Question

assume the spring constant in this lab is 10 N/m. if a mass of 1 lb is attached and the mass pulls for 5 inches. when the mass is released what will be its max minim speed going up in m/s,

and what will be it’s maximum acceleration in m/s?

Answer 1

If we choose downward as the positive Y direction, then Newton's second law gives

$$\sum{F_y }=ma_y$$

where

$$\sum{F_y }=-ky+mg=-k(y_1+y_0)+mg$$ $$ma_y=m\frac{d^2y}{dt^2}=\big\{ y_0=Const\big\}=m\frac{d^2y_1}{dt^2}$$ $$ky_0=mg- equilibrium \space position$$

The equation of motion of the mass takes the form:

$$m\frac{d^2y_1}{dt^2}=-ky_1$$

It has the solution:

$$y_1(t)=y_{1max} cos\bigg (\sqrt{\frac{k}{m}}t\bigg)$$

Then, the speed of movement of the mass:

$$v_1(t)=\frac{dy_1}{dt}=-y_{1max}\sqrt{\frac{k}{m}}sin{\bigg(\sqrt{\frac{k}{m}}t}\bigg)$$

The acceleration:

$$a_1(t)=\frac{d^2y_1}{dt^2}=-y_{1max}\bigg(\frac{k}{m}\bigg)cos{\bigg(\sqrt{\frac{k}{m}}t}\bigg)$$

in SI units:

$$m=1lb=0.4536kg, \space y_{1max}=5in=0.0254m$$

So, the maximum acceleration and maximum speed of the mass upward, taking into account the direction of the Y-axis:

$$v_{1max}=-y_{1max}\sqrt{\frac{k}{m}}=-(0.0254m)\sqrt{\frac{10 \frac N m}{0.4536kg}}\approxeq-0.119 \frac m s$$ $$a_{1max}=-y_{1max}\bigg({\frac{k}{m}}\bigg)=-(0.0254m)\bigg({\frac{10 \frac N m}{0.4536kg}}\bigg)\approxeq-0.56 \frac {m^2} s$$