Solving SHM: Displacement at t=1/80

[hav0c]

Homework Statement

A particle is subjected to two simple harmonic motions given by
x₁=2[sin(100πt)]
x₂=2[sin(120πt+{π/3})] where x is in centimetre and time is in seconds.
Find the displacement of the particle at t=1/80

Homework Equations

The Attempt at a Solution

The answer given in my book is found by simply putting the value of time in both equations and adding the displacements..
But I am confused. How can we add the displacements even though the angular frequencies of both SHM's are different?

 

[rude man]

Homework Statement

A particle is subjected to two simple harmonic motions given by
x₁=2[sin(100πt)]
x₂=2[sin(120πt+{π/3})] where x is in centimetre and time is in seconds.
Find the displacement of the particle at t=1/80

Homework Equations

The Attempt at a Solution

The answer given in my book is found by simply putting the value of time in both equations and adding the displacements..
But I am confused. How can we add the displacements even though the angular frequencies of both SHM's are different?

Why would you NOT be able to add them? The frequencies or anytrhing else about the two motions have nothing to do with the validity of adding them.

 

[hav0c]

(d²x₁)/dt²=-ω₁²x₁
(d²x₂)/dt²=-ω₂²x₂
Isn't the property of addition of the SHM's based on the addition of forces? If ω is same we can directly add the displacements.(taking -ω² common from the two equations) But how will i prove that the addition of SHM's in case of different ωs will result in direct addition of displacements?

 

[rude man]

(d²x₁)/dt²=-ω₁²x₁
(d²x₂)/dt²=-ω₂²x₂
Isn't the property of addition of the SHM's based on the addition of forces? If ω is same we can directly add the displacements.(taking -ω² common from the two equations) But how will i prove that the addition of SHM's in case of different ωs will result in direct addition of displacements?

You are not solving the differential equation for shm. You are not adding forces. You are merely adding two displacements.

If you for example had a mass-spring system with given applied forces at two different frequencies, you would have to solve the shm equation with two separate forcing functions. You would merely solve the equation for each forcing function by itself and then add the results. This is the concept of superposition and is valid if the system is linear & time-invariant.

 

[hav0c]

Thanks a lot. Can you please guide me to a rigorous proof of the concept of superposition? The result doesn't seem readily apparent to me. Also that is the significance of the system being linear?

 

[rude man]

You should look up 'linear' and 'superposition'.

I'll give you an example of what would have been the case had your system not been linear. Assume if you put x in the system you get kx^2 out instead of kx. kx is linear, kx^2 is not.

So for your x1 you'd get k x1^2 by itself, and for your x2 you'd get k x2^2 by itself. But they wouldn't add: the output would not be k x1^2 + k x2^2 but k(x1 + x2)^2. You can see the two expressions are not the same.

 

[Chestermiller]

This is simply a poorly posed problem. They don't tell you why the displacement is the linear sum of two shm displacements. They just say that it is. The overall combined motion is obviously not a simple harmonic motion. If the particle is exhibiting this motion in a system that behaves linearly, it cannot be the only mass involved in the system. There must be another mass and at least another spring. Let me guess: this problem is from a math course, and not from a physics course.

Chet