Oscillatory motion - Car driving on bumpy road

[Quesadilla]

Homework Statement

This is an exercise on classical mechanics, filed under the section on oscillatory motion (according to the lecture notes).

A car is driven with constant speed 30 km/h along a bumpy road. The height of the road may be described as y = y(x) = H_0 \sin(kx), x>0. Now set H_0 = 0.15 m and k = 2 m^{-1}. Describe the car's vertical movement.

Homework Equations

Equation of motion for SHM :
\ddot{x} + {\omega}_0^2 x = 0 which has solution
x(t) = A \cos(\omega_0 t + \phi)
could be of relevance, I suppose.

The Attempt at a Solution

My interpretation of the question is that I should find the height y as a function of time t. I attempted to find x(t) after which y(t) would follow from the given relationship between y and x.

Using the chain rule, I got

\dot{y} = \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} = kH_0 \cos(kx) \dot{x}.

Given constant speed 30 km/h, which we could convert to 30 / 3.6 = 25/3 m/s, we can use the Pythagorean identity to get

\left(\frac{25}{3}\right)^2 = \dot{y}^2 + \dot{x}^2 which implies

\dot{x} = \frac{\frac{25}{3}}{\sqrt{1 + k^2H_0^2 \cos(kx)}}.

Separation of variables now yields

\int_0^t dt = \frac{25}{3} \int_0^x \frac{dx}{\sqrt{1 + k^2H_0^2 \cos(kx)}}.
where the latter integral is supposedly an elliptic integral of the first order, which are mentioned in passing in the lecture notes, but with which I do not have any real familiarity.

This would give t(x), so I would have to take some sort of inverse of the elliptic integral to get x(t).

I think I am taking the wrong approach to the problem, or maybe I am making some logical error somewhere in my thought process. Any comments or hints are most welcome. Many thanks in advance.

Remark: This is my first post here and therefore I am not quite sure how to write LaTeX in the posts. I tried looking in other treads and follow their example, but in the preview I only see the "code" written as plain text.

 

[cepheid]

Welcome to PF,

It's not immediately obvious to me how to solve it, but one thing that I thought I would point out is that:$$\dot{x}^2 + \dot{y}^2 = \dot{x}^2[1+k^2 H_0^2\cos^2(kx)]$$Your cosine factor should be squared, but in what you have written above, it is not.

 

[Ibix]

Likewise, I can't do the integral, but could it be you're overthinking it? The wavelength of the bumpiness is about twenty times its amplitude and the car will be crossing somewhat under three waves per second. Is it plausible that the answer they expect is just x=(25/3)t, y=H_0\sin(\omega t+\phi)?

 

[Infinitum]

\ Is it plausible that the answer they expect is just x=(25/3)t, y=H_0\sin(\omega t+\phi)?

Hi Ibix,

The expression for x seems wrong to me. The car has a constant speed of (25/3)m/s which includes both the horizontal and vertical components, and they vary...

 

[Quesadilla]

Likewise, I can't do the integral, but could it be you're overthinking it? The wavelength of the bumpiness is about twenty times its amplitude and the car will be crossing somewhat under three waves per second. Is it plausible that the answer they expect is just x=(25/3)t, y=H_0\sin(\omega t+\phi)?

Thank you for your reply.The integral cannot be expressed in terms of elementary functions, why I would have to use some conventional notation for it, e.g.
F(\theta,R):= \int_0^{\theta} \frac{d\theta'}{\sqrt{1+R^2 \cos^2(\theta')}} and then express x(t) using F^{-1}, to mean the inverse of F, in some sense.

I disregarded interpreting the 30 km/s to be meant to be in the horizontal direction since it seemed to make the question too easy, but if one can argue why it would yield a decent approximation, it could be plausible. The question actually says "velocity 30 km/s" and not "constant speed 30 km/s", if that matters. The question was not originally given in English, but I tried to translate it as precisely as I could.

Welcome to PF,

It's not immediately obvious to me how to solve it, but one thing that I thought I would point out is that:$$\dot{x}^2 + \dot{y}^2 = \dot{x}^2[1+k^2 H_0^2\cos^2(kx)]$$Your cosine factor should be squared, but in what you have written above, it is not.

Thank you for pointing this out. I had it right in my notes, but wrote it down incorrectly in the post.