Initial conditions in simple harmonic motion

In summary: Indeed. I was thinking about the more general SHO equation ##\ddot x(t)=-\omega^2 x(t)~##that does not necessarily involve a spring-mass system.
  • #1
Homework Statement
A particle of ##100 g## moves with simple harmonic motion. When the particle is ##8 cm## away from its position of equilibrium its velocity is ##45 \frac{m}{s}## and its acceleration is ##300 \frac{m}{s^2}## opposing the motion. Find the amplitude and frequency of oscilation.
Relevant Equations
##A=\sqrt (x_0^2 + \frac{\dot x_0 ^2}{\omega^2})##
I have the formula for amplitude ##A=\sqrt (x_0^2 + \frac{\dot x_0 ^2}{\omega^2})##.
But ##x_0## and ##\dot x_0## refers to the initial conditions, and the information that I'm given is not related to the initial conditions, or at least I'm not told so.
 
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  • #2
You can treat the numbers you were given as initial conditions. Initial conditions doesn't mean when the motion started, it means at the start of the period you are analysing.
 
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  • #3
Also, your formula is a relationship between amplitude and frequency if you know the position and speed. Since you have two unknowns, you are going to need one more relation in order to solve for both.
 
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  • #4
Like Tony Stark said:
Homework Statement: A particle of ##100 g## moves with simple harmonic motion. When the particle is ##8 cm## away from its position of equilibrium its velocity is ##45 \frac{m}{s}## and its acceleration is ##300 \frac{m}{s^2}## opposing the motion. Find the amplitude and frequency of oscilation.
Homework Equations: ##A=\sqrt (x_0^2 + \frac{\dot x_0 ^2}{\omega^2})##

I have the formula for amplitude ##A=\sqrt (x_0^2 + \frac{\dot x_0 ^2}{\omega^2})##.
But ##x_0## and ##\dot x_0## refers to the initial conditions, and the information that I'm given is not related to the initial conditions, or at least I'm not told so.

In what fundamental way do "initial conditions" differ from the state of the system at any other time?
 
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  • #5
Thanks
I used the formula but I realized that I didn't have ##\omega##. So I used the equations for SHM with ##t=0##
##8=A.cos(\phi)##
##45=-A\omega sin(\phi)##
##-300=-A \omega ^2 cos (\phi)##.

So I can find ##\phi##, ##\omega## and ##A##, but I get a negative value for ##\omega## and the frequency. Is that possible?
 
  • #6
Like Tony Stark said:
Thanks
I used the formula but I realized that I didn't have ##\omega##. So I used the equations for SHM with ##t=0##
##8=A.cos(\phi)##
##45=-A\omega sin(\phi)##
##-300=-A \omega ^2 cos (\phi)##.

So I can find ##\phi##, ##\omega## and ##A##, but I get a negative value for ##\omega## and the frequency. Is that possible?

I'm not sure how you got that. The first and third equations ought to give youn##\omega^2##.
 
  • #7
Like Tony Stark said:
Thanks
I used the formula but I realized that I didn't have ##\omega##. So I used the equations for SHM with ##t=0##
##8=A.cos(\phi)##
##45=-A\omega sin(\phi)##
##-300=-A \omega ^2 cos (\phi)##.

So I can find ##\phi##, ##\omega## and ##A##, but I get a negative value for ##\omega## and the frequency. Is that possible?
Careful with the units!
 
  • #8
Like Tony Stark said:
Thanks
I used the formula but I realized that I didn't have ##\omega##. So I used the equations for SHM with ##t=0##
##8=A.cos(\phi)##
##45=-A\omega sin(\phi)##
##-300=-A \omega ^2 cos (\phi)##.

So I can find ##\phi##, ##\omega## and ##A##, but I get a negative value for ##\omega## and the frequency. Is that possible?
You don't need all that to find ##\omega##. In simple harmonic motion the acceleration is always proportional to the displacement from equilibrium, the constant of proportionality being ##-\omega^2##. The minus sign is there because the acceleration vector is always in the opposite direction as the displacement vector; just strip the negative sign to get a relation between magnitudes. Here, you are given displacement and acceleration so ...
 
  • #9
kuruman said:
You don't need all that to find ##\omega##. In simple harmonic motion the acceleration is always proportional to the displacement from equilibrium, the constant of proportionality being ##-\omega^2##. The minus sign is there because the acceleration vector is always in the opposite direction as the displacement vector; just strip the negative sign to get a relation between magnitudes. Here, you are given displacement and acceleration so ...
Do you mean ##-k\Delta x=m.a##?
 
  • #10
Like Tony Stark said:
Do you mean ##-k\Delta x=m.a##?
No. I mean what I said, ##a(t)=-\omega^2x(t)##. The second derivative of a sine or cosine is the negative of itself and the chain rule applied twice brings the square of the frequency up front. This works with the general solution ##x(t)=A\sin(\omega t)+B\cos(\omega t)##. Try it.
 
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  • #11
kuruman said:
No. I mean what I said, ##a(t)=-\omega^2x(t)##. The second derivative of a sine or cosine is the negative of itself and the chain rule applied twice brings the square of the frequency up front. This works with the general solution ##x(t)=A\sin(\omega t)+B\cos(\omega t)##. Try it.

You can get that also from ##F = ma = -kx##, with ##\omega^2 = k/m##.
 
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  • #12
PeroK said:
You can get that also from ##F = ma = -kx##, with ##\omega^2 = k/m##.
Indeed. I was thinking about the more general SHO equation ##\ddot x(t)=-\omega^2 x(t)~##that does not necessarily involve a spring-mass system.
 

1. What are initial conditions in simple harmonic motion?

Initial conditions in simple harmonic motion refer to the starting position, velocity, and acceleration of an object that is undergoing periodic motion. These initial conditions determine the shape and amplitude of the resulting oscillation.

2. How do initial conditions affect simple harmonic motion?

The initial conditions, specifically the starting position and velocity, determine the amplitude and phase of the oscillation in simple harmonic motion. The acceleration is determined by the net force acting on the object, which is typically a restoring force that is proportional to the displacement from the equilibrium position.

3. What is the role of initial conditions in determining the period of simple harmonic motion?

The period of simple harmonic motion is determined by the mass and the force constant of the system. The initial conditions, however, can affect the time it takes for the object to complete one full oscillation or cycle. The amplitude of the oscillation also affects the period, with larger amplitudes resulting in longer periods.

4. Can initial conditions change during simple harmonic motion?

In an ideal system, initial conditions will not change during simple harmonic motion. However, in real-world systems, factors such as air resistance and friction can cause changes in initial conditions over time. These changes can result in a decrease in amplitude and an increase in the period of the oscillation.

5. How can initial conditions be manipulated in simple harmonic motion?

The initial conditions in simple harmonic motion can be manipulated by changing the starting position, velocity, or acceleration of the object. This can be achieved by applying external forces or by altering the properties of the system, such as the mass or force constant. By manipulating the initial conditions, the resulting motion can be altered to fit specific needs or desired outcomes.

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