Conversation started Dec 4, 2018 at 16:12.
Dec 4, 2018 16:12
@sammygerbil A particle of mass m is executing oscillations about the origin on the x axis. Its potential energy is $U(x) = k|x|^3$. where x is a positive constant. If the ampltiude of oscillation is A then its time period T is proportional to:

a) $1/\sqrt a$

b) a^0 (independent of a)

c) $\sqrt a$

d) $a^{3/2}$
1st question
@Abcd I am trying to find a question on the main site which addressed this issue.
Maybe we can solve by reasoning. For harmonic oscillator $U=kx^2$ period is independent of amplitude.
there is a finite limit.
$x_o$
See the solution above
@sammygerbil option a is eliminated
When $a= 0, T =0$
Not infinite.
So option A gone.
Dec 4, 2018 16:19
Hmm, what's the question exactly?
@Dante i just have solution and answer key as I said
Oh, I see
@Abcd That is good thinking. But not necessarily true. ... If we have $U=k|x|^n$ then we might expect period to increase with amplitude for n > 2 and decrease with amplitude for n < 2.
@sammygerbil But option A gives infinity man thats plain wrong
@sammygerbil how to solve the problem but
@Abcd Period could tend to infinity for small amplitudes if the curve is flat at the origin.
Dec 4, 2018 16:24
@sammygerbil hmm
@Abcd From my argument above I would guess that for n=3 the period decreases as the amplitude increases. Option A!
2
A: Frequency of small oscillation of particle under gravity constrained to move in curve $y=ax^4$

Emilio PisantyThe motion in such a curve is quite hard to calculate, and even more so if you do not want to get into the messy details of Jacobi elliptic functions like $\text{sn}(u|k)$. However, for the case of small oscillations there is a simple scaling argument that lets you calculate the dependence of the...

This is not the question I was thinking of, but it shows that for n=4 the period is proportional to 1/a.
Which agrees with $1/\sqrt{a}$ for n=3.
@sammygerbil i dont know what the answer is saying , we hvent been taught all that in maths
@sammygerbil you mean its increasing in powers of $-1/4$
$x -> a^0$
$x^2 -> a^{(-1/4)}$
$x^3 -> a^{(-1/2)}$
@sammygerbil Can we not write acceleration of particle as $-3kx *\mod{x}$ and find time required to reach equilibrium from extreme using suitable integration process and then multiply it by 4 to get the answer?
@Dante why dont you try and tell if you get the answer?
@Dante \times not *
I tried it, could do it
Dec 4, 2018 16:32
@Abcd The answers say that $T \propto a^{1-n/2}$.
@sammygerbil do you remember a similar problem before and I had done that using dimensional anlsysis?
couldn't* maybe sammy knows the proper way to integrate
@Dante lol what a typo
Hehe
My friend's asking where you got the key from?
1 min ago, by Abcd
@sammygerbil do you remember a similar problem before and I had done that using dimensional anlsysis?
@Dante Its posted in my fiitjee batch group after the exam
@sammygerbil this:
Aug 7 at 4:57, by Abcd
@JohnRennie In exam, I did this question using dimensional analysis. I checked which of the 4 options has dimensions of time and got the right answer ;).
Aug 7 at 15:03, by Abcd
@sammygerbil Wow! Amazing method. Thanks
1
Q: Period $T$ of oscillation with cubic force function

Colonel Thirty TwoHow would I find the period of an oscillator with the following force equation? $$F(x)=-cx^3$$ I've already found the potential energy equation by integrating over distance: $$U(x)={cx^4 \over 4}.$$ Now I have to find a function for the period (in terms of $A$, the amplitude, $m$, and $c$), b...

Dec 4, 2018 16:36
4
Q: Non-SHM oscillatory motion

ABC How to solve these kind of questions , where $|F| \propto x^2$? How to find time period and velocity type related things to the oscillatory motion? $$m\dfrac{d^2x}{dt^2}=F=-\dfrac{dU}{dx}=-3kx|x|.$$ But after this $$m\dfrac{d^2x}{dt^2}=-3kx|x|.$$ What is general solution of this ODE? I thin...

6
A: Finding the period of an anharmonic oscillation by substituting the solution for SHM

sammy gerbilYour doubts about the solution given are justified. The method of solution seems to be invalid and misguided - but see my footnote. However, the correct answer choice is still (A). If the potential energy is $V=k|x|^3$ then (as you observe) the motion is not simple harmonic and cannot be descri...

@sammygerbil please tell how to do
@sammygerbil More questions, leave that if you are not getting it right now.
I cant give more than 5 or max 7 minutes to a question.
@Abcd There is a solution using dimensional analysis by zkf in the question Non-SHM Oscillatory Motion.
However, I think you can reason as I suggested. As n increases the curve becomes flatter at the origin, so you should expect the period to increase.
(The curve $y=x^n$ looks more and more like a "square well" for $n\to \infty$.)
@sammygerbil zkf solution takes the assumption that time period will be of form $1/\sqrt k$
 
Conversation ended Dec 4, 2018 at 16:50.