A tuning fork of unknown frequency makes 5 beats per second with another tuning fork which can cause a closed organ pipe of length 40 cm to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded with wax. Find its original frequency. The speed of sound in air is 320 m s-1.

Question

Navjyot Kalra · Answer

Sol. Given that, l = 25 cm = 25 × 10^–2 m By shortening the wire the frequency increases, [f = (1/ 2l) √(TB /M) ] As the vibrating wire produces 4 beats with 256 Hz, its frequency must be 252 Hz or 260 Hz. Its frequency must be 252 Hz, because beat frequency decreases by shortening the wire. So, 252 = ½ * 25 * 10^-2 √T/M …(1) Let length of the wire will be l, after it is slightly shortened, ⇒ 256 = ½ * l base 1 √T/M Dividing (1) by (2) we get 252/256 = l base 1/2 * 25 * 10^-2 ⇒ l base 1 252 * 2 * 25 * 10^-2/260 = 0.2431 m So, it should be shorten by (25 – 24.61) = 0.39 cm.

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