Two metallic rods, A and B , are both clamped at their midpoints and set into longitudinal resonance at their fundamental frequencies. Rod B is made of a material whose Young's modulus is 1.69 times that of rod A and whose density is 1.44 times that of rod A . If the length of rod B is 1.3 times the length of rod A , and the fundamental frequency of rod A is f₀ , then the fundamental frequency of rod B is:

Step-by-step solution

The speed of longitudinal waves in a rod is given by v = Y . For rod B , the wave speed is: v_B = 1.69 Y_A 1.44 _A = 1.3 1.2 Y_A _A = 1.3 1.2 v_A For a rod clamped at its midpoint, the fundamental mode of longitudinal vibration has a node at the center and antinodes at the ends. The length of the rod corresponds to half a wavelength ( L = 2 ), so the fundamental frequency is f = v 2L . The fundamental frequency of rod A is f₀ = v_A 2L_A . The fundamental frequency of rod B is: f_B = v_B 2L_B = 1.3 1.2 v_A 2(1.3 L_A