Respuesta :
The motion can be modeled as
[tex]x(t) =9 \, cos( \frac{2 \pi t}{T} )[/tex]
where
x = the position below the equilibrium position
T = the period of the motion
t = time
The weight returns to its equilibrium position when x = 0.
This occurs when
[tex] \frac{2 \pi t}{T} = \frac{ \pi }{2} [/tex]
That is,
[tex]t= \frac{T}{4} [/tex]
Because the weight returns to the equilibrium position when t = 3 s, therefore
T = 12 s
The motion is
[tex]x(t) = 9 \, cos( \frac{ \pi t}{6} )[/tex]
When t = 1 s, the position of the weight from equilibrium is
[tex]d = 9 \, cos( \frac{ \pi }{6} ) = 9( \frac{1}{2} ) = 4.5 \, in[/tex]
Answer:
[tex]d = 9 \, cos( \frac{ \pi }{6} ) = 4.5 \, in[/tex]
[tex]x(t) =9 \, cos( \frac{2 \pi t}{T} )[/tex]
where
x = the position below the equilibrium position
T = the period of the motion
t = time
The weight returns to its equilibrium position when x = 0.
This occurs when
[tex] \frac{2 \pi t}{T} = \frac{ \pi }{2} [/tex]
That is,
[tex]t= \frac{T}{4} [/tex]
Because the weight returns to the equilibrium position when t = 3 s, therefore
T = 12 s
The motion is
[tex]x(t) = 9 \, cos( \frac{ \pi t}{6} )[/tex]
When t = 1 s, the position of the weight from equilibrium is
[tex]d = 9 \, cos( \frac{ \pi }{6} ) = 9( \frac{1}{2} ) = 4.5 \, in[/tex]
Answer:
[tex]d = 9 \, cos( \frac{ \pi }{6} ) = 4.5 \, in[/tex]
