To solve this exercise we will use the expression of the displacement in the aromatic movement in the two given points. Thus we will find the value of time as a function of frequency. In the end we will replace that value and find the time taken,
The expression for the displacement of the object in simple harmonic motion is
[tex]x = Acos(2\pi ft)[/tex]
At the initial position the value of x is zero,
[tex]0 = Acost (2\pi ft_1)[/tex]
[tex]cost (2\pi ft_1) = 0[/tex]
[tex]2\pi ft_1 = \frac{\pi}{2}[/tex]
[tex]t _1 = \frac{1}{4f}[/tex]
Now the expression for the displacement of the object in simple harmonic motion is
[tex]x = Acos (2\pi ft)[/tex]
At the final position the value of x is -1.8cm
Replacing we have that
[tex]-1.8=1.8cos(2\pi ft_2)[/tex]
[tex]-1 = cos(2\pi ft_2)[/tex]
[tex]t_2 = \frac{1}{2f}[/tex]
The total change in time will be
[tex]t = t_2-t_1[/tex]
[tex]t = \frac{1}{2f}-\frac{1}{4f}[/tex]
[tex]t = \frac{1}{4f}[/tex]
Replacing the value of the frequency we have that
[tex]t = \frac{1}{4*5}[/tex]
[tex]t = 0.05s[/tex]
Therefore the time taken for the displacement in the interval of x=0 to x=-1.8 is 0.05s
