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Learning object 6
- 1. LO6 – Beats LyndonWon 1. 2 tones are played at the same time, the frequency you hear is 753 Hz, with the beat frequency being 6 Hz. Find the frequency of the 2 original tones. 2. Explain why there is fluctuation in the intensity of a sound when 2 tones with similar frequencies are played simultaneously.
- 2. LO6 – Beats Answer Key 1. 2 tones are played at the same time, the frequency you hear is 753 Hz, with the beat frequency being 6 Hz. Find the frequency of the 2 original tones. The frequency at which you hear result of 2 tones being played together is the mean angular frequency, ω = (ω1 + ω2) / 2, where ω1 and ω2 are the respective frequencies of the 2 tones. The beat frequency is equal to the different between the 2 frequencies. To find the original 2 frequencies we can use a little algebra. Since the difference in the frequencies is 6 Hz, we can let ω1 = ω2 + 6 Hz We let ω = 753 Hz 753 Hz = ((ω2 + 6 Hz) + ω2) /2 2(753 Hz) – 6 Hz = 2ω2 1500 = 2 ω2 ω2 = 750 Hz ω1 = ω2 + 6 Hz = 756 Hz 2. Explain why there is fluctuation in the intensity of a sound when 2 tones with similar frequencies are played simultaneously. The difference in frequencies causes interference, both constructive and destructive, causing the amplitude of the resulting wave to increase and decrease, which causes the change in intensity.
- 3. The graph above shows the how the resulting wave acts dues to the interference between the original waves. At the point at which the 2 original waves are in phase and at the max point of their crests the superposition of the resulting graph is at its greatest amplitude. At the point at which the waves are out of phase, where one is at the maximum positive amplitude, and the other is at its most negative amplitude, the resulting superposition is at the equilibrium position.