A weight is attached to an elastic spring that is suspended from a ceiling. If the weight s pulled 1 inch below its rest position and released its displacement in inches after t seconds is given by x(t)=(5pt+p/3. Find the first two times for which the displacement s 1.5 inches.
PLEASE HELP !
looking back at your earlier attempts at typing this question I am going to guess you meant
x(t)= 2 cos (5πt+π/3)
you want this to be 1.5
2cos (5πt + π/3) = 1.5
cos (5πt + π/3) = 1.5/2 = 3/4
I know cos( .722734..) = 3/4
so 5πt + π/3 = .722734
5πt = -.3244633
t = -.020655975
now the period of 2cos(5πt + π/3) is 2π/(5π) = 2/5 or .4
so adding .4 to any answer will produce a new answer.
so t = -.20655.. + .4 = .379344
t = .379344 + .4 = .77934..
check:
let's take t = .77934...
x(t) = 2 cos(5π(.77934 + π/3)
= 2 cos 13.289..
= 1.5
YEAHHHHH
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To find the first two times for which the displacement is 1.5 inches, you need to set up the equation x(t) = 1.5 and solve for t. Let's go through the steps to solve this problem:
Step 1: Start with the given equation x(t) = 5pt + p/3.
Step 2: Set x(t) equal to 1.5 to get the equation 1.5 = 5pt + p/3.
Step 3: Simplify the equation by multiplying both sides by 3 to eliminate the fraction: 4.5 = 15pt + p.
Step 4: Rearrange the equation to isolate t: 15pt = 4.5 - p.
Step 5: Divide both sides of the equation by 15p to solve for t: t = (4.5 - p) / (15p).
Now that we have the equation for t, we can plug in different values for p to find the corresponding t values when the displacement is 1.5 inches.
For the first time when the displacement is 1.5 inches, let's set p = 1:
t = (4.5 - 1) / (15 * 1) = 0.3 seconds.
For the second time when the displacement is 1.5 inches, let's set p = 2:
t = (4.5 - 2) / (15 * 2) = 0.2 seconds.
Therefore, the first two times for which the displacement is 1.5 inches are t = 0.3 seconds and t = 0.2 seconds.