# Maths help

• Jun 6th 2006, 02:02 PM
mathsisfun
Maths help
Hi,

Please can you help me with this.

An oscillating mechanism has a maximum displacement of 3.2 m and a frequency of 50 Hz. At time t = 0 the displacement is 150 cm. Express the displacement in the general form A sin (ωt ± φ).

Newton’s law of cooling is given by θ = θo e-kt , where the excess of temperature at zero time is θo oC and at time t seconds is θ oC . Given that θo = 15 oC and
k = - 0.02, find the time when the rate of change of temperature is 1 oC.

• Jun 6th 2006, 02:10 PM
ThePerfectHacker
Quote:

Originally Posted by mathsisfun
Hi,
Newton’s law of cooling is given by θ = θo e-kt ,

If by that you mean,
$\displaystyle \theta=\theta_0e^{-kt}$
Then you are wrong.
• Jun 7th 2006, 04:22 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
If by that you mean,
$\displaystyle \theta=\theta_0e^{-kt}$
Then you are wrong.

Actually, it is correct if the temperature of the surrounding environment is 0 C.

Regardless, let $\displaystyle \theta = \theta _0 e^{-kt}$

Thus $\displaystyle \frac{d \theta}{dt}=\theta _0 (-k) e^{-kt}$

Solving for t we get:
$\displaystyle t = -\frac{1}{k} ln \left ( -\frac{1}{k \theta _0}\frac{d \theta}{dt} \right )$

Thus t = 60.1986 s.

-Dan
• Jun 7th 2006, 09:36 AM
mathsisfun
thankyou topsquark.

For the other one above iam stuck on, I have attached a picture as it didnt come out well.

Cheers
• Jun 12th 2006, 04:49 AM
topsquark
Quote:

Originally Posted by mathsisfun
thankyou topsquark.

For the other one above iam stuck on, I have attached a picture as it didnt come out well.

Cheers

Sorry it's taken me so long to get to this!

Let's do this. Since you have x(t)=150 cm, let's convert the maximum displacement to cm as well. So A = 3.2 m = 320 cm.

We know that the frequency is f = 50 Hz. We need to convert this to an angular frequency: $\displaystyle \omega = 2 \pi f$ is about 314 rad/s. (This unit is occasionally also given the lable "Hz." Be warned!)

To find the phase angle ($\displaystyle \phi$) we can now plug in some values: (dropping the units for convenience)
$\displaystyle x(t)=A \, sin( \omega t + \phi)$

$\displaystyle 150 = 320 \, sin( 314*0 + \phi) = 320 \, sin(\phi)$

$\displaystyle sin( \phi ) = \frac{150}{320} = 0.46875$

One solution is $\displaystyle \phi \approx 0.488$ rad.

So $\displaystyle x(t) = 320 \, sin( 314t+0.488)$ cm.

-Dan
• Jun 13th 2006, 02:10 AM
mathsisfun
many thanks again Dan.

Cheers