Hello first post here. These problems aren't for a specific course just working through an old textbook specifically Vibrations and Waves by A.P. French. I hope it's alright that I include two problems in one thread. 1. The problem statement, all variables and given/known data Q 3-7) A wire of unstretched length L0 is extended by 10^-3L0 when a mass m is hung from it's end. If this same wire is connected between two points A,B horizontally and then the same mass is hung from the middle of the wire, how large is the depression y of the midpoint? The answer given is L0/20. 2. Relevant equations x^2 + y^2 = h^2 3. The attempt at a solution I assume the total length of the wire will still be (1 + 0.001)L0. So dividing the length in two and the unstretched length in two you get a right angle triangle. (L0/2)^2 + y^2 = (L0(1+0.001)/2)^2 y = (L0(1+0.001)/2)^2 - (L0/2)^2))^0.5 = ((0.5005^2 - 0.25)L0^2)^0.5 = 0.022 L0 1. The problem statement, all variables and given/known data Q 3-8b) A 0.5kg mass is hung from a 2m long steel wire with diameter 0.5mm. What is the maximum height h the mass can be dropped from if the wire is not to break? Young's modulus of steel: 2x10^11 N/m^2 Ultimate strength of steel: 1.1x10^9 N/m^2 The given answer is 0.23m 2. Relevant equations Y = (F/A) / (ΔL/L) US = F/A F = ma v0^2 = v - 2aΔx KE = 0.5mv^2 GPE = mgh 3. The attempt at a solution F = US x A = 1.1x10^9 x ∏(0.0005/2)^2 = 216 N The thing that's throwing me off is that if I have a falling mass I have a energy and not a force. So I'm thinking maybe look at the acceleration over the distance the wire stretches. a = 216/0.5 = 432 m/s^2 ΔL = FL/YA = (216 x 2) / ((∏(0.0005/2)^2)(2x10^11)) = 0.011m v0^2 = 0 - 2(9.8)(0.011) = 9.5 m/s 0.5mv^2 = mgh 0.5(9.5) = 9.8h h = 0.48m Any help with this problems is much appreciated, thanks.