Answer :
The glass tube will produce 5 resonance points for the tuning fork frequency of 606 Hz, with specific resonance lengths determined by the tube's length and the properties of the sound wave.
Resonance in a Water-Filled Glass Tube
To solve this problem, we need to determine the number of resonance points for a tuning fork of frequency 606 Hz in a tube of length 1.5m, considering that the sound's speed in air is 340 m/s.
Resonance occurs at specific lengths of the air column in the tube. For a tube open at one end and closed at the other, resonance happens at odd multiples of one-quarter the wavelength ( /4).
First, we determine the wavelength of the sound using the formula:
λ = v / f
Where:
- v is the speed of sound (340 m/s)
- f is the frequency (606 Hz)
λ = 340 m/s / 606 Hz ≈ 0.561 m
Next, resonance will occur at lengths equivalent to (2n + 1) /4, for n = 0, 1, 2, ... until the length exceeds 1.5m:
- L = λ/4 ≈ 0.14 m
- L = 3λ/4 ≈ 0.42 m
- L = 5λ/4 ≈ 0.70 m
- L = 7λ/4 ≈ 0.98 m
- L = 9λ/4 ≈ 1.26 m
Since the tube is 1.5m long, these values represent all possible resonance lengths. Hence, there will be a total of 5 resonance points
