Assume that simple harmonic motion of a spring is described by the equation S=4 cos πt/ 2 where s is in centimetres and t is in seconds. When during the time 0<=t<=8 is the spring passing through the origin?
I attempted this question but the graph doesn't pass through the origin.
you are correct.
However, S=0 when t=1,3,5,7
That may have been what they intended to ask.
you are right,
S = 4 cos (πt/2) does not pass through the origin.
Perhaps you meant, where does it cross the t - axis?
in that case 4cos(πt/2) = 0
we know cos(π/2) = 0
so πt/2 = π/2
t = 1 or t = -1
we also know that the period of your function is
2π/(π/2) = 4
so adding or subtracting 4 from any answer will yield another solution
1+4 = 5,
-1+4 = -3
1 - 4 = -3
So it will cross again at 5 , 9 , -3 , 3 etc
Also because of the symmetry of the function about the S axis, the times it crosses the t-axis for your given intervals are:
-5, -3, -1, 1, 3, and 5
confirmation:
http://www.wolframalpha.com/input/?i=y+%3D+4cos%28%CF%80t%2F2%29+from+-8+to+8
oops, forgot the 7, thanks Steve
To find when the spring is passing through the origin, we need to set the equation to zero and solve for t.
The equation given is S = 4 cos(πt/2)
Setting S to zero: 0 = 4 cos(πt/2)
Now, we'll solve for t by finding the angle whose cosine is zero.
cos(πt/2) = 0
We know that the cosine function is zero at odd multiples of π/2. So, we can set the argument of cos to any odd multiple of π/2.
Let's start with t = π/2.
cos(π/2) = 0
This is not a solution, as t = π/2 does not lie within the given range of 0 <= t <= 8.
Next, let's try t = 3π/2.
cos(3π/2) = 0
This is a valid solution, as t = 3π/2 lies within the given range.
Therefore, when t = 3π/2, the spring passes through the origin.
Note that this is the only solution within the given range 0 <= t <= 8. The graph of the equation S = 4 cos(πt/2) does not pass through the origin.