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.
Answers
Since with the wax loading of the first tuning-fork decreases the beat frequency, the difference of two frequencies decreases. It means when the frequency of the first tuning fork is reduced the difference is reduced. It can happen if the frequency of the first tuning fork is greater than the second. Let the frequency of the second tuning fork = ν. This frequency causes a closed organ pipe of 40 cm ( = 0.40 m) to vibrate in its fundamental mode, hence ν = V/4L =320/(4*0.40) =80/0.40 Hz = 200 Hz. Beat frequency = 5 per second, hence the frequency of the first tuning-fork = 200 + 5 Hz = 205 Hz.Read more on Sarthaks.com - https://www.sarthaks.com/325873/tuning-unknown-frequency-makes-beats-second-another-tuning-which-cause-closed-organ-length