Guess the acceleration from the graph
A particle moves in 1-dimension. If we plot its velocity and displacement over time, the trajectory forms a circle that's centered at the origin.
Which of the following relations is true regarding its acceleration ( a ) , velocity ( v ) , and displacement ( x ) ?
Note: In the options, k is a positive constant.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
Differentiating to time might be easier to understand: 2 v a + 2 x v = 0 , by applying the chain rule. Hence, a = − x .
Log in to reply
What if the graph was an ellipse instead of the circle?
Log in to reply
Considering graphs equation as l 2 x 2 + m 2 y 2 = 1 then differentiating with respect to x gives a = − l 2 m 2 x
Log in to reply
@Akshat Sharda – Yes, this can also be directly seen from the simple harmonic motion where the velocity-position graph is an ellipse, which follows the equation a = − ω 2 x
@Rohit Gupta Just take the derivative w.r.t. time, and you also get a = − k x , since the factor v still drops out.
We can always adjust the unit of time to make it a circle, so it would still have the same form a = − k x .
I believe if you traverse the circle clockwise, a = -kx (k>0) is correct. If, however, you traverse the circle in a counterclockwise manner, I believe a = kx.
Log in to reply
The particle moves to a maximum x and then returns and then repeats its motion. If acceleration and displacement were in the same direction then why will it have a limitation on the maximum x coordinate?
Log in to reply
Excellent point. Thank you. I had a year of college physics in 1966-67, using a 2-volume set of textbooks by Resnick and Halladay. I enjoyed the course. I think some of it is coming back to me. Perhaps I should have kept those books.
Dfferentiating the given relationship x 2 + v 2 = C 2 to time yields 2 x v + 2 v a = 0 (keeping in mind the chain rule). Hence a = − x .
This approach looks simplest to follow and demonstrate we are dealing with oscillations.
It is characteristic of simple harmonic motion. In which force and acceleration are directly proportional to distance from mean position and always directed towards mean position.
Yes, it is a typical graph for a simple harmonic motion. It can be further generalized for an elliptical curve as well.
yeah this is the phase space of a simple harmonic oscillator
v 2 + x 2 = C 2
Differentiating with respect to x ,
2 v d x d v + 2 x = 0 a = − x