# Checking very simple problems of motion

1. Jan 19, 2008

### end3r7

Just wanting to double check the reasoning and the work for these. They are fairly simple, so I didn't use latex, but I can re-edit at any time if someone so desires (I'm online more often than I like to admit hehehe).

Thanks.

1. The problem statement, all variables and given/known data
1) A mass 'm' is rolled of a table at height 'h' with horizontal speed 'Vx'. Where does the mass land? What trajectory did the mass take?

2) If x = -cos(t) + 3*sin(t), what is the amplitude and phase?

3) Show that an equivalent expression for x = C1 * cos(wt) + C2 * sin(wt) is B*cos(wt + P). How do B and P depend on C1 and C2.

2. Relevant equations
1) F = ma
2) sin(a +- b) = sin(a)cos(b) +- cos(a)sin(b)
3) cos(a +- b) = cos(a)cos(b) -+ sin(a)sin(b)

3. The attempt at a solution

1) I solved the equations of motion for both vertical and horizontal directions.
With y(0) = h and Vy(0) = 0, I got y(t) = h - mg/2 * t^2

With x(0) = 0 and Vx(0) = Vx, I got x(t) = Vx * t

Solving for y = 0, t_land = sqrt(2/[mg]), so I said the mass will have fallen and displaced Vx * t_land away from table.

Plugging x/Vx for 't', I have that the path is y(t) = h - mg/2 * (x/Vx)^2, which is a parabola with downward concavity.

2) If we rewrite as A * sin(wt + P), then A * sin(P) = C1 and A * cos(P) = C2
Therefore A is simply sqrt(C1^2 + C^2) = sqrt(10).
and P = arctan(C1/C2) = arctan(-1/3).

3) If we rewrite as B * cos(wt + P), then B * cos(P) = C1 and - B * sin(P) = C2
Therefore B is simply sqrt(C1^2 + C^2)
and P = arctan(-C2/C1)

2. Jan 19, 2008

### end3r7

Err, I know it's not exactly challenging, but if anybody could take a look and tell me if it's right, that would be cool (especially for #1, the others I'm 100% are correct, not even sure why I posted them).

3. Jan 19, 2008

### Staff: Mentor

That "m" doesn't belong there. Resolve for t and you're good to go.

The other two problems look OK.

4. Jan 19, 2008

### end3r7

Wow, thanks a lot.

After the proper fixing, I got that the time for landing is T = sqrt(2h/g)
x will have displaced by Vx * T.

The equation, in terms of x is y(x) = h - g/2 * (x/Vx)^2, with x ranging from 0 to Vx * T. The equation describing the right half of a parabola with downward concavity.

5. Jan 19, 2008

### Staff: Mentor

Looks good.

6. Jan 19, 2008

### end3r7

I have a question though, for 3, say I use x = -cos(2t) + 3*sin(2t) and try to write it as y = B*cos(2t+P)

Then I use B is simply sqrt(10)
and P = arctan(3)

This is giving me an error though. x = -y.

Why is that?

I know it's because B is in fact -sqrt(10), but how can we determine that?

Last edited: Jan 19, 2008
7. Jan 19, 2008

### Staff: Mentor

Oops. I think you made an error in #3 before:

Redo your answer for P. (One difference between #2 & #3 is that one uses Sin(wt +P) while the other uses Cos(wt + P) )

8. Jan 19, 2008

### end3r7

I thought the P was right...

The B was wrong by a factor of -1.

This is because when we solve for B we get +or-sqrt(C1^2 + C2^2).

For the first case, we got the positive square root, but this time around we need the negtavie, I'm wondering why.

9. Jan 19, 2008

### Staff: Mentor

A change in sign is equivalent to a change in P. If you choose B as positive (amplitudes are always positive), that determines P.

10. Jan 19, 2008

### end3r7

Which step was wrong for the solving for P then? I'm a little confused.

I understand that I could as easily keep my B positive and shift the argument inside the cosine function by adding pi, but I would rather not, unless there is a clear motivation in the equations.

Last edited: Jan 19, 2008
11. Jan 19, 2008

### Staff: Mentor

I think the problem is that (as you pointed out) this method gives a choice of +/- for B. To see which fits a particular case, you must plug in the values. (Interesting!)

Note that:
$$\tan(\theta + \pi) = \tan\theta$$

But:
$$\cos(\theta + \pi) = -\cos\theta$$

12. Jan 20, 2008

### end3r7

I figured out why. When we foud P, we took the arctan which returns an angle between -pi /2 and pi/2, where cosine is never negative.

Thus if we have c1 = Bcos(P), then a negative c1 would imply that we need to take the negative square root.

13. Jan 20, 2008

### Staff: Mentor

Good. Another way of looking at it is what I tried to explain above. Arctan(x) gives an angle which is ambiguous by an additive term of $\pm\pi$ radians.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?