Question

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).

Answer 1

Answer:

$u(t)=\frac{1}{5} sin\ (25t)$

Explanation:

Given:

• mass of the body stretching the spring, $m=150\ g$

• extension in spring, $\Delta x=1.568\ cm$

• velocity of oscillation, $u'(0)=20\ cm.s^{-1}$

• initial displacement position of equilibrium, $u(0)=0$

According to given:

$m.g=k.\Delta x$

$150\times 980=k\times 1.568$

$k=93750\ dyne.cm^{-1}$

we know frequency:

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{93750}{150} }$

$\omega=25$

Now, for position of mass in oscillation:

$u= A.sin\ (\omega.t)+B.cos\ (\omega.t)$

$u= A.sin\ (25.t)+B.cos\ (25.t)$

at $t=0;\ u(0)=0\ \Rightarrow A=0$

∴$u(t)=B.sin\ (25.t)$

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

$u(t)=\frac{1}{5} sin\ (25t)$