# Thread: Further Maths

1. ## Further Maths

An oscillating mechanism has a maximum displacement of 3.2 meters and a frequency of 50HZ. At time T=0 the displacement is 150cm. express the displacement in the general form $A sin (Wt +or-theta)$.
(W= omega)

Can anyone help me out please?

2. Originally Posted by batman121
An oscillating mechanism has a maximum displacement of 3.2 meters and a frequency of 50HZ. At time T=0 the displacement is 150cm. express the displacement in the general form $A sin (Wt +or-theta)$.
(W= omega)

Can anyone help me out please?
$x = A \sin (\omega t + \theta)$.

"maximum displacement of 3.2 meters": A = 3.2.

"frequency of 50HZ: Period = $\frac{2 \pi}{\omega} \Rightarrow f = \frac{\omega}{2 \pi} = 50 \Rightarrow \omega = 100 \pi$.

"At time T=0 the displacement is 150cm": Solve $1.5 = 3.2 \sin (\theta)$ for $\theta$. A decimal approximation will be required.

3. $1.5 = 3.2 \sin \theta$
so $1.5/sin \theta = 3.2$

$sin \theta = 3.2/1.5$
$sin \theta = 0.46875$
$sin^-1 (0.46875) = \theta = 27.953$

That looks about right...

4. Originally Posted by batman121
$1.5 = 3.2 \sin \theta$
so $1.5/sin \theta = 3.2$

$sin \theta = 3.2/1.5$ Mr F says: 1.5/3.2

$sin \theta = 0.46875$
$sin^-1 (0.46875) = \theta = 27.953$

That looks about right...
The problem with your answer is that it's in degrees. It needs to be in radians.

5. 27.953 to radians = $27.953/57.2957= 0.48787r$