Sorry, I didn't get it properly, Sir. The differential equation may be framed as [itex]\frac{d\vec x}{y\hat i-x\hat j}=5dt[/itex]. Is it what you are suggesting? Please clarify.Orodruin said:
OK, we take [itex]\vec x=x\hat i+y\hat j[/itex]. Also, [itex]x[/itex] comp. of velocity, [itex]v_x=5y[/itex], and, [itex]v_y=-5x[/itex]. What may I do next? It follows that [itex]\frac{dx}{y}=5dt[/itex]. Again, [itex]\frac{dy}{x}=-5dt[/itex]. It yields, after some manipulations, [itex]x^2+y^2=constant[/itex]. That means that the trajectory is a circle in fact, centred at the origin.Orodruin said:
To stop you going round in circles, why not differentiate the equations again.PKM said:
Which eqns do you mean? Is it [itex]d\vec x/dt=5(y\hat i-x\hat j)[/itex]?PeroK said:
Yes, but it is hard to work with that because two different notations are being used for vectors. Write ##\vec x## in terms of ##\hat i## and ##\hat j##.PKM said:
Yeah, I've hit this method. Please see post #5. Is it worth what I've done?haruspex said:
Ah, I didn't read PeroK's remark with sufficient context.PKM said:
Let's see. I get [itex]\ddot x=-25\dot x[/itex], and [itex]\ddot y=-25\dot y[/itex].haruspex said:
Please show how you got there and not just the final result you got. It is impossible to identify where you have gone wrong if you don't.PKM said:
You have made a mistake somewhere. If you cannot find it, please post your working.PKM said:
PeroK said:
Does it? The circular motion looks right to me.Orodruin said:
Yes, I deleted that already.haruspex said:
Okay, I better show the way I've reached the conclusion of Post #5.Orodruin said:
Oh, I had blotched it! It should have been [itex]\ddot x=-25x[/itex], and [itex]\ddot y=-25y[/itex]. It represents SHM.haruspex said:
The parametric eqns of the circle may be [itex]x=R\cos \theta, y=R\sin \theta[/itex].Orodruin said:
Of course, [itex]\theta =\omega t[/itex]. If the particle has constant speed, then [itex]\omega[/itex] is the angular velocity. Therefore the eqn of the x-coordinate of the particle is [itex]x=R\cos (\omega t)[/itex], or simple harmonic, isn't it?Orodruin said:
You need to show that this is the case. You cannot just assume it.PKM said:
What makes you think that the particle has constant speed? (It has, but you need to argue for this!)
I find it quite simple to account for.Orodruin said:
Trajectory refers to the path or trajectory that an object follows through space or time.
To calculate the trajectory of a particle, you need to know its initial position, velocity, and acceleration. You can use equations of motion, such as the kinematic equations, to calculate the position of the particle at any given time.
The trajectory of a particle can be affected by several factors, including its initial velocity, acceleration due to external forces, and any changes in the environment, such as air resistance or gravity. Additionally, the shape and mass of the particle can also impact its trajectory.
Yes, there is a difference. In a vacuum, there is no air resistance, so the particle will follow a parabolic path. In a non-vacuum environment, the presence of air resistance will cause the particle to deviate from its original trajectory.
The trajectory of a particle can be predicted accurately if all the relevant factors, such as initial conditions and external forces, are known and accounted for in calculations. However, unpredictable factors, such as wind or turbulence, can also affect the trajectory and make it difficult to predict with absolute accuracy.