1. Maximum 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

2. Originally Posted by nath_quam
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
for x to be maximum
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.

3. Originally Posted by nath_quam
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
Hello,

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

${dx\over dt}={3\over 8}\cos\left({\pi\over 8}\cdot t \right)$

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

${d^2x\over dt^2}={3\over 64}\left(-\sin\left({\pi\over 8}\cdot t \right) \right)$

dx/dt = 0 if $\left({\pi\over 8}\cdot t \right)={{2k+1}\over 2}\cdot \pi$

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

4. Originally Posted by earboth
Hello,

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

${dx\over dt}={3\over 8}\cos\left({\pi\over 8}\cdot t \right)$

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

${d^2x\over dt^2}={3\over 64}\left(-\sin\left({\pi\over 8}\cdot t \right) \right)$

dx/dt = 0 if $\left({\pi\over 8}\cdot t \right)={{2k+1}\over 2}\cdot \pi$

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
will t=12 also not correspond to maximum displacement?

5. Originally Posted by malaygoel
will t=12 also not correspond to maximum displacement?
Hello,

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

6. 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.

7. 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

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maximum and minimum displacement

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