A 50 cm long hollow glass cylinder is open at both ends and is suspended in air. A source of sounds that produces a pure frequency is placed close to one end of the tube. The frequency is placed close to one end of the tube. The frequency of the sound source is slowly increased from a very low value. The tube is first observed to resonate when the source frequency is 320Hz.

(I) The speed of the sound in the tube must be:

(II) The frequency of the source is slowly increased until resonance occurs again FOR THE THIRD TIME. The frequency at which this happens is:

(III)The frequency when resonance occurs for the FIFTH time is:

1. d = V*t,

V = d/t = 0.5m / (1/320) = 160m/s.

2. F = 3 * 320Hz = 960Hz.

3. F = 5 * 320Hz = 1600Hz.

Find the momentum of a bullet of mass 1.34x10-3slug traveling at 735 ft/s

To find the answers to these questions, we need to use the concepts of resonance and the fundamental frequency of a closed tube.

(I) The speed of sound in the tube can be determined using the formula:

v = f * λ

where v is the speed of sound, f is the frequency, and λ is the wavelength.

In a closed tube, the fundamental frequency (the first resonance) occurs when the length of the tube is equal to one-fourth of the wavelength. So, in this case, the length of the tube (50 cm) is equal to one-fourth of the wavelength, which means the wavelength is 4 times the length of the tube (λ = 4 * 50 cm = 200 cm).

Now, we can find the speed of sound in the tube by substituting the values into the formula:

v = f * λ
v = 320 Hz * 200 cm
v = 64,000 cm/s

Therefore, the speed of sound in the tube is 64,000 cm/s.

(II) To find the frequency at which the resonance occurs for the third time, we need to understand the concept of the fundamental frequency and harmonics.

The fundamental frequency (f1) is the lowest frequency at which the tube can resonate and is given by:

f1 = v / λ

Using the speed of sound (64,000 cm/s) and the length of the tube (50 cm), we can calculate the wavelength:

λ = v / f1
λ = 64,000 cm/s / 320 Hz
λ = 200 cm

Now, to find the frequency at which the resonance occurs for the third time, we need to consider the relationship between the fundamental frequency and harmonics. The frequency at which the resonance occurs for the third time (f3) is three times the fundamental frequency:

f3 = 3 * f1
f3 = 3 * 320 Hz
f3 = 960 Hz

Therefore, the frequency at which the resonance occurs for the third time is 960 Hz.

(III) Similar to the previous calculations, the frequency at which the resonance occurs for the fifth time (f5) is five times the fundamental frequency:

f5 = 5 * f1
f5 = 5 * 320 Hz
f5 = 1600 Hz

Therefore, the frequency at which the resonance occurs for the fifth time is 1600 Hz.