Answer
0.44 m/s
Work Step by Step
Please see the attached image first.
Here we use the principle of conservation of energy to find the initial speed.
$E_{kinetic}^{initial}+E_{potential}^{initial}=E_{kinetic}^{final}+E_{potential}^{final}$
So, we can write,
$\frac{1}{2}mV^{2}+0+\frac{1}{2}kx_{1}^{2}=0+0+\frac{1}{2}kx_{2}^{2}$
$V^{2}=\frac{k}{m}x_{2}^{2}-\frac{k}{m}x_{1}^{2}=\frac{k}{m}(x_{2}^{2}-x_{1}^{2})-(1)$
From equation 10.11 $\omega=\sqrt {\frac{k}{m}}-(2)$
(2)=>(1),
$V^{2}=\omega^{2}(x_{2}^{2}-x_{1}^{2})=>V=\omega\sqrt {(x_{2}^{2}-x_{1}^{2})}$
Let's plug known values into this equation.
$V=7\space rad/s\sqrt {(0.08\space m)^{2}-(0.05\space m)^{2}}=0.44\space m/s$
