7107
/www/nrich/html/content/id/7107/
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2011-02-01T00:00:01
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A particle moves in a straight line under the action of a unknown
force and its motion is observed in various situations.<br></br>
<br></br>
The position $x$ of the particle is plotted against its velocity
$v$ for the motion in the following 6 cases.<br></br>
<br></br>
In each case, what can you say about the motion of the
particle?<br></br>
<br></br>
In which way does each travel along the curve?<br></br>
<br></br>
Can you think of a plausible physical interpretation for
each?<br></br>
<br></br>
What can you deduce about points corresponding to equal time
intervals of the motion?<br></br>
<br></br>
<mdo:image width="648" height="426" src="xvgraphs.png" alt=""></mdo:image><br></br>
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Next consider the velocity-position diagrams for the following
situations:<br></br>
<br></br>
1. A rubber ball dropped perpendicular to a hard concrete floor (if
you know about coefficients of restitution, you could draw this
motion accurately for $e=0.7$, neglecting all friction and air
resistance)<br></br>
2. A large parcel dropped from a stationary and very high
helicopter.<br></br>
3. An air hockey puck struck hard and perpendicular to a side on an
air hockey table.<br></br>
<br></br>
<br></br>
<span style="font-style: italic;">Extension:</span> Think more
generally about the meaning of velocity-position diagrams and the
relationship to speed-time diagrams.<br></br>
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<mdoxml version="1.0"><p><span class="editorial">We were very pleased to see Niharika from Leicester devise a method for solving the equations in certain cases. You can read her account <a href="/content/id/7107/Niharika%20-%207107.pdf">here</a>.</span></p>
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<div class="col_layout">In general, the gradient of a curve in phase space is $\frac{dv}{dx}$. </div>
<div class="col_layout">For the first example, we have $$\frac{dv}{dx} \equiv 0 $$ As $v\equiv k >0$, this graph could represent the motion of a particle moving with constant velocity in the positive $x$-direction. Suppose the particle was at the origin at time $t=t_0$, then the motion would be given by $x = k(t-t_0)$. The particle would move from the left of the curve to the right.</div>
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<div class="col_layout">In the second example, $x= x_0\sin(t)$, where $x_0$ is the maximum distance from the origin, would have the required phase space graph as $v = \frac{dx}{dt} =x_0\cos(t)$. The points $(x_0\sin(t), x_0\cos(t)$ form a circle as $t$ ranges from $0$ to $2\pi$. The particle would travel clockwise around the curve. </div>
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<div class="col_layout">The third example could represent a particle travelling with large velocity from the positive $x$-direction and being repelled by a particle of a charge with the same sign positioned at $x=-a, a>0$, a point more negative than any reached by the particle. The particle would travel along the bottom section of the graph from right to left and along the top section
from left to right. </div>
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<div class="col_layout">The fourth example could be a particle starting with a positive velocity moving with constant deceleration from x=0 to x=k and then back to x=0. At this point, the sign of the acceleration suddenly changes, so it reaches x=-k with zero velocity and then accelerates to x=0 again. We can solve the equations to find the position in terms of time. </div>
<div class="col_layout">Consider one of the four parts of the curve, suppose with $k>x,v>0$. The equation is $v= k-x$. We can write $v=\frac{dx}{dt}$ which implies $$\frac{dx}{dt} = k-x \Rightarrow t-t_0 = - \ln(k-x)$$ Inverting the formula gives $x = k - \exp(t_0 - t)$ and $v = exp(t_0 - t)$.</div>
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<div class="col_layout">Consider the next section, with $x>0, v<0$. We have $v = x-k$, with $t_1- t = -\ln(k-x) \Rightarrow x = k - \exp(t-t_1)$.</div>
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<div class="col_layout">The next section with $x<0, v<0$ we have $x+v = -k \Rightarrow \frac{dx}{dt} = -(k+x) \Rightarrow \ln(k+x) = t_2 - t \Rightarrow x = - k + \exp(t_0-t)$ and in the final section, $x = -k + \exp(t-t_3)$. $t_i$ are constants that give the correct boundary conditions. Can you draw a graph showing the approximate behaviour? I think it would look something like this,
where the red lines indicate where the sections start and end. It would repeat for ever as the path in the phase space is a closed loop!</div>
<div class="col_layout" style="text-align: center;"><mdo:image src="phase%20graph.png"></mdo:image></div>
<div class="col_layout">Graph E represents a particle moving from the negative-$x$ direction towards the origin with ever decreasing speed. It eventually reaches the origin, but takes a very long time.</div>
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<div class="col_layout">Graph F shows a particle travelling from the origin with initially extremely high velocity, and the velocity decreases exponentially as it gets further from the origin. </div>
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<div class="col_layout">In general, all curves in the top half of the plane, with $v>0$, will travel from left to right, and all those in the bottom half of the plane, with $v<0$, will travel from right to left. A path which will take unit time for a particle to travel along will vary in length according to the values of $v$ along it. For small values of $v$, the path would be longer than
it would be larger values of $v$. </div>
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<div class="col_layout">For the bouncy ball, the graph would look something like this, where the velocities $v_i$ are determined by the coefficient of restitution. </div>
<div class="col_layout"><mdo:image src="bounce2.png"></mdo:image></div>
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<p>For the parcel falling from the helicopter, the parcel would accelerate constantly until it reached its terminal velocity, at which point the velocity would remain constant until it hit the ground. </p>
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Phase space
Explore the issues surrounding the concept of Phase Space.
Mechanics
Displacement, velocity and acceleration
Pre-Calculus and Calculus
1st and 2nd order linear differential equations with constant coefficients
Mechanics
Newton's laws of motion