# a spot of light undergoing simple harmonic motion on a computer screen has a period of pi seconds. when the spot is 3 rt3 cm to the left of its equilibrium point, it has a velocity of 6cm per...

a spot of light undergoing simple harmonic motion on a computer screen has a period of pi seconds. when the spot is 3 rt3 cm to the left of its equilibrium point, it has a velocity of 6cm per second towards its equilibrium point.

Given x(t)=Asin(nt)+Bcos(nt) is a soloution of the simple harmonic motion equation x''(t)=-n^2 x(t)

use the intial conditions of the spot light find values for n, A and B to show that the function x(t)=3sin(2t)-3 rt3 cos (2t) describes the motion of the spot

oldnick | (Level 1) Valedictorian

Posted on

`x''(t)=-n^2x(t)`

we can  re. write it as a homogeneous linear differential equation of second order:

`x''(t)+n^(2)x(t)=0`

with associated equation:   `lambda^2+n^2=0`  that has complex root:

`lambda=+-"i"n`

thus, general  solution is :

`x(t)=A sin(nt)+ B cos(nt)`

On the other side, speed is described by first derivative of x(t), so that:

`x'(t)= nAcos(nt)-nBsin(nt)`

Now we have that when the spot is `3sqrt(3)` from start, its speed is 6.

So:

`Asin(nt)+Bcos(nt)=3sqrt(3)`

`nAcos(nt)-nBsin(nt)=6`

Is a sytstem of A, B as unknows

`Delta=[[sin(nt),cos(nt)],[ncos(nt),-nsin(nt)]]=-n`

`Delta_A=[[3sqrt(3),cos(nt)],[6,-nsin(nt)]]=-(3nsqrt(3)sin(nt)+6cos(nt))`

`Delta_B=[[sin(nt),3sqrt(3)],[ncos(nt),6]]=6sin(nt)-3nsqrt(3)cos(nt)`

`A=(Delta_a)/(Delta)=3sqrt(3) sin(nt)+(6cos(nt))/n`

``

` B=(Delta_b)/(Delta)=3sqrt(3)cos(nt)-(6sin(nt))/n`

`"So:"x=pi `

`A=-6/n`

`B=-3sqrt(3)`

`x(t)=(6/n)sin(nt)-3sqrt(3)cos(nt)`

`For n=2`

`x(t)=3sin(2t)-3sqrt(3)cos(2t)`

``

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

Given x(t)=Asin(nt)+Bcos(nt) is a soloution of the simple harmonic motion equation x''(t)=-n^2 x(t)

use the intial conditions of the spot light find values for n, A and B

Initial conditions.

When t=0 ,x=-3sqrt(3)

and    t=0   ,  `(dx)/(dt)=6`

`x(t)=Asin(nt)+Bcos(nt)`

`-3sqrt(3)=B`                               (i)

`(dx)/(dt)=nAcos(nt)-nBsin(nt)`

`6=nA`                   (ii)

Thus

`x(t)=(6/n)sin(nt)-3sqrt(3)cos(nt)`       (iii)

(iii) satisfy the given initial conditions. Differentiate (iii) ,two times,

with respect to t

`(d^2x)/(dx^2)=-n^2x`                              (iv)

`x(t)=3sin(2t)-3sqrt(3)cos(2t)`    (v)

differentiate (v) ,with respect to t , two times ,we have

`(d^2x)/(dt^2)=-4x(t)`                        (vi)

compare (iv) and (vi),we have

`n^2=4`

`n=2`

Thus with given initial conditions,

A=3,B=3sqrt(3) and n=2