#digipen/phy250 Homework 3 #### 2 A fisherman's scale stretches 3.9 cm when a 2.9-kg fish hangs from it. $ \begin{align} -3.9 k + 2.9g &= 0 \\ \frac{2.9g}{3.9} &= k\\ \frac{2.9 \cdot 9.81}{3.9} &= k \\ k &= 7.3 \,{\op{N}\op{cm}^{-1}} \\ &= 730 \,{\op{N}\op{m}^{-1}} \\ \end{align}$ ![[../00 Asset Bank/Pasted image 20250922132510.png]] What will be the frequency of oscillation if the fish is pulled down 3.4 cm more and released so that it oscillates up and down? $\begin{align} y(0) &= 3.9 + 3.4=7.3\\ y'' & = 2.9g - 7.3y \\ y &= A\cos( \omega t+\phi) + B \\ y&= 3.4 \cos(\omega t) + 3.9 \\ y'&= 3.4 \omega (-\sin \omega t) \\ y'' &= -3.4 \omega^2 \cos(\omega t)\\ -3.4 \omega^2 \cos(\omega t) &= 2.9g - 7.3\pa{ 3.4 \cos(\omega t)+3.9} \\ -y-3.9 &= 2.9g - 7.3y \\ 6.3y &= 2.9g + 3.9 \\ y&= 2.9g + 3.9 \\ \end{align}$