Particle Simple harmonic motion particle problem

AI Thread Summary
For a particle in simple harmonic motion, the relationship between maximum speed (vmax) and average speed (vaverage) during one cycle is established as vmax = (pi/2) vaverage. To derive this, one must first calculate the total distance traveled in one period, which is a function of the amplitude (A). The average speed is then found by dividing this total distance by the period (T). It's important to note that average velocity is zero, which does not contribute to this calculation. Understanding the distinction between average speed and average velocity is crucial for solving the problem correctly.
<3Science
Messages
3
Reaction score
0

Homework Statement



For a particle in simple harmonic motion, show that vmax = (pi/2) vaverage, where vaverage is the average speed during one cycle of the motion.

Homework Equations



v(t) = -vmaxsin(wt + phi)
vmax = 2piA/ T = wA

The Attempt at a Solution



I'm really not sure at all how to start this, but I am guessing there really wouldn't be a numerical value just theory based?
 
Physics news on Phys.org


<3Science said:
I'm really not sure at all how to start this, but I am guessing there really wouldn't be a numerical value just theory based?
Normally I would insist that show more of your efforts first, but this one gave me a little trouble at the beginning too.

To start with, determine the average speed over one period. The only tricky part is figuring out what "average speed" really means. The average velocity is zero, so that certainly doesn't help. It's also not the root mean square (rms) velocity. Don't go that way.

To find the average speed, first determine the total distance the object travels in one period (not to be confused with displacement). It's a simple function of the amplitude A in,
x = Acos(ωt + Φ). But don't over-think it. How many As are traversed during a single period? Once you have the total distance, divide that by the period T. That's the average speed.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top