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

Explanation:

Data

mass m= 100g

Length L= 5cm

we can use:

gm-kL= 0

divide both side by m

g - =0

where

=

^{2}

so now

=

square both side

We can apply:

u(t)=Acoswt +Bsinwt

u(t)=Acos14t +Bsin14t

u(0)=0  where A=0

therefore

u(0) = Bsin14t  

(0) = 10 ⇒ 10=14B ⇒ B= B=

so now u(t)=sin14t

so t will be:

t=

t=

t=0.22 seconds

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)