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Sagot :
The motion of the pendulum is a repetitive motion and the time at which
the displacement is maximum, can be found by differentiation.
[tex]The \ times \ in \ seconds \ are; -1\dfrac{2}{3}, \ -1, \ 0, \ \dfrac{1}{3}, \ \dfrac{2}{3}, \ 1\dfrac{1}{3} \\[/tex]
Reasons:
The given function for the angular displacement from the rest location, θ(t),
is presented as follows;
θ(t) = 15·cos(3·π·t)
Where;
t = The time of displacement of the pendulum
Required:
To find the times, t, when θ(t) is greatest
Solution:
When θ(t) is greatest, θ'(t) = 0
Therefore;
[tex]\dfrac{d}{dt} \theta(t) = \dfrac{d}{dt} \left(15 \cdot cos(3 \cdot \pi \cdot t) \right) = -15 \cdot 3 \cdot sin (3 \cdot t \cdot \pi ) = 0[/tex]
When sin(3·t·π) = 0, we have;
[tex]t = -\dfrac{2 \cdot n_1 - 1}{3}[/tex] or [tex]t = \dfrac{2 \cdot n_1 }{3}[/tex]
Where;
n₁ = An integer
Therefore, the times, t, in seconds are;
[tex]t = \dfrac{1}{3}, \ -1, \ -1\dfrac{2}{3}, ...[/tex]
[tex]t = 0, \ \dfrac{2}{3}, \ 1\dfrac{1}{3}, ...[/tex]
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