Question

135 sin(2t) . Questions Q1 (0/10) Q2 (0/10) Q3 (0/10) A9 kg weight is attached to a spring with constant k 81 kg/m and subjected to an external force F(t) = The weight is initially displaced 4 meters above equilibrium and given an upward velocity of 3 m/s. Find its displacement for t> 0. v(t) Preview Q4 (0/10) Q5 (0/10) Q6(0/10) License Pnints nneeihle: 10

Answer 1

The equation of motion of the system is

.......(i)

&y dt2 ky f(t)

where
f(tFosin(at)
and
FO 135, a 2

'm' is the mass of the spring and 'k' is the spring constant and 'y' is the displacement of the spring at time 't'

Equn (i) has two solutions . One is Complementary Function (y_CF) and other is Particular Integral.(y_PI)

To find CF, we set

&y ky 0 dt2

Or

y wy 0 dt2 2
[ where
wk/m 9
]

The trial solution of the above equation is

..............(ii)

yCF= Acoswot+Bsinwot

To find PI, we set

&y ky Fosinat dt2

Or

[ where f₀ = F₀/m =15]........................(iii)

y 2 y fosinat dt2

To Solve (iii) we take the trial function as

......................(iv)

yScos(at +

Putting (iv) in (iii) we get

$-\alpha^2Scos( \alpha t+\phi)+\omega_0^2Scos( \alpha t+\phi)=f_0sin( \alpha t+\phi-\phi)$

Or

$-\alpha^2Scos( \alpha t+\phi)+\omega_0^2Scos( \alpha t+\phi)=f_0[sin( \alpha t+\phi)cos\phi-cos( \alpha t+\phi)sin\phi]$

compairing the coefficients of we get

sin(at+), cos(at +

$f_0cos\phi=0\Rightarrow \phi =\pi/2$

and

fosing= Swa

Or

fo lwf-aAs, T/2]

So

$y_{PI}=\frac{f_0}{[\omega_0^2-\alpha^2]} cos (\alpha t +\pi/2)$..........................(v)

So the complete solution is at time t>0 is

$y(t)=y_{PI}+y_{CF}=\frac{f_0}{[\omega_0^2-\alpha^2]} cos (\alpha t +\pi/2)+ Acos \omega_0 t+Bsin \omega_0 t$

Given at t=0, x=4. Putting this boundary condition in the above equation we get

$4=\frac{f_0}{[\omega_0^2-\alpha^2]} cos (\pi/2)+ A \Rightarrow A=4$

The velocity of the system is

$\dot y =\frac{dy}{dt}= -\alpha\frac{f_0}{[\omega_0^2-\alpha^2]} sin (\alpha t +\pi/2)-\omega_0Asin \omega_0 t+\omega_0Bcos \omega_0 t$

Given at t=0, $\dot y =3$, we get

$3 = -\alpha\frac{f_0}{[\omega_0^2-\alpha^2]}+ \omega_0B$

Or

$B=\frac{1}{\omega_0}[3 +\frac{\alpha f_0}{[\omega_0^2-\alpha^2]}]=3$

From the problem statement we find the displacement  at time t>0 is

$y(t)=3 cos (2 t +\pi/2)+ 4cos 3 t+3sin 3 t$