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An air column at [tex]\(25^{\circ} C\)[/tex] in a closed tube is set into vibration by a tuning fork with a frequency of [tex]\(512 \, \text{Hz}\)[/tex]. Calculate:

(i) The velocity of sound at [tex]\(25^{\circ} C\)[/tex].

(ii) The length of the closed tube at the first resonance, neglecting end correction. [Speed of sound in air at [tex]\(0^{\circ} C = 330 \, \text{m/s}\)[/tex]]

Sagot :

GinnyAnsweridn
Sure, let's solve this question step-by-step.

### Given Data

- Temperature [tex]\( T = 25^{\circ}C \)[/tex]
- Frequency [tex]\( f = 512 \)[/tex] Hz
- Speed of sound at [tex]\( 0^{\circ}C \)[/tex], [tex]\( V_0 = 330 \, \text{m/s} \)[/tex]

### (i) Velocity of Sound at [tex]\( 25^{\circ}C \)[/tex]

The speed of sound in air increases by approximately [tex]\( 0.6 \, \text{m/s} \)[/tex] for each degree Celsius increase in temperature. Therefore, the speed of sound at [tex]\( 25^{\circ}C \)[/tex] can be calculated as follows:

[tex]\[ V_{25} = V_0 + (0.6 \times T) \][/tex]

Substituting the given values:

[tex]\[ V_{25} = 330 \, \text{m/s} + (0.6 \times 25) \, \text{m/s} \][/tex]
[tex]\[ V_{25} = 330 \, \text{m/s} + 15 \, \text{m/s} \][/tex]
[tex]\[ V_{25} = 345 \, \text{m/s} \][/tex]

So, the velocity of sound at [tex]\( 25^{\circ}C \)[/tex] is [tex]\( 345 \, \text{m/s} \)[/tex].

### (ii) Length of the Closed Tube at First Resonance

The first resonance in a closed tube occurs when the length of the tube is equal to one-quarter of the wavelength of the sound wave.

First, we need to find the wavelength ([tex]\( \lambda \)[/tex]) of the sound wave using the speed of sound and the frequency. The relation between wavelength, speed of sound, and frequency is given by:

[tex]\[ \lambda = \frac{V_{25}}{f} \][/tex]

Substituting the known values:

[tex]\[ \lambda = \frac{345 \, \text{m/s}}{512 \, \text{Hz}} \][/tex]
[tex]\[ \lambda = 0.673828125 \, \text{m} \][/tex]

The length of the tube at the first resonance corresponds to a quarter of the wavelength:

[tex]\[ \text{Length} = \frac{\lambda}{4} \][/tex]

Substituting the wavelength value:

[tex]\[ \text{Length} = \frac{0.673828125 \, \text{m}}{4} \][/tex]
[tex]\[ \text{Length} = 0.16845703125 \, \text{m} \][/tex]

So, the length of the closed tube at the first resonance is [tex]\( 0.16845703125 \, \text{m} \)[/tex].

### Summary

1. Velocity of sound at [tex]\( 25^{\circ}C \)[/tex]: [tex]\( 345 \, \text{m/s} \)[/tex]
2. Length of the closed tube at the first resonance neglecting end correction: [tex]\( 0.16845703125 \, \text{m} \)[/tex]
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