Question

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 10 cm/s, and if there is no damping, determine the position u of the mass at any time t. When does the mass first return to its equilibrium position

Answer 1

Answer:

$w_{0}=14$

$t=\frac{\pi }{14}$

Explanation:

Data

mass m= 100g

Length L= 5cm

we can use:

gm-kL= 0

divide both side by m

g - $\frac{kL}{m}$=0

where

$\frac{k}{m}$ = $\frac{g}{L}$

$\frac{k}{m}=w_{0}$^{2}

so now

$w_{0}^{2}$ = $\frac{9.8*100}{5}$

$w_{0}^{2}=\frac{980}{5}$

$w_{0}^{2}=196$

square both side

$w_{0}=\sqrt{196}$

$w_{0}=14$

We can apply:

u(t)=Acoswt +Bsinwt

u(t)=Acos14t +Bsin14t

u(0)=0  where A=0

therefore

u(0) = Bsin14t  

$u^{'}$(0) = 10 ⇒ 10=14B ⇒ B=$\frac{14}{10}$ B=$\frac{5}{7}$

so now u(t)=$\frac{5}{7}$sin14t

so t will be:

t=$\frac{\pi }{14}$

t=$\frac{3.14}{14}$

t=0.22 seconds