Question

(1 point) A frictionless spring with a 7-kg mass can be held stretched 0.6 meters beyond its natural length by a force of 70 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1 m/sec, find the position of the mass after t seconds. 0.244sin(4.08t) meters

Answer 1

mass is m=7

displacement is x=0.6

force is F=70

.

from Hooke's law

F = kr

70 = k(0.6)

350

.

there is no damping so damping coefficient is c=0

DE is given by

my' + cy + ky = 0

350 7y" + Oy' + y = 0

350 7y + ZY

find roots

350 7.2 + = ="

7.x² = ತಾಂ

$x^2=-\frac{50}{3}$

gg?+ = 2

for complex roots the general solution is

$y=e^0\left(c_1\cos \left(\frac{5\sqrt{6}}{3}t\right)+c_2\sin \left(\frac{5\sqrt{6}}{3}t\right)\right)$

$y=c_1\cos \left(\frac{5\sqrt{2}t}{\sqrt{3}}\right)+c_2\sin \left(\frac{5\sqrt{2}t}{\sqrt{3}}\right)$.........................(1)

here mass is initially released from the equilibrium position.

so y(0)=0

5V20 0 = Cicos + C2 sin in 5v2.0

0 = ci cos (0) + C2 sin (0)

0= c +0

.....................put it back in equation 1

0 = 15

.

$y=0\cdot \cos \left(\frac{5\sqrt{2}t}{\sqrt{3}}\right)+c_2\sin \left(\frac{5\sqrt{2}t}{\sqrt{3}}\right)$

.........................(2)

Y = sin 52t V3

take derivative

() o༠ , - -

here initial velocity is 1m/s so y'(0)=1

$1=\frac{5\sqrt{2}}{\sqrt{3}}c_2\cos \left(\frac{5\sqrt{2}\cdot 0}{\sqrt{3}}\right)$

15V2 3 cos (0)

${\color{Blue} c_2=\frac{\sqrt{6}}{10}}$.....................put it back in equation 2

.

Y = sin 52t V3

비서 lo sin 52

5161