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.

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Solution

Let the piston resonates at length $I_{1}$ , and $I_{2} .$

Here I =32 cm, v =? n=512 Hz

Now, $v = 512 \times 0.64 = 328 m / s$

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Let the piston resonates at length I1, and I2. Here I =32 cm, v =? n=512 Hz Now, v=512×0.64=328m/s

Let the piston resonates at length $I_{1}$ , and $I_{2} .$

Here I =32 cm, v =? n=512 Hz

Now, $v = 512 \times 0.64 = 328 m / s$