What is the physical significance of A and B in mass spring SHM solutions?

  • Thread starter Thread starter Glenn G
  • Start date Start date
  • Tags Tags
    Mass Shm Spring
AI Thread Summary
In the discussion on mass spring simple harmonic motion (SHM), the solutions x = Acos(wt) and x = Bsin(wt) are explored, highlighting their roles in representing oscillatory motion. A and B correspond to the initial conditions of the system, influencing the amplitude and phase of the oscillation. The general solution can be expressed as x = Ccos(wt - φ), where C represents the resultant amplitude derived from A and B. The conversation clarifies that the values of A and B are not fixed but can vary based on the system's initial position and velocity. Ultimately, these constants help define the specific behavior of the oscillator at any given time.
Glenn G
Messages
113
Reaction score
12
Hi community,
I've been looking at solutions for mass spring shm (undamped for now) ie that

x = Acoswt and x = Bcoswt work as solutions for dx2/dt2 = -(k/m)x

and that the general solution is the sum of these that with a trig identity can be written as

x = C Cos(wt - φ) where C is essentially the amplitude (and is given by √(A2 + B2)

My question is the physical significance of A and B in the two separate solutions above (before this youtube video) I've always gone for the solutions as either the Acoswt or Asinwt (with A being the amplitude) depending on where the mass is in its oscillating cycle at time t=0, i.e. would have gone with the coswt one if

x = +A at t=0.

If I let A and B both be A then my factor C (amplitude) comes out as √(2)A where I want it to represent the Amplitude A.

Would really appreciate help.
regards,
Glenn.
 
Last edited by a moderator:
Physics news on Phys.org
Glenn G said:
(before this youtube video)
Doesn't look like the video link is working...?
 
Glenn G said:
x = Acoswt and x = Bcoswt work as solutions
and is the 2nd term supposed to be sin()?
 
berkeman said:
and is the 2nd term supposed to be sin()?
Sorry yes I did mean Bsinwt for the second solution...

In fact I think it is fine, because if you state that x = 0 after quarter of a cycle (so ∏/2 radians) then you get 0 = Bx1 therefore B = 0 (in x = Acoswt + Bsinwt) so the sine term disappears from the general solution anyway and you end up with C = √A^2 = A for the coefficient which is what I wanted.

thanks,
G.
 
  • Like
Likes berkeman
Yeah, it just means that the spring doesn't necessarily have zero phase angle at t=0. It's a general way to express a sinusoid that has amplitude and phase information... :smile:
 
Note that because we start out with a 2nd order differential equation the general solution will contain two constants. They can be A and B, or A and ##\phi##, or whatever. Their physical interpretation is that they set the boundary conditions, for example the initial position and velocity of the oscillator.
 
  • Like
Likes berkeman
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top