Answer:
The displacement is [tex]A_r = 6.071 \ cm[/tex]
Explanation:
From the question we are told that
The initial displacement is [tex]A_o = 10 \ cm[/tex]
The displacement at the end of first oscillation is [tex]A_d = 9.05 \ cm[/tex]
Generally the damping constant of this damped oscillator is mathematically represented as
[tex]\eta = \frac{A_d}{A_o}[/tex]
substituting values
[tex]\eta = \frac{9.05}{10}[/tex]
[tex]\eta = 0.905[/tex]
The displacement after 4 more oscillation is mathematically represented as
[tex]A_r = \eta^4 * A_d[/tex]
substituting values
[tex]A_r = (0.905)^4 * (9.05)[/tex]
[tex]A_r = 6.071 \ cm[/tex]
Answer:
Displacement reached is 6.0708 cm
Explanation:
Formula for damping Constant "C"
[tex]C^n=\frac{A_2}{A_1}[/tex] where n=1,2,3,........n
Where:
[tex]A_2[/tex] is the displacement after first oscillation
[tex]A_1\\[/tex] is the initial Displacement
[tex]A_1=10\ cm\\A_2=9.05\ cm\\[/tex]
In our case, n=1.
[tex]C=\frac{9.05}{10}\\C=0.905[/tex]
After 4 more oscillation, n=4:
[tex]C^4=\frac{A_6}{A_2}[/tex]
Where:
[tex]A_6[/tex] is the final Displacement after 4 more oscillations.
[tex]A_6=(0.905)^4*(9.05)\\A_6=6.0708\ cm[/tex]
Displacement reached is 6.0708 cm
