Significace of i*sin(wt) in SHM [cos(wt)+i*sin(wt)]

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confusing title i suspect but didnt know what else to call it.

so i'm considering a spring and mass with the displacment of the mass being x and the stiffness of the spring being k

force on the spring = -k*x =m*a

or rearranged etc

d2x/dt2 + (k/m)*x = 0 ...............................................[1]

then i assume a soln of the form e^(at)
and find a= i*(k/m)^.5 [then let sqrt(k/m)= w)

so i get x= e^(i*w*t)

which using the euler identity gives

x= cos(wt)+i*sin(wt)

great so i managed to show that the thing will move sinusoidly
but what about that i*sin(wt) term

if i ignore it, and let x=cos(wt) this still satisfies [1] above
so am i justified in ignoring it? or does it have some physical significance?
or is it something that just gets "thrown up" as a result of me assuming a soln of the form e^at ?

and more generally
if i find a soln to a diff eqn, and find that by dropping some terms in the soln that it still satifies the eqn am i still justified in doing that?
thanks jim
 

Answers and Replies

dst
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You need to simplify a bit further. I can't remember how but eventually the imaginary part disappears when you substitute a certain complex number which IIRC was [tex]\sqrt{\frac{-k}{m}}[/tex].

You will eventually get to something of the form x = a sin (wt + y). where a is the max displacement, wt is the angular frequency, and y is the phase shift.
 
Last edited:
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I suspect that you'll find that your second order differential equation has two linearly independent solutions. (you somehow threw away the -w root of w^2). So the other solution you discarded was
[tex]x= cos(wt)-i*sin(wt)[/tex]
Now, the most general solution to your DE is a linear combination of these solutions, and you can pick the coefficients so that your initial conditions are satisfied. Likely you'll just pick
[tex]x = e^{i w t} + e^{-i \omega t}[/tex]
and all your problems will go away.
 
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nice one!"
so i get x(t)=cos(wt)+i*sin(wt)
and x(t)=cos(wt)- i*sin(wt)

then adding i get

2*x(t)=2*cos(wt)+0

and x(t)=cos(wt)

so i dont have to worry about the complex


so is this called superposition then?
 
550
2
nice one!"
so i get x(t)=cos(wt)+i*sin(wt)
and x(t)=cos(wt)- i*sin(wt)

then adding i get

2*x(t)=2*cos(wt)+0

and x(t)=cos(wt)

so i dont have to worry about the complex


so is this called superposition then?
This is not quite right. What you have there are two different solutions, which are independent of each other. So, let's write these like this:

[tex]x_1(t) = e^{i\omega t}[/tex]
[tex]x_2(t) = e^{-i\omega t}[/tex]

However, because these are solutions of a linear differential equation, any linear superposition of the solutions will, itself, be a solution. This superposition will take the form:

[tex]x(t) = Ax_1(t) + Bx_2(t)[/tex], where A and B are, in general, any complex numbers.

However, since x_2 is the complex conjugate of x_1, the only way to get real-valued solutions for x(t) is to set [itex]B = A^*[/itex].

Now, let's let [itex]A = a e^{-i\phi}[/itex].

In this case:

[tex]x(t) = a e^{i(\omega t - \phi)} + a e^{-i(\omega t - \phi)}[/tex]

With a little manipulation, it is clear that this means that [itex]x(t) = 2a\cos (\omega t - \phi)[/itex]. Now we can recognize that 2a will be the largest possible value of x. So:

[tex] x(t) = x_{max} \cos (\omega t - \phi)[/tex]

[itex]x_{max}[/itex] are not specified by the original differential equation. This essentially means that the solution can have any amplitude and phase. In other words, the oscillations can have any size and can start at any time.

If you recall your trig identities, you'll recognize that we could also write this as:

[tex]x(t) = x_{max}(\cos \phi \cos (\omega t) + \sin \phi \sin (\omega t))[/tex]

This can also be written:

[itex]x(t) = a_1 \cos (\omega t) + a_2 \sin (\omega t)[/itex], where we can pick [itex]a_1[/itex] and [itex]a_2[/itex] to have any values we like, including 0. So, in this way, we can see that both [itex]\cos (\omega t)[/itex] and [itex]\sin (\omega t)[/itex] are valid solutions.
 
if i ignore it, and let x=cos(wt) this still satisfies [1] above
so am i justified in ignoring it?
Yes, you can ignore it. A commonly used trick is to use the identity [itex]cos(x)=Re(e^{ix})[/itex] to convert a trig function problem into an exponential problem.

or does it have some physical significance?
The significance is that [itex]sin(\omega t)=Im(e^{i \omega t})[/itex] is also a solution.

or is it something that just gets "thrown up" as a result of me assuming a soln of the form e^at ?

and more generally
if i find a soln to a diff eqn, and find that by dropping some terms in the soln that it still satifies the eqn am i still justified in doing that?
thanks jim
A solution to the equation is a solution to the equation. How you derive it is irrelevant.
 
Last edited:
Andy Resnick
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confusing title i suspect but didnt know what else to call it.

so i'm considering a spring and mass with the displacment of the mass being x and the stiffness of the spring being k

force on the spring = -k*x =m*a

<snip>

so i get x= e^(i*w*t)

which using the euler identity gives

x= cos(wt)+i*sin(wt)

great so i managed to show that the thing will move sinusoidly
but what about that i*sin(wt) term

<snip>
thanks jim
Others have pointed out that [tex]e^{i \omega t}[/tex] and [tex]e^{-i \omega t}[/tex] are general solutions, the specific form of which can be determined by initial conditions.

However, for your original question, the term i * sin(wt) does have a physical relevance- in mechanics and electromagnetics, the linear response of a system to a periodic disturbance (a fancy way of thinking about the mass-spring problem) has a component that is in phase to the disturbance (the real cos (wt) part) and a component that is in phase quadrature (i *sin(wt)). In mechanics, the phase quadrature component arises due to viscous or dissipative effects; in electromagnetics, the phase quadrature component corresponds to the reactance, IIRC.
 

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