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

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In summary, the conversation revolved around solving a differential equation for a spring and mass system, where the displacement of the mass was represented by x and the stiffness of the spring by k. The conversation explored various solutions to the equation, including using Euler's identity to simplify the solution to the form of x=cos(wt)+i*sin(wt). The significance of the i*sin(wt) term was discussed, with the conclusion that it represents a phase quadrature component in the system's response to a periodic disturbance. The conversation also delved into the concept of superposition and the justification for dropping terms in a solution to a differential equation.
  • #1
phlegmy
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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 displacement 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
 
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  • #2
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.
 
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  • #3
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.
 
  • #4
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 don't have to worry about the complex


so is this called superposition then?
 
  • #5
phlegmy said:
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 don't 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.
 
  • #6
phlegmy said:
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.
 
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  • #7
phlegmy said:
confusing title i suspect but didnt know what else to call it.

so I'm considering a spring and mass with the displacement 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.
 

1. What is the significance of i*sin(wt) in SHM?

The term i*sin(wt) represents the imaginary part of the complex number in the expression for simple harmonic motion (SHM), which is given by cos(wt)+i*sin(wt). It plays a crucial role in describing the motion of a particle in SHM, as it represents the vertical displacement of the particle from its equilibrium position.

2. How does i*sin(wt) affect the motion of a particle in SHM?

The term i*sin(wt) adds a vertical component to the motion of the particle, which results in an elliptical motion rather than a purely horizontal motion seen in cos(wt). This means that the particle not only moves back and forth along a straight line, but also oscillates up and down, creating a circular or elliptical path.

3. Can i*sin(wt) have a negative value in SHM?

Yes, i*sin(wt) can have a negative value in SHM. This indicates that the particle is moving downwards from its equilibrium position, as opposed to upwards when i*sin(wt) has a positive value. The negative value of i*sin(wt) also affects the amplitude of the oscillation, making it larger or smaller depending on the value of i*sin(wt).

4. Is the term i*sin(wt) present in all types of SHM?

Yes, the term i*sin(wt) is present in all types of SHM, as it is a fundamental component of the complex number used to describe the motion. Whether the SHM is simple, damped, or forced, the term i*sin(wt) will always be present in the expression for the motion of the particle.

5. How does i*sin(wt) relate to the concept of phase in SHM?

In SHM, the term i*sin(wt) is directly related to the phase of the motion. The phase represents the position of the particle in its oscillation cycle, and the value of i*sin(wt) at a given time determines the phase angle. This means that the value of i*sin(wt) changes as the particle moves through different phases, providing a way to track the position of the particle in its oscillation.

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