How Do You Calculate Oscillation Frequency for a Quadratic-Cubic Potential?

In summary, the equilibrium points for a particle moving in 1D in a potential of the form $$U=Ax^2+Bx^4$$ are obtained by setting the derivative of U equal to 0. If A is positive, the equilibrium point is 0 and the frequency of small oscillations is ##\sqrt{2A/m}##. If A is negative, the equilibrium points (stable equilibrium) are ##\pm\sqrt{-A/2B}##, and the frequency of oscillations is ##\sqrt{-4A/m}##, with a factor of 2 appearing in the frequency calculation due to the presence of a factor of 2 in the potential function. However, it is possible that there
  • #1
Silviu
624
11

Homework Statement


A particle moves in 1D in a potential of the form $$U=Ax^2+Bx^4$$ where A can be either positive or negative. Find the equilibrium points and the frequency of small oscillations.

Homework Equations

The Attempt at a Solution


So the equilibrium points are obtained by setting the derivative of U equal to 0. If A is positive the equilibrium point is 0 and the frequency is ##\sqrt{2A/m}##. If A is negative the equilibrium points (stable equilibrium) are ##\pm\sqrt{-A/2B}##. Now in the book, they say that the frequency of the oscillations here is equal to ##\sqrt{-A/m}## but I don't get this answer. This is what I did: $$\ddot{x}=-2Ax-4Bx^3$$ I will call the equilibrium point of the case when A is negative, simply ##x## and the small deviation ##a## so the equation becomes: $$\ddot{x}+\ddot{a}=-2A(x+a)-4B(x+a)^3$$ $$\ddot{x}+\ddot{a}=-2Ax-2Aa-4Bx^3(1+a/x)^3$$ $$\ddot{x}+\ddot{a}=-2Ax-2Aa-4Bx^3(1+3a/x)$$ and using the fact that x is the equilibrium point, we get: $$\ddot{a}=-2Aa-12aBx^2$$ and we know that here ##x=\pm\sqrt{-A/2B}## so we get $$\ddot{a}=-2Aa-12aB(-A/2B)$$ $$\ddot{a}=4Aa$$ so the osciallation frequency would be ##\sqrt{-4A/m}## (I ignored the mass in the equations). What am I doing wrong, why do I get that 4 there? Thank you!
 
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  • #2
I agree with your answers; perhaps there is a mistake in the book.
(Strictly, what you have calculated is the angular frequency ω (rad/s). The frequency of oscillation (cycles/s) is ω/2π.)
 
  • #3
Hi,

So you replace ##U(x)## by ##U(a) = 2A a^2##

Who says you do anything wrong ? The factor 2 appearing in ##U(a)## makes it logical you get a factor ##\sqrt 2## in the frequency.

SO I suspect an error in the book answer...

The red parabolas (this is for (##A=-2,\ \ B= 1##) at ##\pm 1## are sharper than the green (for ##A=+2,\ \ B= 0##) at 0

upload_2018-8-2_15-2-48.png
 

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Related to How Do You Calculate Oscillation Frequency for a Quadratic-Cubic Potential?

What is frequency of oscillations?

Frequency of oscillations refers to the number of cycles or repetitions of a periodic motion per unit of time. It is measured in Hertz (Hz).

What factors affect the frequency of oscillations?

The frequency of oscillations is affected by the amplitude, mass, and stiffness of the oscillating object. It is also affected by external factors such as friction and air resistance.

How is frequency of oscillations related to period?

The frequency of oscillations and period are inversely proportional. This means that as frequency increases, the period decreases, and vice versa. The relationship can be represented by the equation: f = 1/T, where f is frequency and T is period.

Why is frequency of oscillations important in science and engineering?

Frequency of oscillations is important because it helps us understand and analyze various phenomena such as sound, light, and mechanical vibrations. It is also crucial in designing and optimizing systems and structures that involve periodic motion, such as bridges and musical instruments.

How is frequency of oscillations measured?

Frequency of oscillations can be measured using various methods such as a stopwatch, oscilloscope, or frequency counter. The method used will depend on the type of oscillation being measured and the accuracy required.

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