Answer :
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]
### 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]