Vtyjkyuk

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 $$d=100+10cos(500t^circ)$$

Find:

1. the greatest and least depths of the ball,


2. the time at which the ball first reaches its highest position,


3. the time taken for a complete oscillation,


4. 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):

$90=100+10cos(500t)$

$90-100=10cos(500t)$

$-10=10cos(500t)$

$-1=cos(500t)$

$180=500t$

$frac180500=t$

$t=0.36s$

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 $d = 100 + 10cos 500t^circ$ is $d=100$.

The mean line for $d = 10cos 500t^circ$ 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 $d=100 + 10cos 500t^circ$ and $d=99$ in one complete oscillation $(0.72s)$. For $d=cos t$, where $t$ is in radians, the line required is $d= -frac 110$.

$d=d$

$cos t= -frac 110$

$t= arccos (-frac 110)$ and $t= 2pi-arccos (-frac 110)$

So of a possible $2pi$ seconds, the ball is below $-0.01$ (counter-intuitive, I know) for $2pi-arccos (-frac 110)-arccos (-frac 110)=2pi-2arccos (-frac 110)$.

This means that the proportion of time for when the depth of the ball is below $99$ centimeters is $frac 2pi-2arccos (-frac 110)2pi=1-frac arccos (-frac 110)pi approx 0.4681157196$ 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.