# Math Help - Maths help

1. ## 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.

2. Originally Posted by mathsisfun
Hi,
Newton’s law of cooling is given by θ = θo e-kt ,
If by that you mean,
$\theta=\theta_0e^{-kt}$
Then you are wrong.

3. Originally Posted by ThePerfectHacker
If by that you mean,
$\theta=\theta_0e^{-kt}$
Then you are wrong.
Actually, it is correct if the temperature of the surrounding environment is 0 C.

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

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

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

Thus t = 60.1986 s.

-Dan

4. thankyou topsquark.

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

Cheers

5. 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: $\omega = 2 \pi f$ is about 314 rad/s. (This unit is occasionally also given the lable "Hz." Be warned!)

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

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

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

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

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

-Dan

6. many thanks again Dan.

Cheers