Question

How to calculate the displacement of a wave which undergoes damped oscillation due to an external periodic force?

Answer 1

Please post full question.

Answer 2

i want to know how to derive the eq for a general case..

Answer 3

For ex: if there is a spring-block system and \[ F(t)= F _{0}\cos \omega _{d}t\] is the external periodic force and f=-bv is the damping force then how to find displacement as a function of time..?

Answer 4

@ParthKohli @samigupta8 @Vincent-Lyon.Fr

Answer 5

@inkyvoyd @ganeshie8

Answer 6

can u tell me the proper expression that we get at the end of this derivation?

Answer 7

D2x/dt^2

Answer 8

yes ofcourse!

Answer 9

atleast tell me what will be the end result..

Answer 10

if you're given a bunch of options, just plug all of them and see which satisfies your equation.

Answer 11

what if i am given values..

Answer 12

\[m\frac{ d^2x }{ dt^2 } = -kx-bv+F _{0}\cos \omega _{d}t\]

Answer 13

then x=?

Answer 14

x will be a sine function after a few oscillations , so plug in something like:
\(x(t) = A\cos \omega _{d}t\)

Answer 15

Then you will have to use a phasor diagram to add all the terms of the equation.

Answer 16

solve it as you would any 2nd order DE

forst complementary solution for the homogeneous DE, ie where LHS = 0

then go after particular solution which you can get in a number of ways, but Vincent's suggestions is easiest and reflects practical reality, so guess a sinusoid form and plug it in

Answer 17

I should have written : \(x(t) = A\cos (\omega _{d}t+\phi)\)

Answer 18

u told x will be a sine function but u have written cos funct. ..why? and what will be A in terms of \[A _{o}\] (highest amplitude/ the amplitude when there is no damping force..)

Answer 19

the particular solution to \(m\ddot x+ b\dot x+ kx = F _{0}\cos \omega _{d}t\) can be obtained in a number of ways, with probs the most basic being to assume a solution in the form \(y_p = \alpha \cos \omega _{d}t + \beta \sin \omega _{d}t\), with \(\alpha, \beta\) = const. so you stuff that into the actual DE and then get \(\alpha, \beta\) from balancing terms on LHS and RHS

or you can combine the \( \alpha \cos \omega _{d}t + \beta \sin \omega _{d}t\) into a single cos or sin function with a phase shift. so \(x_p= A\cos (\omega _{d}t+\phi)\) or \(x_p= A\sin (\omega _{d}t+\theta)\)

it's just this: http://www.sosmath.com/trig/addform/addform.html

and can be seen here: http://hyperphysics.phy-astr.gsu.edu/hbase/oscdr.html

Answer 20

@IrishBoy123 can u pls tell me what is 'w' in that eq in the second link?

Answer 21

http://hyperphysics.phy-astr.gsu.edu/hbase/oscdr.html#c2

Answer 22

@IrishBoy123 i saw that link already..

Answer 23

i have a doubt in that... in the final expression what is 'w' ?

Answer 24

\[m(-Aw _{d}^2\sin(w _{d}t+\theta)) +b(Aw _{d}\cos(w _{d}t+\theta)) +kx = F _{0}cosw _{d}t\]

Answer 25

how to solve further? @Vincent-Lyon.Fr @IrishBoy123

Answer 26

i think it will be simpler if you substitute \(x_p = \alpha \cos \omega _{d}t + \beta \sin \omega _{d}t\)...

substitute it into the differential equation, and re-arrange it so you get something like
\( ( ~~) \cos \omega _{d}t + (~~) \sin \omega _{d}t=F_0cos\omega_dt\)

compare the coefficients of sin and cos terms of LHS and RHS to find alpha and beta

Answer 27

Here is a way to do it if you are not familiar with complex numbers.

Let's look for a solution as \(x(t)=A\cos(\omega t)\) responding to a driving force \(F(t)=F_o\cos(\omega t+\phi)\)

We will need velocity and acceleration.
\(v(t)=-\omega A\sin(\omega t)=\omega A\cos(\omega t+\pi /2)\)
\(a(t)=-\omega ^2 A\cos(\omega t)=\omega ^2A\cos(\omega t+\pi )\)

The differential equation can be drawn as a phasor:


Hence \(F_0=A\sqrt {(k-m\omega ^2)^2+(b\omega)^2}\)

\(\tan \phi =\dfrac{b\omega}{k-m\omega ^2}\)

Answer 28

Slight mistake corrected here:

Answer 29

How did u draw that phasor diagram..? can u teach me how to draw one for the below equations..?
\[F(t)=F _{0}\cos(\omega _{d}t)\]\[x(t)= Asin(\omega _{d}t +\phi)\]\[v(t)=A \omega _{d} cos(\omega _{d}t+\phi) \]\[a(t) = -A \omega _{d}^2\sin(\omega _{d}t+\phi) \]