Question

A damped oscillator is released from rest with an initial displacement of 10.00 cm. At the end of the first complete oscillation, the displacement reaches 9.05 cm. When 4 more oscillations are completed, what is the displacement reached

Answer 1

The displacement is
`A_r = 6.071 \ cm`

Step-by-step explanation:

From the question we are told that

The initial displacement is
`A_o = 10 \ cm`

The displacement at the end of first oscillation is
`A_d = 9.05 \ cm`

Generally the damping constant of this damped oscillator is mathematically represented as

`\eta = (A_d)/(A_o)`

substituting values

`\eta = (9.05)/(10)`

`\eta = 0.905`

The displacement after 4 more oscillation is mathematically represented as

`A_r = \eta^4 * A_d`

substituting values

`A_r = (0.905)^4 * (9.05)`

`A_r = 6.071 \ cm`

Answer 2

Displacement reached is 6.0708 cm

Step-by-step explanation:

Formula for damping Constant "C"

`C^n=(A_2)/(A_1)` where n=1,2,3,........n

Where:

`A_2` is the displacement after first oscillation

`A_1\\` is the initial Displacement

`A_1=10\ cm\\A_2=9.05\ cm\\`

In our case, n=1.

`C=(9.05)/(10)\\C=0.905`

After 4 more oscillation, n=4:

`C^4=(A_6)/(A_2)`

Where:

`A_6` is the final Displacement after 4 more oscillations.

`A_6=(0.905)^4*(9.05)\\A_6=6.0708\ cm`

Displacement reached is 6.0708 cm