One end of a piece of elastic is attached to a point at the top of a door frame and the other end hangs freely. A small ball is attached to the free end of the elastic. When the ball is hanging freely it is pulled down a small distance and then released,so that the ball oscillates up and down on the elastic. The depth $d$ centimetres of the ball from the top of the door frame after $t$ seconds is given by $$d=100+10\cos(500t^\circ)$$
Find:
- the greatest and least depths of the ball,
- the time at which the ball first reaches its highest position,
- the time taken for a complete oscillation,
- the proportion of the time during a complete oscillation for which the depth of the ball is less than $99$ centimetres.
Answer for (1):
I did a graph and found the maximum and minimum points: $110$ and $90$
Answer for (2):
$90=100+10\cos(500t)$
$90-100=10\cos(500t)$
$-10=10\cos(500t)$
$-1=\cos(500t)$
$180=500t$
$\frac{180}{500}=t$
$t=0.36s$
Answer for (3):
$0.36 \times 2 = 0.72s$
I need help with answer (4) please!!
https://www.desmos.com/calculator/bqsaqjqym3
This will help you understand what I will now explain.
The mean line for $d = 100 + 10\cos {500t^{\circ}}$ is $d=100$.
The mean line for $d = 10\cos {500t^{\circ}}$ is $d=0$.
Now, since we are told to find the proportion of time, degree or radian does not matter, nor does the value in front of $\cos$ or the number with which $t$ is multiplied. So we consider the base case: $d=\cos t$ and the mean line is still $d=0$.
$99$ is $1$ less than $100$. So we need to find the coordinates of the intersections of $d=100 + 10\cos {500t^{\circ}}$ and $d=99$ in one complete oscillation $(0.72s)$. For $d=\cos t$, where $t$ is in radians, the line required is $d= -\frac 1{10}$.
$d=d$
$\cos t= -\frac 1{10}$
$t= \arccos (-\frac 1{10})$ and $t= 2\pi-\arccos (-\frac 1{10})$
So of a possible $2\pi$ seconds, the ball is below $-0.01$ (counter-intuitive, I know) for $2\pi-\arccos (-\frac 1{10})-\arccos (-\frac 1{10})=2\pi-2\arccos (-\frac 1{10})$.
This means that the proportion of time for when the depth of the ball is below $99$ centimeters is $\frac {2\pi-2\arccos (-\frac 1{10})}{2\pi}=1-\frac {\arccos (-\frac 1{10})}{\pi} \approx 0.4681157196$ which in percentage is approximately $46.81157196$%.
Of course you could do it without simplification and in degrees but working in radians is simpler for me. At the end, the proportion can be calculated using either method ($\frac 12=\frac 24$) so the answer should be correct unless I have made any mistake. Hope my complicated style helped.
P.S. $\arccos x=\cos^{-1} x$ if you did not know that.