If by that you mean,Originally Posted by mathsisfun
Then you are wrong.
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.
Thanks for your help.
Sorry it's taken me so long to get to this!Originally Posted by mathsisfun
(Warning! Make sure your calculator is in "radian" mode!)
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: is about 314 rad/s. (This unit is occasionally also given the lable "Hz." Be warned!)
To find the phase angle ( ) we can now plug in some values: (dropping the units for convenience)
One solution is rad.