If we choose downward as the positive Y direction, then Newton's second law gives
∑ F y = m a y \sum{F_y }=ma_y ∑ F y = m a y where
∑ F y = − k y + m g = − k ( y 1 + y 0 ) + m g \sum{F_y }=-ky+mg=-k(y_1+y_0)+mg ∑ F y = − k y + m g = − k ( y 1 + y 0 ) + m g m a y = m d 2 y d t 2 = { y 0 = C o n s t } = m d 2 y 1 d t 2 ma_y=m\frac{d^2y}{dt^2}=\big\{ y_0=Const\big\}=m\frac{d^2y_1}{dt^2} m a y = m d t 2 d 2 y = { y 0 = C o n s t } = m d t 2 d 2 y 1 k y 0 = m g − e q u i l i b r i u m p o s i t i o n ky_0=mg- equilibrium \space position k y 0 = m g − e q u i l ib r i u m p os i t i o n The equation of motion of the mass takes the form:
m d 2 y 1 d t 2 = − k y 1 m\frac{d^2y_1}{dt^2}=-ky_1 m d t 2 d 2 y 1 = − k y 1 It has the solution:
y 1 ( t ) = y 1 m a x c o s ( k m t ) y_1(t)=y_{1max} cos\bigg (\sqrt{\frac{k}{m}}t\bigg) y 1 ( t ) = y 1 ma x cos ( m k t )
Then, the speed of movement of the mass:
v 1 ( t ) = d y 1 d t = − y 1 m a x k m s i n ( k m t ) v_1(t)=\frac{dy_1}{dt}=-y_{1max}\sqrt{\frac{k}{m}}sin{\bigg(\sqrt{\frac{k}{m}}t}\bigg) v 1 ( t ) = d t d y 1 = − y 1 ma x m k s in ( m k t ) The acceleration:
a 1 ( t ) = d 2 y 1 d t 2 = − y 1 m a x ( k m ) c o s ( k m t ) a_1(t)=\frac{d^2y_1}{dt^2}=-y_{1max}\bigg(\frac{k}{m}\bigg)cos{\bigg(\sqrt{\frac{k}{m}}t}\bigg) a 1 ( t ) = d t 2 d 2 y 1 = − y 1 ma x ( m k ) cos ( m k t ) in SI units:
m = 1 l b = 0.4536 k g , y 1 m a x = 5 i n = 0.0254 m m=1lb=0.4536kg, \space y_{1max}=5in=0.0254m m = 1 l b = 0.4536 k g , y 1 ma x = 5 in = 0.0254 m So, the maximum acceleration and maximum speed of the mass upward, taking into account the direction of the Y-axis:
v 1 m a x = − y 1 m a x k m = − ( 0.0254 m ) 10 N m 0.4536 k g ≊ − 0.119 m s v_{1max}=-y_{1max}\sqrt{\frac{k}{m}}=-(0.0254m)\sqrt{\frac{10 \frac N m}{0.4536kg}}\approxeq-0.119 \frac m s v 1 ma x = − y 1 ma x m k = − ( 0.0254 m ) 0.4536 k g 10 m N ≊ − 0.119 s m a 1 m a x = − y 1 m a x ( k m ) = − ( 0.0254 m ) ( 10 N m 0.4536 k g ) ≊ − 0.56 m 2 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 a 1 ma x = − y 1 ma x ( m k ) = − ( 0.0254 m ) ( 0.4536 k g 10 m N ) ≊ − 0.56 s m 2
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