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Answer :
a) The first resonant frequency in the human ear canal is approximately 3400 Hz.
b) The wavelength at the second resonance is approximately 0.03 meters (or 3 centimeters).
Given:
Speed of sound (v) = 340 m/s
Ear canal length (L) = 2.5 cm = 0.025 m (converted to meters)
a. First Resonant Frequency
For a closed-one-end organ pipe (like the ear canal), the first resonance occurs at the fundamental frequency. We can calculate it using the following formula:
f1 = v / (4 * L)
where:
f1 is the first resonant frequency (in Hz)
v is the speed of sound (in m/s)
L is the length of the ear canal (in meters)
Calculation:
f1 = 340 m/s / (4 * 0.025 m)
f1 ≈ 3400 Hz
Therefore, the first resonant frequency is approximately 3400 Hz.
b. Wavelength at Second Resonance
The second resonance occurs at three times the fundamental frequency (f2 = 3 * f1). To find the wavelength (λ) at the second resonance, we can use the following formula:
λ = v / f
where:
λ is the wavelength (in meters)
v is the speed of sound (in m/s)
f is the frequency (in Hz)
Using the second resonant frequency (f2):
λ = 340 m/s / (3 * 3400 Hz)
λ ≈ 0.03 m (or 3 cm)
Therefore, the wavelength at the second resonance is approximately 0.03 meters (or 3 centimeters).
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Rewritten by : Brahmana