AP Physics 2- 14.6 Wave Interference and Standing Waves- Study Notes- New Syllabus
AP Physics 2- 14.6 Wave Interference and Standing Waves – Study Notes
AP Physics 2- 14.6 Wave Interference and Standing Waves – Study Notes – per latest Syllabus.
Key Concepts:
- Wave Interference
- Wave Superposition and Beats
- Standing Waves
Wave Interference
Wave interference is the phenomenon that occurs when two or more waves overlap in space, resulting in a new wave pattern formed by the superposition of the individual waves.
Principle of Superposition:
When two or more waves overlap, the resultant displacement at any point is the algebraic sum of the displacements of the individual waves:
$ \mathrm{y_{total} = y_1 + y_2 + \dots} $
Constructive Interference:
Occurs when waves meet in phase (\(\mathrm{\Delta \phi = 0, 2\pi, 4\pi, \dots}\)), resulting in maximum amplitude:
$ \mathrm{A_{max} = A_1 + A_2} $
Destructive Interference:
Occurs when waves meet out of phase (\(\mathrm{\Delta \phi = \pi, 3\pi, 5\pi, \dots}\)), resulting in minimum amplitude:
$ \mathrm{A_{min} = |A_1 – A_2|} $
Path Difference and Phase Difference:
Path difference is the physical distance between two wavefronts or the difference in distance traveled by two waves, measured in units of length like meters. Phase difference is the angular separation between the oscillations of two waves at a given point, measured in radians or degrees.
$ \mathrm{\Delta \phi = \dfrac{2\pi \Delta x}{\lambda}} $
where
\(\mathrm{\Delta x}\) = path difference, \(\mathrm{\lambda}\) = wavelength.
Interference Pattern:
Depending on the path difference, points of constructive and destructive interference appear. This is commonly observed in double-slit experiments and other wave superposition setups.
Example :
Two coherent waves of equal amplitude \(\mathrm{A = 0.1 \, m}\) interfere. If they meet in phase, find the resultant amplitude. If they meet out of phase, find the resultant amplitude.
▶️ Answer/Explanation
Step 1: Constructive interference (in phase): \(\mathrm{A_{max} = A_1 + A_2 = 0.1 + 0.1 = 0.2 \, m}\)
Step 2: Destructive interference (out of phase): \(\mathrm{A_{min} = |A_1 – A_2| = |0.1 – 0.1| = 0 \, m}\)
Final Answer: \(\mathrm{A_{max} = 0.2 \, m}, \; A_{min} = 0 \, m\)
Example :
In a double-slit experiment, light of wavelength \(\mathrm{600 \, nm}\) produces bright fringes on a screen. The path difference between the waves arriving at a point is \(\mathrm{900 \, nm}\). Determine whether this point corresponds to constructive or destructive interference.
▶️ Answer/Explanation
Step 1: Path difference for constructive interference: \(\mathrm{\Delta x = n \lambda, \; n = 0, 1, 2, \dots}\)
Step 2: Compare given path difference: \(\mathrm{\Delta x / \lambda = 900/600 = 1.5}\)
Step 3: Fractional value (1.5) indicates half-integer multiple → destructive interference.
Final Answer: The point corresponds to destructive interference.
Wave Superposition and Beats
Graphs of wave pulses or periodic waves are useful to determine the resultant wave when two waves interact. By adding the displacements point by point (algebraic sum), we can predict regions of constructive or destructive interference.
Beats:
Beats occur when two waves of slightly different frequencies interfere. The superposition produces a wave whose amplitude varies periodically in time, producing a fluctuation in loudness (for sound waves) or intensity (for other waves).
Mathematical Description:
$ \mathrm{y_1 = A \sin(2\pi f_1 t), \quad y_2 = A \sin(2\pi f_2 t)} $
$\mathrm{y = y_1 + y_2 = 2A \cos\Big(\pi (f_1 – f_2) t \Big) \sin\Big(\pi (f_1 + f_2) t \Big)} $
- The term \(\mathrm{2A \cos(\pi(f_1-f_2)t)}\) represents the envelope that varies slowly in time (beat pattern).
The frequency of amplitude modulation (beat frequency) is:
$\mathrm{f_{beat} = |f_1 – f_2|}$
Applications of Beats:
- Tuning musical instruments by listening to beats.
- Measuring small frequency differences between two sources.
- Interference analysis in acoustics and electronics.
Example :
Two tuning forks of frequencies \(\mathrm{256 \, Hz}\) and \(\mathrm{260 \, Hz}\) are sounded together. Determine the beat frequency.
▶️ Answer/Explanation
Step 1: Beat frequency: \(\mathrm{f_{beat} = |f_1 – f_2|}\)
Step 2: \(\mathrm{f_{beat} = |260 – 256| = 4 \, Hz}\)
Final Answer: Beat frequency = \(\mathrm{4 \, Hz}\)
Example :
Two waves on a string have frequencies \(\mathrm{f_1 = 100 \, Hz}\) and \(\mathrm{f_2 = 102 \, Hz}\). Sketch the amplitude envelope and calculate the beat frequency.
▶️ Answer/Explanation
Step 1: Beat frequency: \(\mathrm{f_{beat} = |f_1 – f_2| = |102 – 100| = 2 \, Hz}\)
Step 2: Amplitude varies as \(\mathrm{2A \cos(\pi(f_1-f_2)t) = 2A \cos(2\pi t)}\), producing 2 beats per second.
Step 3: The resultant wave oscillates at average frequency \(\mathrm{f_{avg} = (f_1 + f_2)/2 = 101 \, Hz}\) with slowly varying amplitude (envelope).
Standing Waves
A standing wave is formed when two waves of the same frequency and amplitude travel in opposite directions in a confined region and interfere, producing a stationary pattern of nodes and antinodes.
Nodes and Antinodes
Node (N):
A point where the amplitude is always zero.
$ \mathrm{y_{node} = 0} $
Antinode (A):
A point where the amplitude reaches its maximum.
$ \mathrm{y_{antinode} = 2A} $ if each wave has amplitude \(\mathrm{A}\).
Spacing:
Distance between adjacent nodes or antinodes:
$ \mathrm{\Delta x = \dfrac{\lambda}{2}}$
Wavelength and Boundary Conditions
The possible wavelengths depend on the size of the region (\(\mathrm{L}\)) and the boundary conditions:
String fixed at both ends:
$ \mathrm{\lambda_n = \dfrac{2L}{n}}, \quad n = 1, 2, 3, \dots $
Pipe open at both ends: Same as above.
Pipe closed at one end: Only odd harmonics exist:
$ \mathrm{\lambda_n = \dfrac{4L}{n}}, \quad n = 1, 3, 5, \dots $
Frequency and Harmonics
Frequency of the nth harmonic:
$ \mathrm{f_n = \dfrac{v}{\lambda_n}} $
- Fundamental (first harmonic): longest wavelength, lowest frequency.
- Second harmonic: second-longest wavelength (or next allowed harmonic).
- Third harmonic: third-longest wavelength, and so on.
- Pipe closed at one end: only odd harmonics appear.
Visual diagrams of standing waves show the location of nodes and antinodes. These representations help in determining the relationship between:
- Length of the medium (\(\mathrm{L}\))
- Wavelength (\(\mathrm{\lambda}\))
- Harmonic number (\(\mathrm{n}\))
- Frequency (\(\mathrm{f_n}\))
- Wave speed (\(\mathrm{v}\))
Applications
- Vibrating strings in musical instruments
- Sound in organ pipes, flutes, and other wind instruments
- Microwave and optical cavities
Example :
A string of length \(\mathrm{L = 1.2 \, m}\) is fixed at both ends. The wave speed on the string is \(\mathrm{v = 240 \, m/s}\). Find the wavelength and frequency of the second harmonic.
▶️ Answer/Explanation
Step 1: Wavelength of the second harmonic (\(n = 2\)): $ \mathrm{\lambda_2 = \dfrac{2L}{n} = \dfrac{2 \times 1.2}{2} = 1.2 \, m} $
Step 2: Frequency: $ \mathrm{f_2 = \dfrac{v}{\lambda_2} = \dfrac{240}{1.2} = 200 \, Hz} $
Final Answer: \(\mathrm{\lambda_2 = 1.2 \, m}, \; f_2 = 200 \, Hz\)
Example :
A pipe closed at one end has length \(\mathrm{L = 0.85 \, m}\). The speed of sound is \(\mathrm{v = 340 \, m/s}\). Determine the frequency of the third harmonic.
▶️ Answer/Explanation
Step 1: For a pipe closed at one end, odd harmonics only: \(n = 1, 3, 5, \dots\)
Step 2: Wavelength of the third harmonic (\(n = 3\)): $ \mathrm{\lambda_3 = \dfrac{4L}{n} = \dfrac{4 \times 0.85}{3} \approx 1.133 \, m} $
Step 3: Frequency: $ \mathrm{f_3 = \dfrac{v}{\lambda_3} = \dfrac{340}{1.133} \approx 300 \, Hz} $
Final Answer: Frequency of third harmonic = \(\mathrm{300 \, Hz}\)