In the arrangement depicted in the figure, a wire with a mass of 4.00 g oscillates in its second harmonic. This sets the air column in the tube into vibrations in its fundamental mode. Provided that the speed of sound in air is 340 m s ⁻¹ , determine the tension in the wire.

Step-by-step solution

For the air column in the tube, it is closed at one end (water surface) and open at the other. It vibrates in its fundamental mode. The fundamental frequency of the closed organ pipe is given by: f_a = v_s 4L_a where v_s = 340 m s ⁻¹ is the speed of sound and L_a = 1.0 m is the length of the air column. f_a = 340 4 1.0 = 85 Hz The wire oscillates in its second harmonic and sets the air column into resonance, so the frequency of the wire is equal to the frequency of the air column. f_w = f_a = 85 Hz The frequency of