Question

Suppose a mass weighing 32 lb stretches a spring 2 ft. If the mass is released from rest at the equilibrium position, find the equation of motion x(t) if an impressed force f (t) - sin t acts on the system for 0 t 2π and is then removed

Answer 1

a mass weighing 32 lbs
stretches a spring 2 ft
.
by hooke's law
mg=ks
32=k*2
16=k
k=16
.
mass weight by 32 pound
divide by gravity 32 ft/s^2 , so here we got m=1 slugs
the weight is released from equilibrium point, it represent x(0)=0
and the weight is release from rest , it represent x'(0)=0
.

sin (t) 0 if 0 < t < 2π ift>2n f(t) =

f(t) = sin (t)-sin (t) H (t-2n)

.

mx" + kx = f(t)


1-x" +16xsin (t)-sin (t) H (t 2T)

r" + 162 sin (t) _ sin (t) H (t-2π )

$x''+16x=\:\sin \left(t\right)-\sin \left(t - 2 \pi\right)\text{H}\left(t-2\pi \:\right)$

take laplace

$L[x'']+16L[x]=\:L[\sin \left(t\right)]-L[\sin \left(t - 2 \pi\right)\text{H}\left(t-2\pi \:\right)]$

s2r(s) - sr(0) (0) + 16r(s)- -

here we have  x(0)=0 and x'(0)=0

一2TS - 0 + 16r(s) -

一2ns 16x(s)

$x(s) (s^2+16)= \frac{1}{s^2+1} - \frac{e^{-2\pi s}}{s^2+1}$

e-273 r(s)(s2 + 16) = (A) (s2 + 1)

$x(s) = \frac{1}{(s^2+1)(s^2+16)} - \frac{e^{-2\pi s}}{(s^2+1)(s^2+16)}$

take inverse laplace

17

take partial fraction

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

(s2 + 1) (s21)s2+1s2 +16

= (als + ao ) (s2 + 16) + (a3s+ a2) (s2 +1)

$1=a_1s^3+16a_1s+a_0s^2+16a_0+a_3s^3+a_3s+a_2s^2+a_2$

$1=s^3\left(a_1+a_3\right)+s^2\left(a_0+a_2\right)+s\left(16a_1+a_3\right)+\left(16a_0+a_2\right)$

compare coefficient both sides

a1 + a30

by solving this we get $a_3=0,\:a_1=0,\:a_2=-\frac{1}{15},\:a_0=\frac{1}{15}$

put this constant in equation 1

$\frac{1}{\left(s^2+1\right)\left(s^2+16\right)}=\frac{0\cdot \:s+\frac{1}{15}}{s^2+1}+\frac{0\cdot \:s+\left(-\frac{1}{15}\right)}{s^2+16}$

.

(s2 +1) (s21) 15 (s2+1) 15 (s2+ 16)

.

our last step was

.

17

$x(t)=\:L^{-1}\left\{\frac{1}{15\left(s^2+1\right)}-\frac{1}{15\left(s^2+16\right)}\right\}+L^{-1}\left\{\frac{e^{-2\pi s}}{\left(s^2+1\right)\left(s^2+16\right)}\right\}$

apply inverse laplace rule

$L^{-1}\left\{e^{-as}F\left(s\right)\right\}=H\left(t-a\right)f\left(t-a\right)$

$\mathrm{For\:}\frac{e^{-2\pi s}}{\left(s^2+1\right)\left(s^2+16\right)}:\quad F\left(s\right)=\frac{1}{\left(s^2+1\right)\left(s^2+16\right)},\:\quad \:a=2\pi$

$x(t)=\:L^{-1}\left\{\frac{1}{15\left(s^2+1\right)}-\frac{1}{15\left(s^2+16\right)}\right\}+L^{-1}\left\{\frac{e^{-2\pi s}}{\left(s^2+1\right)\left(s^2+16\right)}\right\}$$x(t)=\:L^{-1}\left\{\frac{1}{15\left(s^2+1\right)}-\frac{1}{15\left(s^2+16\right)}\right\}+L^{-1}\left\{e^{-2\pi s}\left(\frac{1}{15\left(s^2+1\right)}-\frac{1}{15\left(s^2+16\right)}\right)\right\}$

.

r(t) = L- 15 (s2+ 1) J 15 (s 1) 15 (s2 16)J

.

$x(t)=\frac{1}{15}\sin \left(t\right)-\frac{1}{60}\sin \left(4t\right)+\text{H}\left(t-2\pi \right)\left(\frac{1}{15}\sin \left(t-2\pi \right)-\frac{1}{60}\sin \left(4\left(t-2\pi \right)\right)\right)$

here we use H=u notation

${\color{Red} x(t)=\frac{1}{15}\sin \left(t\right)-\frac{1}{60}\sin \left(4t\right)+\text{u}\left(t-2\pi \right)\left(\frac{1}{15}\sin \left(t-2\pi \right)-\frac{1}{60}\sin \left(4\left(t-2\pi \right)\right)\right)}$

and also here wen can write  $\sin \left(t-2\pi \right)= \sin \left(t\right)$

${\color{Red} x\left(t\right)=\frac{1}{15}\sin \left(t\right)-\frac{1}{60}\sin \left(4t\right)+\text{u}\left(t-2\pi \:\right)\left(\frac{1}{15}\sin \left(t\right)-\frac{1}{60}\sin \left(4t\right)\right)}$

or

u2π (t ) ( Љ sin (t)-60 sin (4t) sin (t) _ 60 sin (40 r (t) 15 15

all form of answers are correct