Thread: trignometry question

Can someone please help me with this question..
I attempted it and got the correct ans for the first 3 parts but for the last part my ans was wrong, and I couldn't figure out why

One end of a piece of elastic is attatched to a point at the top of a door frame and the other end hangs freely. A small ball is attatched to the free end of the elastic.When the ball is hanging freely it is pulled down a distance and then released, so that the ball oscillated 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 degrees

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 than 99 cm.

I got for
a) 110 cm and 90cm
b) 0.36s
c)0.72s

d)
99-100=10cos500t
500t=95.74
500t=264.3
t=0.191s
t=0.528
the proportion of time where the depth is less than 99 = 0.528-0.191 =0.338

But correct ans for this part is 0.00213s

Any help would be appreciated..

Hey puresoul.

For d) I got

99-100=10cos(500t) which means

-1/10 = cos(500t) which means

> acos(-1/10)
[1] 1.670964 implies that

500t = 1.670964 (approx) which means t is given by
> acos(-1/10)/500
[1] 0.003341927

Checking the answer gives us:

> x = acos(-1/10)/500
> 100 + 10*cos(500*x)
[1] 99

which is correct.

Now you try and obtain the next value of t where this is equal and get the length of the appropriate interval.

Originally Posted by chiro

Hey puresoul.

For d) I got

99-100=10cos(500t) which means

-1/10 = cos(500t) which means

> acos(-1/10)
[1] 1.670964 implies that

500t = 1.670964 (approx) which means t is given by
> acos(-1/10)/500
[1] 0.003341927

Checking the answer gives us:

> x = acos(-1/10)/500
> 100 + 10*cos(500*x)
[1] 99

which is correct.

Now you try and obtain the next value of t where this is equal and get the length of the appropriate interval.

they gave us the equation d= 100+10cos 500t degrees
in degrees not radians so i found the value of acos(-1/10)
in degrees which is 95.74 degrees or 1.670964radians
Won't we get a different ans if we used radians??



Continuing what you did,the other value of t will be..

pi-1.670964 = 1.4706287

500t=1.4706287 so t=0.0029413 s

The interval will be (0.003341927-0.0029413= 0.00040066969 s )


(((And sorry the correct ans that was given in the book is 0.468 not 0.00213s , I copied the ans of another question >< )))
Its very near the ans that I got (0.338) but still not the same,

Using R to validate everything:

> 100+10*cos(500*(acos(-1/10)/500))
[1] 99
> 100+10*cos(500*(2*pi - acos(-1/10)/500))
[1] 99
> (2*pi - acos(-1/10))/500
[1] 0.009224443
> acos(-1/10)/500
[1] 0.003341927
> a = (2*pi - acos(-1/10))/500
> b = acos(-1/10)/500
> a-b
[1] 0.005882516

Originally Posted by chiro

Using R to validate everything:

> 100+10*cos(500*(acos(-1/10)/500))
[1] 99
> 100+10*cos(500*(2*pi - acos(-1/10)/500))
[1] 99
> (2*pi - acos(-1/10))/500
[1] 0.009224443
> acos(-1/10)/500
[1] 0.003341927
> a = (2*pi - acos(-1/10))/500
> b = acos(-1/10)/500
> a-b
[1] 0.005882516

I understand what u ar doing, that's what I did but using degrees..
I still don't understand why u used radians in this situation...
And the ans. Is 0.468 not 0.00588

You need to remember that you are looking for a proportion.

I gave you the time that it was below, now find the proportion of time with respect to the time it takes for a complete oscillation.