An open pipe 0.40m in length is placed vertically in a cylindrical bucket having a bottom area of 0.10 m^2. Water is poured into the bucket until a sounding tuning fork of frequency 440 Hz, placed over the pipe, produces resonance. Find the mass of water in the bucket at this moment.
I've been totally stumped in this, all I got was a very brief overview of standing waves in open and closed air columns. Totally lost.
To solve this problem, we need to understand the concept of resonance in open pipes and how it relates to the length of the pipe and the speed of sound in the medium.
1. Resonance in open pipes: When a tuning fork is held over an open pipe, the sound waves produced by the fork can create standing waves within the pipe. In such a case, the length of the pipe plays a crucial role in determining whether resonance occurs or not.
2. Standing waves in open pipes: In an open pipe, the length of the pipe determines the wavelength of the standing wave that can be established. The fundamental frequency (first harmonic) occurs when the pipe is exactly one-quarter of a wavelength in length, resulting in a wavelength four times the length of the pipe.
3. Relationship between speed of sound and frequency: The speed of sound (v) in a medium is related to the frequency (f) and wavelength (λ) of a sound wave through the equation v = fλ. In our case, the frequency (440 Hz) and the wavelength (4 * length of the pipe) are important.
Now let's proceed with the solution:
Step 1: Calculate the wavelength of the sound wave using the length of the pipe.
Since the length of the open pipe is 0.40m, the wavelength of the sound wave produced at the fundamental frequency is given by λ = 4 * 0.40m = 1.60m.
Step 2: Calculate the speed of sound in air.
The speed of sound in air is approximately 343 m/s at room temperature. Let's assume it as v = 343 m/s.
Step 3: Calculate the frequency using the speed of sound and the wavelength.
v = fλ
443 = f * 1.60m
f = 443 / 1.60 Hz ≈ 276 Hz
Step 4: Compare the frequency obtained with the given frequency (440 Hz).
Since the calculated frequency (276 Hz) is less than the given frequency (440 Hz), the pipe is not resonating at the fundamental frequency. However, it is resonating at a higher harmonic.
Step 5: Determine the harmonic number.
To find the harmonic number when the pipe resonates, we need the formula n = fn / f1, where n is the harmonic number, fn is the frequency of the nth harmonic, and f1 is the first harmonic frequency (fundamental frequency).
In our case, the harmonic number is n = 440 Hz / 276 Hz ≈ 1.59.
Since the harmonic number cannot be a fraction, we can conclude that the pipe is resonating at the second harmonic (n = 2).
Step 6: Calculate the mass of water in the bucket.
The mass of water in the bucket can be determined using the volume of water poured, the density of water, and the formula mass = density * volume.
To find the volume of water, we need to determine the change in volume of the bucket when the water is added. The water will fill the bucket up to the point where the pipe is submerged, forming a closed end. Hence, the change in volume is equivalent to the volume of the pipe.
The volume of the pipe is given by A * h, where A is the cross-sectional area and h is the length of the pipe.
Given that the bottom area of the bucket is 0.10 m², which is equal to the cross-sectional area of the pipe (A = 0.10 m²), and the length of the pipe (h = 0.40 m), the volume of the water is 0.10 m² * 0.40 m = 0.04 m³.
The density of water is approximately 1000 kg/m³.
Finally, we can calculate the mass of water:
mass = density * volume = 1000 kg/m³ * 0.04 m³ = 40 kg.
Therefore, the mass of water in the bucket at the moment of resonance is 40 kg.
The closed pipe is 1/4 lambda long (closed at the bottom due to water).
so the height of the water is .4-1/4 lambda
volume water= area*heighwater
mass water=denstiy water*volume water