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Shane Macguire Grade: upto college level
`        A piston is fitted in a cylindrical tube of small cross section with the other end of the tube open. The tube resonates with a tuning fork of frequency 512 Hz. The piston is gradually pulled out of the tube and it is found that a second resonance occurs when the piston is pulled out through a distance of 32.0 cm. Calculate the speed of sound in the air of the tube. `
3 years ago

Deepak Patra
474 Points
```										Sol. Let n base 0 = frequency of the turning fork, T = tension of the string
L = 40 cm = 0.4 m, m = 4g = 4 × 10^–3 kg
So, m = Mass/Unit length = 10^–2 kg/m
n base 0 = 1/2l√T/m.
So, 2nd harmonic 2n base 0 = (2/ 2l) √T /m
As it is unison with fundamental frequency of vibration in the air column

⇒ 2n base 0 = 340/4 * 1 = 85 Hz
⇒ 85 = 2/2 * 0.4  √T/14 ⇒ T = 85^2 * (0.4)^2 * 10^-2 =- 11.6 Newton.

```
3 years ago
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