Answer: f = 927.55Hz
Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below
L = λ/4 where λ = wavelength.
speed of sound in air = v = 343m/s.
fundamental frequency of open closed tube = 315Hz
λ = 4L.
v = fλ
343 = 315 * 4L
343 = 1260 * L
L = 343/ 1260
L = 0.27m
In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.
The length of tube and wavelength are related by the formulae below
L = λ/4, λ=4L
λ = 4 * 0.27
λ = 1.087m.
v = fλ
1010 = f * 1.087
f = 1010/1.807
f = 927.55Hz
The first choice on the list is the correct one.
Answer:
1 x 10 -10 whisper at 1m distance.
Explanation:
- Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency
