A wire of length 30.0 cm and mass 10.0 g is secured at both ends and vibrates in its fundamental mode. A closed organ pipe measuring 50.0 cm in length is positioned with its open end close to the wire, and it is driven into resonance in its fundamental mode by the vibrating wire. Determine the tension in the wire. The speed of sound in air is 340 m s ⁻¹ .

Step-by-step solution

The fundamental frequency of the closed organ pipe is given by: f = v 4L_c Substituting v = 340 m s ⁻¹ and L_c = 0.5 m : f = 340 4 0.5 = 170 Hz Since the wire is in resonance with the pipe, its fundamental frequency is also 170 Hz . The fundamental frequency of a stretched wire is given by: f = 1 2L_w T Here, the length of the wire is L_w = 0.3 m and its linear mass density is = m L_w = 10 10⁻³ 0.3 = 1 30 kg m ⁻¹ . Substituting the known values into the frequency formula: 170 = 1 2 0.3 T 1/30 170 0.6 = 30T 102 = 30