled by anNin the animation above. The nodes are always located at the same location along the medium, giving the entire pattern an appearance of standing still (thus the name "standing waves"). A careful inspection of the animation will reveal that the nodes are the result of the destructive interference of the two interfering waves. At all times and at all nodal points, the blue wave and the green wave interfere to completely destroy each other, thus producing a node.

Midway between every consecutive nodal point are points which undergo maximum displacement. These points are calledantinodes; the anti-nodal nodal positions are labeled by anAN. Antinodes are points along the medium which oscillate back and forth between a largepositivedisplacement and a largenegativedisplacement. A careful inspection of the above animation will reveal that the antinodes are the result of the constructive interference of the two interfering waves.

Standing Waves on StringThere are a variety of patterns which could be produced by vibrations within a string, slinky, or rope. Each pattern corresponds to vibrations which occur at a particular frequency and is known as aharmonic.

Standing Waves on StringThe lowest possible frequency at which a string could vibrate to form a standing wave pattern is known as the fundamental frequency or the first harmonic. An animation of a string vibrating with the first harmonic is shown below.

The frequency associated with each harmonic is dependent upon the speed at which waves move through the medium and the wavelength of the medium. The speed at which waves move through a medium is dependent upon the properties of the medium (tension of the string, thickness of the string, material composition of the string, etc.). The wavelength of the harmonic is dependent upon the length of the string and the harmonic number (first, second, third, etc.). Variations in either the properties of the medium or the length of the medium will result in variations in the frequency at which the string will vibrate.

Second HarmonicThe second lowest frequency at which a string could vibrate is known as the second harmonic; the third lowest frequency is known as the third harmonic; and so on. An animation of a string vibrating with the second harmonic is shown below.

Third Harmonic

Fourth Harmonic

Fifth Harmonic

HarmonicsThere is a predictability about this mathematical relationship that allows one to generalize and deduce a statement concerning this relationship. To illustrate, consider the first harmonic standing wave pattern for a vibrating rope as shown below.One complete wave in a standing wave pattern consists of twoloops. Thus, one loop is equivalent to one-half of a wavelength.

First HarmonicIn comparing the standing wave pattern for the first harmonic with its single loop to the diagram of a complete wave, it is evident that there is only one-half of a wave stretching across the length of the string. That is, the length of the string is equal to one-half the length of a wave. Put in the form of an equation:

Second HarmonicNow consider the string being vibrated with a frequency that establishes the standing wave pattern for the second harmonic.

The second harmonic pattern consists of two anti-nodes. Thus, there are two loops within the length of the string. Since each loop is equivalent to one-half a wavelength, the length of the string is equal to two-halves of a wavelength. Put in the form of an equation:

Third HarmonicThe same reasoning pattern can be applied to the case of the string being vibrated with a frequency that establishes the standing wave pattern for the third harmonic.

The third harmonic pattern consists of three anti-nodes. Thus, there are three loops within the length of the string. Since each loop is equivalent to one-half a wavelength, the length of the string is equal to three-halves of a wavelength. Put in the form of an equation:

HarmonicsThe number of antinodes in the pattern shown in the first 3 harmonics is equal to theharmonic numberof that pattern. first harmonic has one antinode; the second harmonic has two antinodes; and the third harmonic has three antinodes.

HarmonicsThus, it can be generalized that thenth harmonic hasnantinodes wherenis an integer representing the harmonic number. Furthermore, one notices that there arenhalves wavelengths present within the length of the string. Put in the form of an equation:

Harmonics Length-Wavelength Relationship

HarmonicPattern# of loopsLength-Wavelength Relationship1st1L = 1/2 2nd2L = 2/2

3rd3L = 3/2

4th4L = 4/2

5th5L = 5/2

6th6L = 6/2

nthnL = n/2

Illustrative Example # 1Suppose that a string is 1.2 meters long and vibrates in the first, second and third harmonic standing wave patterns. Determine the wavelength of the waves for each of the three patterns.

SolutionFirst harmonic:L = () = 2(1.2 m) = 2.4 mSecond harmonic: L = (2/2) = 1(1.2 m) = 1.2 mThird harmonic: L = (3/2) = 2/3(1.2 m) = 0.8 m

Illustrative Example # 22. The string is 1.5 meters long and is vibrating as the first harmonic. The string vibrates up and down with 33 complete vibrational cycles in 10 seconds. Determine the frequency, period, wavelength and speed for this wave.

SolutionGiven: L = 1.5 m; no. of waves = 33 cycles; t = 10 sFind: a) f; b) T; c) ; d) va) The frequency refers to how often a point on the medium undergoes back-and-forth vibrations; it is measured as the number of cycles per unit of time. In this case, it isf = (33 cycles) / (10 seconds) =3.3 Hzb) The period is the reciprocal of the frequency.T = 1/f = 1/ (3.3 Hz) =0.303 sThe string is 1.5 meters long and is vibrating as the first harmonic. The string vibrates up and down with 33 complete vibrational cycles in 10 seconds. Determine the frequency, period, wavelength and speed for this wave.SolutionGiven: L = 1.5 m; no. of waves = 33 cycles; t = 10 sFind: a) f; b) T; c) ; d) vc)The wavelength of the wave is related to the length of the rope. For the first harmonic as pictured in this problem, the length of the rope is equivalent to one-half of a wavelength. That is, L = . = 2 L = 2 (1.5 m) =3.0 md) The speed of a wave can be calculated from its wavelength and frequency using the wave equation:v = f = (3.3 Hz) (3. 0 m) =9.9 m/s

The string is 1.5 meters long and is vibrating as the first harmonic. The string vibrates up and down with 33 complete vibrational cycles in 10 seconds. Determine the frequency, period, wavelength and speed for this wave.SolutionGiven: L = 1.5 mFind: a) f; b) T; c) and d) v

a) f = 33 cycles/10 s = 3.3 Hzb) T = 1/f = 1/ 3.3Hz = 0.303 s = 2(L) = 2 (1.5m) = 3.0 m v = f = (3 m)(3.3Hz) = 9.9 m/s The string is 1.5 meters long and is vibrating as the first harmonic. The string vibrates up and down with 33 complete vibrational cycles in 10 seconds. Determine the frequency, period, wavelength and speed for this wave.Solve the following.A string is 6.0 meters long and is vibrating as the third harmonic. The string vibrates up and down with 45 complete vibrational cycles in 10 seconds. Determine the frequency, period, wavelength and speed for this wave.A string is 5.0 meters long and is vibrating as the fourth harmonic. The string vibrates up and down with 48 complete vibrational cycles in 20 seconds. Determine the frequency, period, wavelength and speed for this wave.1. Solution:Given: L = 6.0 m; no. of waves = 45 cycles; t = 10 sFind: a) f; b) T; c) ; d) va) f = (45 cycles) / (10 seconds) =4.5 Hzb) The period is the reciprocal of the frequency.T = 1 / (4.5 Hz) =0.222 sc) = (2 / 3) L = (2 / 3) (6.0 m) =4.0 md) The speed of a wave can be calculated from its wavelength and frequency using the wave equation:v = f = (4.5 Hz) (4. 0 m) =18 m/s

2. SolutionGiven: L = 5.0 m; no. of waves = 48 cycles; t = 20 sFind: a) f; b) T; c) ; d) va. The frequency refers to how often a point on the medium undergoes back-and-forth vibrations; it is measured as the number of cycles per unit of time. In this case, it isf = (48 cycles) / (20 seconds) =2.4 Hzb. The period is the reciprocal of the frequency.T = 1 / (2.4 Hz) =0.417 s

c. The wavelength of the wave is related to the length of the rope. For the fourth harmonic as pictured in this problem, the length of the rope is equivalent to two full wavelengths. That is, L = 2 W where W is the wavelength. Rearranging the equation and substituting leads to the following results: = 0.5 L = 0.5 (5.0 m) =2.5 md. The speed of a wave can be calculated from its wavelength and frequency using the wave equation:v = f = (2.4 Hz) (2.5 m) =6.0 m/s