Particle

**nath_quam** · July 10th 2006, 03:58 AM

$<br /> x = 2 cos(3t+\frac{\pi}{6}\))<br />$

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

**malaygoel** · July 10th 2006, 05:45 AM

Originally Posted by nath_quam

$<br /> x = 2 cos(3t+\frac{\pi}{6}\))<br />$

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

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

Keep Smiling
Malay

**nath_quam** · July 10th 2006, 02:57 PM

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

**dmoran** · July 10th 2006, 03:12 PM

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

**nath_quam** · July 10th 2006, 03:28 PM

Originally Posted by malaygoel

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

Keep Smiling
Malay

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

Thanks Nath

**galactus** · July 10th 2006, 04:07 PM

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

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

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

**malaygoel** · July 10th 2006, 05:55 PM

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

**nath_quam** · July 10th 2006, 05:57 PM

Thanks for your help everyone makes mistakes

**dmoran** · July 10th 2006, 05:57 PM

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

**malaygoel** · July 10th 2006, 06:00 PM

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

**galactus** · July 11th 2006, 04:01 AM

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

I do it all the time.

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