Question:
A buoy of virtual mass 30 kg oscillates in a fluid medium as a single degree of freedom system. If the total damping in the system is set as 188.5 N-s/m, such that the oscillation just ceases to occur, then the natural period of the system is .................... s (round off to one decimal place)
A buoy of virtual mass 30 kg oscillates in a fluid medium as a single degree of freedom system. If the total damping in the system is set as 188.5 N-s/m, such that the oscillation just ceases to occur, then the natural period of the system is .................... s (round off to one decimal place)
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Look for keywords in vibration problems. "Oscillation just ceases," "fastest return to equilibrium without overshoot," or "damping ratio \(\zeta=1\)" all mean critical damping. The formula \(c_c = 2m\omega_n = 2\sqrt{km}\) is central to these problems.
Updated On: Aug 30, 2025
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Solution and Explanation
Step 1: Understanding the Concept:
The phrase "oscillation just ceases to occur" is the definition of a critically damped system. In a critically damped system, the damping coefficient (\(c\)) is equal to the critical damping coefficient (\(c_c\)). The value of \(c_c\) is related to the mass (\(m\)) and the undamped natural frequency (\(\omega_n\)) of the system. The natural period (\(T_n\)) is inversely related to the natural frequency.
Step 2: Key Formula or Approach:
1. The condition for critical damping is that the damping \(c\) is equal to the critical damping coefficient \(c_c\). 2. The formula for the critical damping coefficient is: \(c_c = 2 m \omega_n\). 3. The relationship between natural period and natural frequency is: \(T_n = \frac{2\pi}{\omega_n}\).
Step 3: Detailed Explanation or Calculation:
Given values:
Virtual mass, \(m = 30\) kg
Critical damping coefficient, \(c = c_c = 188.5\) N-s/m
1. Find the undamped natural frequency (\(\omega_n\)): Using the formula for critical damping, we can solve for \(\omega_n\): \[ \omega_n = \frac{c_c}{2m} = \frac{188.5 \text{ N-s/m}}{2 \times 30 \text{ kg}} = \frac{188.5}{60} \approx 3.14167 \text{ rad/s} \] This value is extremely close to \(\pi\). The value \(60\pi \approx 188.495\), so we can confidently assume \(\omega_n = \pi\) rad/s. 2. Find the natural period (\(T_n\)): Using the relationship between period and frequency: \[ T_n = \frac{2\pi}{\omega_n} = \frac{2\pi}{\pi} = 2 \text{ s} \] Step 4: Final Answer:
Rounding to one decimal place, the natural period of the system is 2.0 s.
Step 5: Why This is Correct:
The problem statement defines a critically damped system. By applying the formula for critical damping, we determine the system's natural frequency. The natural period is then easily calculated. The result of 2.0 s matches the provided answer range.
The phrase "oscillation just ceases to occur" is the definition of a critically damped system. In a critically damped system, the damping coefficient (\(c\)) is equal to the critical damping coefficient (\(c_c\)). The value of \(c_c\) is related to the mass (\(m\)) and the undamped natural frequency (\(\omega_n\)) of the system. The natural period (\(T_n\)) is inversely related to the natural frequency.
Step 2: Key Formula or Approach:
1. The condition for critical damping is that the damping \(c\) is equal to the critical damping coefficient \(c_c\). 2. The formula for the critical damping coefficient is: \(c_c = 2 m \omega_n\). 3. The relationship between natural period and natural frequency is: \(T_n = \frac{2\pi}{\omega_n}\).
Step 3: Detailed Explanation or Calculation:
Given values:
Virtual mass, \(m = 30\) kg
Critical damping coefficient, \(c = c_c = 188.5\) N-s/m
1. Find the undamped natural frequency (\(\omega_n\)): Using the formula for critical damping, we can solve for \(\omega_n\): \[ \omega_n = \frac{c_c}{2m} = \frac{188.5 \text{ N-s/m}}{2 \times 30 \text{ kg}} = \frac{188.5}{60} \approx 3.14167 \text{ rad/s} \] This value is extremely close to \(\pi\). The value \(60\pi \approx 188.495\), so we can confidently assume \(\omega_n = \pi\) rad/s. 2. Find the natural period (\(T_n\)): Using the relationship between period and frequency: \[ T_n = \frac{2\pi}{\omega_n} = \frac{2\pi}{\pi} = 2 \text{ s} \] Step 4: Final Answer:
Rounding to one decimal place, the natural period of the system is 2.0 s.
Step 5: Why This is Correct:
The problem statement defines a critically damped system. By applying the formula for critical damping, we determine the system's natural frequency. The natural period is then easily calculated. The result of 2.0 s matches the provided answer range.
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