Respuesta :
Answer:
1. [tex]A=0.0847\ m[/tex]
2. [tex]v=0.050\ m.s^{-1}[/tex]
Explanation:
Given:
- mass of block, [tex]m=0.9\ kg[/tex]
- spring constant, [tex]k=14.5\ N.m^{-1}[/tex]
- maximum velocity of block, [tex]v_m=34\ cm.s^{-1}[/tex]
1.
We know from the energy of oscillating spring:
[tex]E=\frac{1}{2}.k.A^2= \frac{1}{2}.m.v_m^2[/tex]
[tex]\frac{1}{2}\times 14.5\times A^2=\frac{1}{2}\times 0.9\times 0.34^2[/tex]
[tex]A=0.0847\ m[/tex]
2.
Now, given that:
instantaneous position, [tex]x=0.15\times A=0.0127\ m[/tex]
we find angular speed,
[tex]\omega=\sqrt{\frac{k}{m} }[/tex]
[tex]\omega=\sqrt{\frac{14.5}{0.9} }[/tex]
[tex]\omega=4.013\ rad.s^{-1}[/tex]
So we have
[tex]v=x.\omega[/tex]
[tex]v=0.0127\times 4.013[/tex]
[tex]v=0.051\ m.s^{-1}[/tex]
