Question

A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10 cm/s and no damping is applied.

(a) Determine the position u of the mass at any time t. Use 9.8 m/s as the acceleration due to gravity. Pay close attention to the units.
(b) When does the mass first return to its equilibrium position?

Answer 1

`u(t)=1.15 \sin (8.68t)cm`

0.3619sec

Step-by-step explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

`\omega_0=\sqrt{(g)/(L)}\\\\=\sqrt{(980)/(13)}\\\\=8.68 rad/s`

Where
`g=980 cm/s^2`

`u(t)=Acos8.68 t+Bsin 8.68t`

u(0)=0

Substitute the value

`A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t`

Substitute u'(0)=10

`8.68B=10`

`B=(10)/(8.68)=1.15`

Substitute the values

`u(t)=1.15 \sin (8.68t)cm`

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec