Question

A mass weighing 9 lb stretches a spring 8 in. The mass is pulled down an additional 7 in and is then set in motion with an initial upward velocity of 2 ft/s. No damping is applied. a. Determine the position u of the mass at any time t. Use 32 ft/s as the acceleration due to gravity. Pay close attention to the units. u(t) = 5 cos (4 3 t) + sin(4V3 t) 2V3 b. Determine the period, amplitude and phase of the motion. T= Instructions -

Answer 1

weight is w=9 lb

mass is given by

m =

$m=\frac{9}{32}$

spring stretches 8 inches = 8/12 feet

from Hooke's law,

F = kr

$9=k\cdot \frac{8}{12}$

$k=\frac{27}{2}$

there is no damping so c=0

DE is given by

my' + cy + ky = 0

9 27 32"' +0v y' + 0y + y = 0

9 27_ 324" + y = (

find roots

04 +

6

8F-F = 2

I= IV 487

I = +4V3i

for 2 complex roots, a general solution is

4c = eº (cy cos (173t) + c, sin (4v3t))

.................(1)

y = C1 cos (4V3t+ C2 sin ( 4V3t

.

initially, mass is pulled down 7 inches = 7/12 feet. so y(0)=7/12

15 = ci cos (403. 0) +cı sin (4V3.0)

= Cicos (0) + C2 sin (0)

19 = C1 +0

= C1

....................put in an equation 1

C1 =

.

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

y = cos (4/3t) + cəsin (473)

take derivative

v = sin (1v3e) +c2cos (4V3i). 473

here initial upward velocity is y(0)=-2

—2 = asin (4V3.0) +cą cos (473. o). 473

—2 = sin (0) + C2 cos (0) - 403

-2 = -0+04V3

....................put in an equation 2

C2 = -2

.

v=zoe (1/3) + (༧) aa (༣༦)

${\color{Red} y=\frac{7}{12}\cos \left(4\sqrt{3}t\right)-\frac{\sqrt{3}}{6}\sin \left(4\sqrt{3}t\right)}$

here we have

w=4V3

.

a period is given by

$T=\frac{2 \pi}{ \omega }$

T = 27 4V3

узл T =

.

.

amplitude is given by

R= V(C) + (ca)

r(s) (3)

R = 12

.

༥ Av) • - (g) so -

(gn:) "">s (p:)" P ) -

compare with

sin (s) cos (t) - cos (s) sin(t)

761 sin (2) =- 61

x = 1.11125

.

y = VO2 (sin (1.11125) cos (4V3+) – cos (1.11125) sin (473t))

- - (in 1 12-103)

to find the phase take

1.11125 – 4V3t = 0

4V3t = 1.11125

t = 1.11125 473

t = 0.16039

= 0.16039