An unknown tuning fork produces 5 beats per second with a second tuning fork that causes a closed organ pipe of length 40 cm to vibrate in its fundamental mode. When the first tuning fork is loaded slightly with wax, the beat frequency decreases. Determine the original frequency of the first tuning fork. Assume the speed of sound in air is 320 m/s .

Step-by-step solution

The fundamental frequency of a closed organ pipe is given by n₂ = v 4L . Substituting the given values, n₂ = 320 4 0.4 = 200 Hz . Since the unknown tuning fork produces 5 beats per second with the second tuning fork, its frequency is n₁ = n₂ 5 = 200 5 . Thus, n₁ is either 205 Hz or 195 Hz . When a tuning fork is loaded with wax, its frequency decreases. If n₁ = 205 Hz , a decrease in frequency will bring it closer to 200 Hz , resulting in a beat frequency less than 5 Hz . This matches the given condition. If n₁ = 1