1. ## Particle

$\displaystyle x = 2 cos(3t+\frac{\pi}{6}\))$

When does the particle first come to rest after t=0
Thanks Nath

2. Originally Posted by nath_quam
$\displaystyle x = 2 cos(3t+\frac{\pi}{6}\))$

When does the particle first come to rest after t=0
Thanks Nath
At rest, velocity will be zero.
$\displaystyle velocity=\frac{dx}{dt}$
$\displaystyle \frac{dx}{dt}=-6sin(3t + \frac{\pi}{6})$
Since,$\displaystyle sin2\pi =0$
$\displaystyle 3t + \frac{\pi}{6}=2\pi$
$\displaystyle t=\frac{11\pi}{18}$

Keep Smiling
Malay

3. ## Just Quickly

Could someone please check the above as when i graphed the velocity graph i got the first point at approx 0.95 could someone show me how to prove this exact?

Thanks Nath

4. Originally Posted by nath_quam
Could someone please check the above as when i graphed the velocity graph i got the first point at approx 0.95 could someone show me how to prove this exact?

Thanks Nath
Malay's solution of 11pi/18 is approximately 1.92 so I'd say you made an error somewhere. Malay's method looks fine to me.

Edit: Now that I look at it, I wonder if Malay should've set it equal to pi since that's the next time it equals 0.

Dave

5. Originally Posted by malaygoel
At rest, velocity will be zero.
$\displaystyle velocity=\frac{dx}{dt}$
$\displaystyle \frac{dx}{dt}=-6sin(3t + \frac{\pi}{6})$
Since,$\displaystyle sin2\pi =0$
$\displaystyle 3t + \frac{\pi}{6}=2\pi$
$\displaystyle t=\frac{11\pi}{18}$

Keep Smiling
Malay

By using $\displaystyle sin2\pi =0$
doesn't that find the second point
can't we use $\displaystyle sin\pi = 0$
and finish with $\displaystyle t=\frac{5\pi}{18}$

Thanks Nath

6. You're correct Nath, it first comes to rest(assuming it's moving along the positive x-axis) at $\displaystyle t=\frac{5{\pi}}{18}$

Left of the y-axis, it first comes to rest at $\displaystyle t=\frac{-\pi}{18}$

Add or subtract multiples of $\displaystyle \frac{\pi}{3}$ to arrive at any subsequent extrema. The corresponding y values are -2 or 2.

7. Originally Posted by dmoran
Malay's solution of 11pi/18 is approximately 1.92 so I'd say you made an error somewhere. Malay's method looks fine to me.

Edit: Now that I look at it, I wonder if Malay should've set it equal to pi since that's the next time it equals 0.

Dave
It was a silly mistake... Sorry
I always run into problems when dealing with trigonometric values, I am more comfortable with variables.

Keep Smiling
Malay

8. Thanks for your help everyone makes mistakes

9. Originally Posted by malaygoel
It was a silly mistake... Sorry
I always run into problems when dealing with trigonometric values, I am more comfortable with variables.

Keep Smiling
Malay
Not a problem. Happens to all of us!

Dave

10. Originally Posted by nath_quam
Thanks for your help everyone makes mistakes
I am going to have my old signature:
"An expert is one who had made all the possible mistakes in the narrowest field"
---Niels Bohr

Keep Smiling
Malay

11. It's easy to overlook the little things. Anyone who has done math knows that.

I do it all the time.