Question

A single degree-of-freedom spring–mass–damper system has viscous damping ratio $\zeta = 0.1$. The mass has initial displacement of $10$ cm without velocity. After exactly two complete cycles of damped oscillation, find amplitude of displacement (in cm, round off to two decimals).

Hint:

For damped vibrations, amplitude decays exponentially with cycles. The logarithmic decrement $\delta = \dfrac{2\pi \zeta}{\sqrt{1-\zeta^2}}$ is useful for quick calculations.

Answer 1

Step 1: Damped oscillation formula.
Amplitude decays as: \[ x(t) = X_0 e^{-\zeta \omega_n t} \cos(\omega_d t) \] where $\omega_d = \omega_n\sqrt{1-\zeta^2}$.

Step 2: Damped period.
\[ T_d = \frac{2\pi}{\omega_d} = \frac{2\pi}{\omega_n\sqrt{1-\zeta^2}} \]

Step 3: Time after 2 cycles.
\[ t = 2T_d = \frac{4\pi}{\omega_n\sqrt{1-\zeta^2}} \]

Step 4: Amplitude after $t$.
\[ X(t) = X_0 e^{-\zeta \omega_n t} \] \[ = X_0 \exp\left(-\zeta \omega_n \cdot \frac{4\pi}{\omega_n\sqrt{1-\zeta^2}}\right) \] \[ = X_0 \exp\left(-\frac{4\pi \zeta}{\sqrt{1-\zeta^2}}\right) \]

Step 5: Substitute values.
$X_0 = 10 \,\text{cm}, \, \zeta=0.1$ \[ \frac{4\pi \zeta}{\sqrt{1-\zeta^2}} = \frac{4\pi(0.1)}{\sqrt{1-0.01}} = \frac{1.257}{0.995} \approx 1.263 \] So, \[ X = 10 e^{-1.263} = 10 \times 0.819 = 8.19 \,\text{cm} \] \[ \boxed{8.19 \,\text{cm}} \]