Question:

For a dynamical system governed by the equation
\[ \ddot{x}(t) + 2 \zeta \omega_n \dot{x}(t) + \omega_n^2 x(t) = 0, \] the damping ratio is given as \( \zeta = \frac{1}{2\pi} \log_e 2 \). A displacement peak in the positive direction is measured as 4 mm. Neglecting higher powers (\(>1\)) of damping ratio, the displacement at the next peak (positive direction) will be ________________ mm (integer).

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For lightly damped systems, the logarithmic decrement simplifies using
\( \delta \approx 2\pi \zeta \), which makes peak-to-peak amplitude calculations straightforward.
Updated On: Dec 1, 2025
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Correct Answer: 1.95

Solution and Explanation

For an underdamped system undergoing free vibration, the ratio of successive peak amplitudes is given by the logarithmic decrement formula:
\[ \delta = \ln\left(\frac{x_1}{x_2}\right) \] and for small damping,
\[ \delta \approx 2\pi \zeta. \] Given: \[ \zeta = \frac{1}{2\pi}\log_e 2 \] So,
\[ \delta = 2\pi \zeta = 2\pi \left( \frac{1}{2\pi}\log_e 2 \right) = \log_e 2. \] Thus, \[ \frac{x_1}{x_2} = e^{\delta} = e^{\log_e 2} = 2. \] Given first peak: \[ x_1 = 4\ \text{mm} \] So the next peak is: \[ x_2 = \frac{x_1}{2} = \frac{4}{2} = 2\ \text{mm}. \] Therefore, the displacement at the next peak in the positive direction is \(2\) mm.
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