Answer to Question #142524 in Classical Mechanics for khaled mohammed

Question

A 500 g block is attached to a spring on a frictionless horizontal surface. The block is pulled to stretch the spring by 10 cm, then gently released. A short time later, as the block passes through the equilibrium position, its speed is
m/s.

What is the block’s period of oscillation?
What is the block’s speed at the point where the spring is compressed by 5.0 cm?

Answer 1

$E=E_k+E_p=\frac{mV^2}{2}+\frac{kx^2}{2}$

$\text{where }k\text{ is the stiffness of the spring } x\text{ the size of compression (extension)}$

$\text{let } V_m\text{ maximum speed at }x = 0 \text{ and} A =0.1 m\text{ maximum spring tension}$

$\text {for } x=0\ E=\frac{mV_m^2}{2}\ ;\text { for } x=A\ E=\frac{kA^2}{2} (1)\text{ from here}$

$k=\frac{mV^2_m}{A^2}$

$T=2\pi\sqrt{\frac{m}{k}}=2\pi{\frac{A}{V_m}}\approx\frac{0.63}{V_m}$

$\text{for }x=0.05;x=\frac{A}{2}\ E= E_p+E_k= \frac{mV^2}{2}+\frac{kx^2}{2}=\frac{mV^2}{2}+\frac{kA^2}{2*4}$

$\text{from(1) }\frac{mV^2}{2}+\frac{kA^2}{2*4}=\frac{kA^2}{2};$

$\frac{mV^2}{2}=\frac{kA^2}{2}-\frac{kA^2}{8}=\frac{3}{4}*\frac{kA^2}{2}=\frac{3}{4}*\frac{mV_m^2}{2}$

$V=\frac{\sqrt{3}}{2}V_m$

Answer:$T\approx\frac{0.63}{V_m}$ for spring is compressed by 5.0 cm $V=\frac{\sqrt{3}}{2}V_m$