# A Difficult Applied Math Question

**Question**

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 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 centimeters of the ball from the top of the door frame after t seconds is given by:

d = 100 + 10cos500t

Find

a) the greatest and least depths of the ball,

b) the time at which the ball first reaches its highest position,

c) the time taken for a complete oscillation,

d) the proportion of the time during a complete oscillation for which the depth of the ball is less then 99 centimeters.

**Answer**

d = 100 + 10cos500t

For a) the greatest and least depths of the ball,

We need to maximize and minimize the function d. SInce

-1 <= cos500t <= 1 We have,

90 = 100 + 10 ( -1 ) <= d <= 100 + 10 (1) = 110,

That is,

90 <= d <= 110.

That is the ‘greatest and least depths of the ball are 90 and 110 centimetres respectively’.

b) the time at which the ball first reaches its highest position,

At the ‘hightest position’, the ‘distance is 90’ by the inequality above, so… 90 = 100 + 10cos500t, yields,

cos500t = -1 or 500t = 180 degrees = ‘pi’ radian. Hence,

t = pi’ / 500 = 0.0063 sec approximately. Note that we have to get the answer in ‘radian’ and not ‘degrees’ because thats the ‘convention’ on how the cos functions are defined. Lets not get too complicated. hehehe. Also, its 180 degrees and not 360 degrees because

at 0 degrees, the ball is at the ‘lowest position’ for the first time while

at 90 degrees, its ‘at the middle position’ at first time

and at 180 degrees, its at the ‘highest position’ for the first time

c) the time taken for a complete oscillation,

One complete oscilation is how long it takes for ‘360’ and by symmetry its the time taken 2 times the time taken for the ‘highest point’. That is about

2 * 0.0063 = 0.013 seconds

d) the proportion of the time during a complete oscillation for which the depth of the ball is less then 99 centimeters.

For d < 99.

100 + 10cos500t < 99

10cos500t < -1

cos500t < -0.1

From this point onward, a different solver proceeded brilliantly (and correctly) thus:

*10cos500t<-1 cos 500t <-1/10 1,67/500<t<(2pi-1.67)/500 *

*total time= (2pi-3.34)/500 and proportion = (2pi-3.34)/2pi= *

*46.84%*