On the particle displacement-time graph x=3sin(pie/8)t in units of centimetres and seconds. Find the first two times when the displacement is maximum

Thanks for any help

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- Jun 5th 2006, 11:48 PMnath_quamMaximum displacement
On the particle displacement-time graph x=3sin(pie/8)t in units of centimetres and seconds. Find the first two times when the displacement is maximum

Thanks for any help - Jun 5th 2006, 11:56 PMmalaygoelQuote:

Originally Posted by**nath_quam**

sin(pie/8)t = +1 or -1

sin(pie/8)t = +1 = sin(pie/2)

t/8= 1/2

t=4sec.

sin(pie/8)t =-1 =sin(3pie/2)

t/8 =3/2

t=12sec. - Jun 6th 2006, 12:15 AMearbothQuote:

Originally Posted by**nath_quam**

1. Derivate x(t) with respect to t:

You'll get a maximum if dx/dt = 0**and**the 2nd derivative is negative:

dx/dt = 0 if

That means t = 4*(2k+1)

You'll get a maximum if k = 1, 3, 5, ... (Use the 2nd derivative to sort these numbers out!)

So the first 2 maxima ar at t = 4 or t = 20. (See attachment)

Greetings

EB - Jun 6th 2006, 12:31 AMmalaygoelQuote:

Originally Posted by**earboth**

- Jun 6th 2006, 02:35 AMearbothQuote:

Originally Posted by**malaygoel**

you can calculate extreme values: the minimum, that's the smallest, and the maximum, that's the greatest.

So in my opinion the value you get at t = 12 is a minimum and not a maximum.

Greetings

EB - Jun 6th 2006, 02:44 AMnath_quam
Thanks for your help guys but because the time axis is classed as the origin wouldn't the first two be 4 and 12 because its the absolute distance/maximum from the origin.

- Jun 6th 2006, 04:31 AMtopsquark
The solution probably depends on if you are doing Math or Physics. It sounds like a Physics problem, so I'd say t=4, 12. If it's a Math problem I'd say it's t=4, 20. (Physics is a lot more liberal in it's terms: maximum displacement implies largest distance from the origin in any direction, though that's not what the Mathematical definition of it would be.)

-Dan