Kinematics - trajectory formula

[quas]

Homework Statement

given :

i need to find the trajectory formula

Homework Equations

i'm not sure if to use :

The Attempt at a Solution

[/B]
I tried different options with the trigonometric identities that I have written before:

thanks

 

[PeroK]

The Attempt at a Solution

[/B]

If $x$ and $y$ are as above, can you not find a simple relationship? You've already quoted the relevant trig identity somewhere.

 

[quas]

If $x$ and $y$ are as above, can you not find a simple relationship? You've already quoted the relevant trig identity somewhere.

do you mean?

 

[PeroK]

do you mean?

No, I meant:

$\frac{y}{a} = \cos(2 \omega t) =$ something to do with $\sin(\omega t) =$ something to do with $\frac{x}{a}$

You're terribly over-complicating this.

 

[quas]

No, I meant:

$\frac{y}{a} = \cos(2 \omega t) =$ something to do with $\sin(\omega t) =$ something to do with $\frac{x}{a}$

You're terribly over-complicating this.

Ok I can write $\frac{y}{a} =1-2sin^2(\omega t)$, also to $\frac{x^2}{a^2}\ = sin^2(\omega t)$ and then insert $sin^2(\omega t)$
to the equation $\frac{y}{a} =1-2sin^2(\omega t)$ . would it be correct?

 

[PeroK]

Ok I can write $\frac{y}{a} =1-2sin^2(\omega t)$, also to $\frac{x^2}{a^2}\ = sin^2(\omega t)$ and then insert $sin^2(\omega t)$
to the equation $\frac{y}{a} =1-2sin^2(\omega t)$ . would it be correct?

You seem to have a mental block about replacing $\sin(\omega t)$ by $\frac{x}{a}$.

 

[quas]

You seem to have a mental block about replacing $\sin(\omega t)$ by $\frac{x}{a}$.

first of all thanks for your help it's not taken for granted
Last try:
in the end I will get : $\frac{2x^2}{a^2}+ \frac{y}{a}=1$ then : $y= - \frac{2}{a}x^2+a$ and that's parabola figure . right ?

 

[PeroK]

first of all thanks for your help it's not taken for granted
Last try:
in the end I will get : $\frac{2x^2}{a^2}+ \frac{y}{a}=1$ then : $y= - \frac{2}{a}x^2+a$ and that's parabola figure . right ?

It is!