The speed of sound in air at 20 C is 344 m/s. (a) What is the wavelength of a sound wave with a frequency of 784 Hz, corresponding to the note G5 on a piano, and how many milliseconds does each vibration take? (b) What is the wavelength of a sound wave one octave higher than the note in part (a)?

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The speed of sound in air at 20 C is 344 m/s. (a) What is the wavelength of a sound wave with a frequency of 784 Hz, corresponding to the note G5 on a piano, and how many milliseconds does each vibration take?  (b) What is the wavelength of a sound wave one octave higher than the note in part (a)?

giskacu · Accepted Answer

Step 1 Knowns We know that the speed of a periodic wave  υ  with wavelength  λ  and frequency  f  is given by: υ=f⋅λ We also know that the relation between the frequency and the time peiod is given by: T=1f Step 2 Given The frequency of the sound wave (corresponding to G5 on a piano) is f1=784Hz, , and the speed of sound in air at 20∘C is v=344m/s Step 3 (a) Sound is a periodic wave, so we can use equation (1); We plug our values for υ and f1 into equation (1), so we get: 344ms=(784s−1)λ1 ∴λ1=0.439m We know that the time each vibration takes is the definetion of a time period T, so wesubstitute for f1 into equation(2),so we get: T1=1784s−1=1.28×10−3s ∴T1=1.28 Step 4 (b) We know that one octave higher means double the frequency, so the frequency of a sound wave one octave higher than the one in part (a) is: f2=2×f1=2×784=1568Hz Now, we substitute for f2 and υ into equation (1), so we get: 344ms=(1568s−1)λ2 ∴λ2=0.219m Result (a)λ1=0.439m T1=1.28ms (b)λ2=0.219m

The speed of sound in air at 20 C is 344 m/s. (a) What is the wavelength of a so

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