A vertical tube with a tap at the base is filled with water, and a tuning fork vibrates over its mouth. As the water level is lowered in the tube, resonance is heard when the water level has dropped 18 cm, and again after 54 cm of distance exists from the water to the top of the tube. What is the frequency of the tuning fork?
To find the frequency of the tuning fork, we can use the formula for the speed of sound in air: v = fλ.
First, let's find the wavelength (λ) of the sound wave produced by the tuning fork. The resonance occurs when there is a standing wave inside the tube, which means the length of the tube corresponds to half a wavelength.
The first resonance occurs when the water level has dropped 18 cm. Since the tube is vertical, this distance corresponds to half a wavelength. Therefore, the wavelength (λ₁) is twice the distance:
λ₁ = 2 * 18 cm = 36 cm = 0.36 m
The second resonance occurs when the water level has dropped 54 cm. Again, this distance corresponds to half a wavelength. Therefore, the wavelength (λ₂) is twice the distance:
λ₂ = 2 * 54 cm = 108 cm = 1.08 m
Now, we can use these values to find the speed (v) of sound in air. In air, the speed of sound is approximately 343 m/s.
For the first resonance:
v = f₁ * λ₁
343 m/s = f₁ * 0.36 m
Solving for f₁:
f₁ = 343 m/s / 0.36 m
f₁ ≈ 952 Hz
For the second resonance:
v = f₂ * λ₂
343 m/s = f₂ * 1.08 m
Solving for f₂:
f₂ = 343 m/s / 1.08 m
f₂ ≈ 317 Hz
Since the tuning fork produces the sound that produces these resonances, both frequencies must match. Therefore, the frequency of the tuning fork is approximately 952 Hz.