# A 3kg block is attached to an ideal spring with a force constant k=200N/m. The block is given an initial velocity in the positive direction of magnitude u=12m/s and no initial displacement(x.=0). (a) find the amplitude?

Jul 29, 2018

$= \frac{3 \sqrt{6}}{5} \text{ m}$

#### Explanation:

Conservation of total Energy $E$:

$E = T + U$

• At displacement = 0: $q \quad U = 0 , q \quad T = {T}_{\text{max}}$

• At maximum displacement: $q \quad T = 0 , q \quad U = {U}_{\text{max}}$

So:

• ${U}_{\text{max")= T_("max}}$

$\implies \frac{1}{2} k {x}_{\text{max")^2 = 1/2 m v_("max}}^{2}$

$\therefore {x}_{\text{max") = sqrt(m/k) v_("max}} = \sqrt{\frac{3}{200}} \cdot 12$

$= \frac{3 \sqrt{6}}{5} \text{ m}$