Suppose that a weight on a spring has initial position s(0) and period P. s(0)= 1in ;p=0.8

1 answer:

4 0

The question is to find a s function given by s(t) = a cos(ωt) that models the displacement of the weight.

Answer:
The function is given as:
s(t) = a cos(ωt)
This means that to get the exact function modelling the given situation, we will have to get values for ω and a.

1- getting the value of ω:
period = 0.8 sec
frequency = 1/period = 1/0.8 = 5/4 hz
ω = 2πf = 2*π*5/4 = 2.5π rad/sec
The function now became:
s(t) = a cos(2.5πt)

2- getting the value of a:
We are given that:
s(0) = 1 in
This means that:
s(t) = a cos(2.5πt)
1 = a cos(2.5π*0)
1 = a cos(0)
a = 1

Therefore, the function is:
s(t) = cos(2.5πt)

To get the value of the function at any instance (t), substitute with the value of "t" in the above function and get the corresponding s(t)

Hope this helps :)

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Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

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By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .