In a resonance-tube containing air, the source of sound is a small loudspeaker driven by a variable-frequency oscillator. The distance between successive antinodes is 10 cm for a frequency of 1625 Hz. When the frequency is slowly reduced, the standing wave pattern at first collapses and then reapears at a frequency of 1600 Hz. (a) Find the speed of the sound in air. (b) Find the distance between successive antinodes when the frequency is 1600 Hz. (c) Estimate the vibrating length of the tube, if the loudspeaker is close to a displacement node. (d) If the frequency is further reduced, at what frequency will the next pattern be obtained?
To solve this problem, we can use the formula for the speed of sound in air: v = f * λ, where v is the speed of sound, f is the frequency, and λ is the wavelength.
(a) To find the speed of sound in air, we need to find the wavelength at each frequency. The distance between successive antinodes is equal to half the wavelength, so the wavelength at a frequency of 1625 Hz is 2 * 10 cm = 20 cm.
Using the formula v = f * λ, we can substitute the values and solve for v:
v = 1625 Hz * 0.20 m = 325 m/s
Therefore, the speed of sound in air is 325 m/s.
(b) We need to find the distance between successive antinodes when the frequency is 1600 Hz. Using the same formula, we can substitute the new value and solve for λ:
λ = v / f = 325 m/s / 1600 Hz ≈ 0.203125 m ≈ 20.3125 cm
Therefore, the distance between successive antinodes when the frequency is 1600 Hz is approximately 20.3125 cm.
(c) When the standing wave pattern collapses and reappears at a frequency of 1600 Hz, it means that the tube's length is equal to half the wavelength at that frequency. Since the frequency is reduced and the pattern reappears, the vibrating length must be half that of the previous length.
Using the formula λ = 2L, where L is the vibrating length, we can substitute the known values to find L:
0.203125 m = 2L
L ≈ 0.1015625 m ≈ 10.15625 cm
Therefore, the estimated vibrating length of the tube when the loudspeaker is close to a displacement node is approximately 10.15625 cm.
(d) To determine the frequency at which the next pattern will be obtained, we need to find the wavelength and calculate the corresponding frequency. Since the pattern reappears when the vibrating length is decreased, it means the next pattern will appear when the vibrating length is further reduced to one-fourth the original length.
Using the same formula as before, we can find the wavelength at the new length:
(1/4)L = (1/4)(0.1015625 m) ≈ 0.0253906 m
Now, we can calculate the frequency:
v = f * λ
f = v / λ = 325 m/s / 0.0253906 m ≈ 12,770.62 Hz
Therefore, the next pattern will be obtained at a frequency of approximately 12,770.62 Hz.