Solve Forced Oscillation using Differential Equation Method

[angelwentheng]

The differential eqn that governs the forced oscillation is shown below:
0.2(d^2 y)/(dt^2 )+ 1.2dy/dt + 2y = r(t) where r(t) is the external force

Given that r(t) = 5cos4t with y(0) = 0.5 and y'(0) = 0. Find the equation of motion of the forced oscillation..

Please help me to solve by using differential eqn method..thx!urgent

 

[tiny-tim]

welcome to pf!

hi angelwentheng! welcome to pf!

(try using the X² icon just above the Reply box )

show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[angelwentheng]

0.2(d^2 y)/(dt^2 )+ 1.2dy/dt + 2y = r(t)
0.2(d^2 y)/(dt^2 ) + 1.2dy/dt + 2y = 5cos4t
0.2m^2 + 1.2m + 2 = 0
m = (-b±√(b^2-4ac))/2a
= (-(1.2)± √(〖(1.2)〗^2-4(0.2)(2)))/(2(0.2))
= (-1.2 ±0.4i)/0.4
= -3 ± i
y_h(t) = e^(-3t) (Acosx+Bsinx)
r (t) = 5cos4t
y_p (t) = pcos4t + qsin4t ------------ (1)
y_p^' (t)= -4psin4t + 4qcos4t---------- (2)
y_p^'' (t) = -16pcos4t – 16qsin4t ---------- (3)
Substitute (1), (2) and (3) into 0.2 (d^2 y)/(dt^2 ) + 1.2 dy/dt + 2y = 5cos4t
0.2(-16pcos4t – 16qsin4t) + 1.2(-4psin4t + 4qcos4t) + 2(pcos4t + qsin4t) = 5cos4t
-3.2pcos4t – 3.2qsin4t – 4.8psin4t + 4.8qcos4t + 2pcos4t + 2qsin4t = 5cos4t
Equate the coefficients:
For cos4t:
-3.2p + 4.8q + 2p = 5
-1.2p + 4.8q = 5 ---------- (1)
For sin4t:
-3.2q – 4.8p + 2q = 0
-1.2q – 4.8p = 0 ----------- (2)
Equation 1 x4
19.2 q – 4.8p = 20---------- (3)
-20.4q = -20
q = 50/51
Substitute q =50/51 into equation (2)
-1.2(50/51) – 4.8p = 0
- 4.8p = 20/17
P = - 25/102
y_p (t)= - 25/102cos4t + 50/51sin4t
y (t) =y_h(t) + y_p(t)
= e^(-3t)(Acos t + Bsin t) - 25/102cos4t + 50/51sin4t
Given y (0) = 0.5
A – 25/102 = 0.5
A = 38/51
y’(t) = e^(-3t )(-Asin t + Bcos t) +(Acos t + Bsin t)(-3e^(-3t)) + 50/51sin4t +200/51cos4t
Given y’ (0) = 0
B – 3A + 200/51 = 0
B – 3(38/51) + 200/51 = 0
B = -86/51
Therefore, the equation of motion of the forced oscillation is y(t) = e^(-3t)(38/51cos t – 86/51sin t) – 25/102cos4t + 50/51sin4t


pls help me to check am i correct or not.thanks

 

[angelwentheng]

sorry for not noticing the sign

 

[tiny-tim]

(please using the X² icon just above the Reply box, to make your equations readable )

herefore, the equation of motion of the forced oscillation is y(t) = e^(-3t)(38/51cos t – 86/51sin t) – 25/102cos4t + 50/51sin4t

well y(0) = 76/102 - 25/102 = 0.5

and y'(0) = -86/51 - 3*38/51 + 200/51 = 0,

so that looks ok ​

 

[angelwentheng]

Thanks~I'm wondering am I need to do the forced oscillation equation?which means using the spring equation f = kx

 

[tiny-tim]

Thanks~I'm wondering am I need to do the forced oscillation equation?which means using the spring equation f = kx

The differential eqn that governs the forced oscillation is shown below:
0.2(d^2 y)/(dt^2 )+ 1.2dy/dt + 2y = r(t) where r(t) is the external force

Given that r(t) = 5cos4t with y(0) = 0.5 and y'(0) = 0. Find the equation of motion of the forced oscillation..

you've solved the given equation for forced oscillation …

what does some other equation have to do with it? ​

 

[angelwentheng]

oh...i am blur...sorry

 

[angelwentheng]

A simple pendulum of length is oscillating through a small angle θ in a medium for which
the resistance is proportional to the velocity. Obtain the differential equation of its motion
and discuss the motion.
sorry help me with these...I'm totally blank

 

[tiny-tim]

A simple pendulum of length is oscillating through a small angle θ in a medium for which
the resistance is proportional to the velocity. Obtain the differential equation of its motion
and discuss the motion.
sorry help me with these...I'm totally blank

well the differential equation should be easy …

what do you make it?

 

[angelwentheng]

sorry I have no idea for this ques...can u help me?

 

[tiny-tim]

A simple pendulum of length is oscillating through a small angle θ in a medium for which
the resistance is proportional to the velocity.

just translate the English into maths …

you know what the usual equations is (without resistance) …

now add a force proportional to the speed

 

[angelwentheng]

External forces are zero, so the governing equation is homogeneous:
mx"+Bx'+kx=0

x"=d²x/dt².
x'=dx/dt
x=displacement, positive to the right
m=mass
B=resistance proportional to the velocity
k=stiffness (of a spring), which is resistance proportional to the displacement.

 

[tiny-tim]

External forces are zero, so the governing equation is homogeneous:
mx"+Bx'+kx=0

i don't understand what you mean by "External forces are zero" …

both gravity and the resistance are external forces

but anyway, yes that's the correct equation

now the question asks you to "discuss the motion" …

so what are the solutions, and how do they change if you increase B ?

 

[angelwentheng]

external forces means f(t)...cz actually the original eqn will be mx'' + Bx' + kx = f(x)
since there is no external force...therefore f(x)=o
but why increasing the resistance?the question said directly proportional to velocity...

erm...sorry i 'm nt physics students...so i might not understand phy laws...so pls guide me

 

[tiny-tim]

external forces means f(t)...cz actually the original eqn will be mx'' + Bx' + kx = f(x)
since there is no external force...therefore f(x)=o

actually, f(t) would be a time-dependent external force;

kx is an f(x), an external position-dependent force

but why increasing the resistance?the question said directly proportional to velocity...

yes, you can't change B in the middle of a particular motion

but you can have different experiments with different Bs, and the question is asking you to categorise the solutions …

the solutions will be of different characters, depending how large B is (think "damping" )

 

[angelwentheng]

For a pendulum of length L and mass m,
making an angle θ with the vertical, three forces act on the mass m.
The vertical force due to gravity, mg.
A horizontal restoring force (towards the equilibrium position) of mgsin(θ), and the tension, which is equal and opposite to the resultant of the two forces.
The restoring force mgsin(θ) is often approximated by
kx=mgLsin(θ)≅mgLθ for small angles (when θ is in radians).

This is my explanation for this ques...is it correct?

 

[tiny-tim]

A simple pendulum of length is oscillating through a small angle θ in a medium for which
the resistance is proportional to the velocity. Obtain the differential equation of its motion
and discuss the motion.

This is my explanation for this ques...

no it isn't!

its your explanation for a different question …

the question asked you to discuss the motion with a resistance proportional to the velocity

 

[angelwentheng]

then how to explain??

 

[tiny-tim]

well, first you'll have to solve the equation, won't you?

 

[nailteenz]

A simple pendulum of length is oscillating through a small angle θ in a medium for which
the resistance is proportional to the velocity. Obtain the differential equation of its motion
and discuss the motion.
sorry help me with these...I'm totally blank

I am stucked in this same question too!could anyone help?