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A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 20 cms, and if there is no damping, determine the position u of the mass at any time t. Enclose arguments of functions in parentheses. For example, sin(2x).

Respuesta :

Answer:

[tex]u(t)=\frac{1}{5} sin\ (25t)[/tex]

Explanation:

Given:

  • mass of the body stretching the spring, [tex]m=150\ g[/tex]
  • extension in spring, [tex]\Delta x=1.568\ cm[/tex]
  • velocity of oscillation, [tex]u'(0)=20\ cm.s^{-1}[/tex]
  • initial displacement position of equilibrium, [tex]u(0)=0[/tex]

According to given:

[tex]m.g=k.\Delta x[/tex]

[tex]150\times 980=k\times 1.568[/tex]

[tex]k=93750\ dyne.cm^{-1}[/tex]

we know frequency:

[tex]\omega=\sqrt{\frac{k}{m} }[/tex]

[tex]\omega=\sqrt{\frac{93750}{150} }[/tex]

[tex]\omega=25[/tex]

Now, for position of mass in oscillation:

[tex]u= A.sin\ (\omega.t)+B.cos\ (\omega.t)[/tex]

[tex]u= A.sin\ (25.t)+B.cos\ (25.t)[/tex]

at [tex]t=0;\ u(0)=0\ \Rightarrow A=0[/tex]

∴[tex]u(t)=B.sin\ (25.t)[/tex]

∵ at [tex]t=0;\ u'(0)=20\ cm.s^{-1}\ \Rightarrow B=\frac{1}{5}[/tex]

[tex]u(t)=\frac{1}{5} sin\ (25t)[/tex]

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