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A tuning fork of frequency 480 Hz is sounded on a resonance tube. The first and second resonance are observed at 16 cm and 50 cm respectively. The velocity of sound in air is approximately
- a)326 m/s
- b)300 m/s
- c)340 m/s
- d)350 m/s
Correct answer is option 'A'. Can you explain this answer?
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A tuning fork of frequency 480 Hz is sounded on a resonance tube. The ...
Use concept of resonance with correction.
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A tuning fork of frequency 480 Hz is sounded on a resonance tube. The ...
Given:
Frequency of the tuning fork = 480 Hz
Length of the resonance tube at first resonance = 16 cm = 0.16 m
Length of the resonance tube at second resonance = 50 cm = 0.5 m
To find: Velocity of sound in air.
Let's analyze the resonance in a closed-end resonance tube.
Resonance in a closed-end resonance tube:
In a closed-end resonance tube, the closed end acts as a node (point of minimum displacement) and the open end acts as an antinode (point of maximum displacement) for a standing wave to form.
First Resonance:
At the first resonance, the length of the resonance tube is equal to λ/4, where λ is the wavelength of the sound wave produced by the tuning fork.
λ/4 = 0.16 m
λ = 0.16 m × 4
λ = 0.64 m
Second Resonance:
At the second resonance, the length of the resonance tube is equal to 3λ/4.
3λ/4 = 0.5 m
λ = (0.5 m × 4)/3
λ = 0.67 m
Calculating the velocity of sound in air:
The velocity of sound in air can be calculated using the formula:
v = f × λ
where v is the velocity of sound, f is the frequency of the tuning fork, and λ is the wavelength of the sound wave.
For the first resonance:
v₁ = 480 Hz × 0.64 m = 307.2 m/s
For the second resonance:
v₂ = 480 Hz × 0.67 m = 321.6 m/s
The velocity of sound in air is the average of the velocities calculated at the first and second resonance.
v = (v₁ + v₂)/2
v = (307.2 m/s + 321.6 m/s)/2
v = 628.8 m/s/2
v = 314.4 m/s
The approximate velocity of sound in air is 314.4 m/s, which is closest to option A) 326 m/s.
Frequency of the tuning fork = 480 Hz
Length of the resonance tube at first resonance = 16 cm = 0.16 m
Length of the resonance tube at second resonance = 50 cm = 0.5 m
To find: Velocity of sound in air.
Let's analyze the resonance in a closed-end resonance tube.
Resonance in a closed-end resonance tube:
In a closed-end resonance tube, the closed end acts as a node (point of minimum displacement) and the open end acts as an antinode (point of maximum displacement) for a standing wave to form.
First Resonance:
At the first resonance, the length of the resonance tube is equal to λ/4, where λ is the wavelength of the sound wave produced by the tuning fork.
λ/4 = 0.16 m
λ = 0.16 m × 4
λ = 0.64 m
Second Resonance:
At the second resonance, the length of the resonance tube is equal to 3λ/4.
3λ/4 = 0.5 m
λ = (0.5 m × 4)/3
λ = 0.67 m
Calculating the velocity of sound in air:
The velocity of sound in air can be calculated using the formula:
v = f × λ
where v is the velocity of sound, f is the frequency of the tuning fork, and λ is the wavelength of the sound wave.
For the first resonance:
v₁ = 480 Hz × 0.64 m = 307.2 m/s
For the second resonance:
v₂ = 480 Hz × 0.67 m = 321.6 m/s
The velocity of sound in air is the average of the velocities calculated at the first and second resonance.
v = (v₁ + v₂)/2
v = (307.2 m/s + 321.6 m/s)/2
v = 628.8 m/s/2
v = 314.4 m/s
The approximate velocity of sound in air is 314.4 m/s, which is closest to option A) 326 m/s.
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A tuning fork of frequency 480 Hz is sounded on a resonance tube. The first and second resonance are observed at 16 cm and 50 cm respectively. The velocity of sound in air is approximatelya)326 m/sb)300 m/sc)340 m/sd)350 m/sCorrect answer is option 'A'. Can you explain this answer?
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