# Learning object - Interference

asroine
Mar. 25, 2015              1 of 14

### Learning object - Interference

1. Beats Variations in Amplitude
2. What is a beat?  A beat occurs when:  Two waves have SLIGHTLY DIFFERENT FREQUENCIES  Amplitude increases and decreases due to constructive and destructive interference  CONSTRUCTIVE interference occurs when the waves are IN PHASE  DESTRUCTIVE interference occurs when the waves are OUT OF PHASE Red & Blue wave out of phase
3. What is a beat? (Continued)  A BEAT is the resulting VARIATION in AMPLITUDE.  The rate of amplitude variation is proportional to the frequency difference.  When frequency differences are large, sound is heard as two distinct tones. Constructive Interference Destructive Interference Corresponds with “In Phase” Corresponds with “Out of Phase”
4. What is a beat? (Audio/Visual Demo) This video allows one to hear beating, and see its applications in everyday life: https://www.youtube.com/watch?v=IQ1q8XvOW6g
5. Deriving the Beat Equation (Step 1) 𝑠1 𝑥0, 𝑡 = 𝑠 𝑚cos(𝑘1 𝑥0 − 𝜔1 𝑡) 𝑠2 𝑥0, 𝑡 = 𝑠 𝑚cos(𝑘2 𝑥0 − 𝜔2 𝑡) AMPLITUDE: Waves have the same amplitude (Recall: Amplitude only varies with interference) (m). ANGULAR FREQUENCY: Waves have different frequencies (rad/s). WAVE NUMBER: Wave number changes with angular frequency (𝑘 = 𝜔 𝑣 ), so the two differ (1/m). COSINE: The cosine function describes the wave shape. DISPLACEMENT: This is the displacement of the wave from the equilibrium position (m). 𝑥0, 𝑡 : The “x” variable is the position of the wave (m). The “t” variable is the time (s). *This is the general equation of a wave without the phase constant.
6. Deriving the Beat Equation (Step 2) 𝑆𝑡𝑜𝑡𝑎𝑙 𝑥0, 𝑡 = 𝑠 𝑚cos(𝑘1 𝑥0 − 𝜔1 𝑡)+ 𝑠 𝑚cos(𝑘2 𝑥0 − 𝜔2 𝑡)  Principle of Superposition: The two waves are added. 𝑆𝑡𝑜𝑡𝑎𝑙 𝑥0, 𝑡 = 𝑠 𝑚(cos(𝑘1 𝑥0 − 𝜔1 𝑡)+ cos(𝑘2 𝑥0 − 𝜔2 𝑡))  The amplitude variable is isolated. 𝑆𝑡𝑜𝑡𝑎𝑙 𝑥0, 𝑡 = 2𝑠 𝑚cos0.5(𝑘1 𝑥0 − 𝜔1 𝑡 + 𝑘2 𝑥0 − 𝜔2 𝑡) × cos0.5(𝑘1 𝑥0 − 𝜔1 𝑡 − 𝑘2 𝑥0 − 𝜔2 𝑡) 𝑆𝑡𝑜𝑡𝑎𝑙 𝑥0, 𝑡 = 2𝑠 𝑚cos(( 𝑘1+𝑘2 2 )𝑥0 − ( 𝜔1+𝜔2 2 )𝑡) × 𝑐𝑜𝑠(( 𝑘1−𝑘2 2 )𝑥0 − ( 𝜔1−𝜔2 2 )𝑡)  Identity: cos 𝑢 + cos 𝑣 = 2 cos 𝑢+𝑣 2 cos( 𝑢−𝑣 2 )
7. Deriving the Beat Equation (Step 3) 𝜔 = 𝜔1 + 𝜔2 2  This is called the MEAN ANGULAR FREQUENCY.  Think About It: To calculate a mean, add up all the values and divide the total by the number of values. ∆𝜔 = 𝜔1 − 𝜔2 2  This is called the ANGULAR FREQUENCY DIFFERENCE.  Think About It: To calculate a difference, subtract one value from another. Why are these values important?  When two waves are oscillating at close frequencies, the resulting tone has the mean frequency and its amplitude varies with the angular frequency difference. Beating is the result of the different frequencies.
8. Deriving the Beat Equation (Step 4) The equation is simplified: 𝑆𝑡𝑜𝑡𝑎𝑙 0, 𝑡 = 2𝑠 𝑚 cos − 𝜔𝑡 cos(−∆𝜔𝑡) 𝑆𝑡𝑜𝑡𝑎𝑙 0, 𝑡 = 2𝑠 𝑚 cos 𝜔𝑡 cos(∆𝜔𝑡) Amplitude (m) Mean Angular Frequency (rad/s) Angular Frequency Difference (rad/s) The “x” position is 0 because beating only depends on amplitude and time.
9. Calculating Beat Frequencies Angular Frequency Difference: ∆𝜔 = 𝜔1−𝜔2 2 We know that 𝑓 = 𝜔 2𝜋 Therefore, 𝑓 ∝ 𝜔, so: 𝑓 = 𝑓1 − 𝑓2 2  There are absolute value bars because beating can be heard whether a second sound has a higher or lower frequency. 𝑓𝑏𝑒𝑎𝑡 = 2𝑓 = 𝑓1 − 𝑓2  The beat frequency is two times the previous equation because a single period contains two maxima in intensity.
10. Band Practice Your orchestra is getting ready to play in the annual spring concert. Before going out, you need to tune all of your instruments. The last section you need to tune are the violins. As a physicist, you decide that this is a perfect time to investigate frequencies and beating in instruments. a) The first two violinists players play at tones of 210Hz and 207Hz. What beat frequency do you hear? b) After adjusting their tones, the players tune again. Their frequencies are now 210Hz and 208Hz. How many beats do you hear over a period of 8 seconds? c) Another violinist plays at a frequency of 208Hz. When a second violinist plays, one hears 3 beats per second. The violinist increases the tension of the string, and the two violins now play at a beat frequency of 1.0Hz. What was the original frequency of the un-tuned string? By what percentage was the tension of the string increased?
11. Band Practice: Solution A  The first two violinists players play at tones of 210Hz and 207Hz. What beat frequency do you hear? 1. Use the equation given in “Calculating Beat Frequencies”: 𝑓𝑏𝑒𝑎𝑡 = 𝑓1 − 𝑓2 2. Insert values: 𝑓𝑏𝑒𝑎𝑡 = 210𝐻𝑧 − 207𝐻𝑧 = 3𝐻𝑧 ∴ 𝑌𝑜𝑢 ℎ𝑒𝑎𝑟 𝑎 𝑏𝑒𝑎𝑡 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 3𝐻𝑧.
12. Band Practice: Solution B  After adjusting their tones, the players tune again. Their frequencies are now 210Hz and 208Hz. How many beats do you hear over a period of 8 seconds? 1. Use the equation given in “Calculating Beat Frequencies”: 𝑓𝑏𝑒𝑎𝑡 = 𝑓1 − 𝑓2 2. Insert values to find beat frequency: 𝑓𝑏𝑒𝑎𝑡 = 210𝐻𝑧 − 208𝐻𝑧 = 2𝐻𝑧 3. Use the frequency equation: 𝑓𝑏𝑒𝑎𝑡 = 1 𝑇 4. Replace “1” with “x” to determine the number of beats to be heard: 𝑓𝑏𝑒𝑎𝑡 = 𝑥 𝑇 5. Rearrange for “x”: x = 𝑓𝑏𝑒𝑎𝑡 𝑇 6. Insert values to find the number of beats: x = 2𝑠−1 8𝑠 = 16 𝑏𝑒𝑎𝑡𝑠 ∴ 𝑌𝑜𝑢 ℎ𝑒𝑎𝑟 16 𝑏𝑒𝑎𝑡𝑠 𝑜𝑣𝑒𝑟 𝑎 𝑝𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 8 𝑠𝑒𝑐𝑜𝑛𝑑𝑠.
13. Band Practice: Solution C  Another violinist plays at a frequency of 208Hz. When a second violinist plays, one hears 3 beats per second. The violinist increases the tension of the string, and the two violins now play at a beat frequency of 1Hz. What was the original frequency of the un-tuned string? By what percentage was the tension of the string increased? 1. From the equation given in “Calculating Beat Frequencies”: 𝑓𝑏𝑒𝑎𝑡 = 𝑓1 − 𝑓2 3Hz = 208𝐻𝑧 − 𝑓2 𝑜𝑟 3𝐻𝑧 = 𝑓1 − 208𝐻𝑧 ∴ 𝑇ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑤𝑎𝑠 𝑒𝑖𝑡ℎ𝑒𝑟 205𝐻𝑧 𝑜𝑟 211𝐻𝑧. 2. Because the tension is increasing, frequency is increasing, so the original frequency must have been lower (205Hz). This same relationship indicates that the new frequency must be 207Hz. (The tension is rising the frequency to 208Hz, there is a beat frequency of 1Hz). 𝑣 = 𝑇𝑠 𝜇 = 𝜆𝑓 𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑎𝑛𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑤𝑖𝑡ℎ 𝑒𝑎𝑐ℎ 𝑜𝑡ℎ𝑒𝑟. 3. Because tension and frequency increase with one another, their changes can be represented by a ratio: 𝑣2 = 𝑇𝑠 𝜇 = (𝜆𝑓) 2 (Derived from equation in Step 2) 𝑇𝑠2 𝑇𝑠1 = 𝑓2 2 𝑓1 2 = (207𝐻𝑧)2 (205𝐻𝑧)2 × 100% =101.96% ∴ 𝑇ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑤𝑎𝑠 205𝐻𝑧 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑖𝑛𝑔 𝑤𝑎𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑 𝑏𝑦 1.96%.
14. References  Hawkes, R., Iqbal, J., Mansour, F., Milner-Bolotin, M., & Williams, P. (2015). Interference and Sound. In Physics for Scientists and Engineers An Interactive Approach (p. 246-251). Toronto: Nelson Education.  Beats Physics [Motion picture]. (2014). UCLA Department of Physics and Astronomy. https://www.youtube.com/watch?v=IQ1q8XvOW6g  Beats Physics Questions. (n.d.). Retrieved March 14, 2015. https://answers.yahoo.com/question/index?qid=20100131003842AAMIziM  Interference and Beats. (n.d.). Retrieved March 14, 2015. http://www.physicsclassroom.com/class/sound/Lesson-3/Interference-and-Beats  Table of Trigonometric Identities. (n.d.). Retrieved March 14, 2015. http://www.sosmath.com/trig/Trig5/trig5/trig5.html  16.5 The Superposition of Waves. (n.d.). Retrieved March 14, 2015. https://curricula2.mit.edu/pivot/book/ph1605.html?acode=0x0200