In a resonance tube experiment, a tube of fixed length is closed at one end and several tunning forks of increasing frequency used to obtain resonance at the open end. If the tunning fork with the lowest frequency which gave resonance had a frequency f1 and the next tunning fork to give resonance had a frequency f2, find the ratio f2:f1

Hey Chioma, good to see you again. Step 1: Determine the formula for resonant frequencies in a closed-end tube. For a resonance tube closed at one end and open at the other, resonance occurs when the length of the tube L is an odd multiple of a quarter wavelength. The resonant frequencies f_n are given by: f_n = (nv)/(4L) where n = 1, 3, 5, (odd integers) and v is the speed of sound in air. Step 2: Find the lowest resonant frequency (f_1). The lowest frequency corresponds to n=1. f_1 = (1 · v)/(4L) = (v)/(4L) Step 3: Find the next resonant frequency (f_2). The next resonant frequency corresponds to the next odd integer, which is n=3. f_2 = (3 · v)/(4L) Step 4: Calculate the ratio f_2:f_1. To find the ratio (f_2)/(f_1), we divide the expression for f_2 by the expression for f_1: (f_2)/(f_1) = (3v)/(4L)(v)/(4L) (f_2)/(f_1) = (3v)/(4L) × (4L)/(v) (f_2)/(f_1) = 3 So, the ratio f_2:f_1 is 3:1. The ratio f_2:f_1 is 3:1. Send me the next one 📸