Let's break down your questions step by step, starting with the temperature probe and then moving on to the second-order system. Both problems involve fundamental concepts in control systems and dynamics, so we'll tackle them methodically.
Temperature Probe Response
The temperature probe has a first-order response characterized by a time constant (τ) of 1 second. When a step input of 50°C is applied, we can use the first-order system response formula to find the temperature indicated after 0.6 seconds.
First-Order System Response Formula
The response of a first-order system to a step input can be described by the equation:
T(t) = T_final + (T_initial - T_final) * e^(-t/τ)
Where:
- T(t) = temperature at time t
- T_final = final steady-state temperature (50°C)
- T_initial = initial temperature (0°C)
- t = time after the input is applied (0.6 seconds)
- τ = time constant (1 second)
Calculating the Temperature
Plugging in the values:
T(0.6) = 50 + (0 - 50) * e^(-0.6/1)
Now, calculate the exponential term:
e^(-0.6) ≈ 0.5488
Substituting this back into the equation:
T(0.6) = 50 - 50 * 0.5488
T(0.6) = 50 - 27.44 = 22.56°C
So, the temperature indicated by the probe 0.6 seconds after the input is applied is approximately 22.56°C.
Natural Frequency and Damping Constant
Now, let's move on to the second part of your question regarding the linear second-order system. We have a mass (m) of 8 grams and a stiffness (k) of 1000 N/m. First, we need to calculate the natural frequency (ω_n) of the system.
Natural Frequency Calculation
The formula for the natural frequency of a mass-spring system is given by:
ω_n = √(k/m)
Before we calculate, we need to convert the mass from grams to kilograms:
m = 8 gm = 0.008 kg
Now, substituting the values into the formula:
ω_n = √(1000 / 0.008) = √125000 = 353.55 rad/s
Damping Constant for No Overshoot
To prevent overshoot in a second-order system, we need to achieve critical damping. The damping ratio (ζ) for critical damping is 1. The relationship between the damping constant (c), mass (m), and natural frequency (ω_n) is given by:
c = 2 * m * ω_n
Substituting the values we have:
c = 2 * 0.008 * 353.55 ≈ 5.65 N·s/m
Thus, the damping constant necessary to just prevent overshoot in response to a step input of force is approximately 5.65 N·s/m.
Summary
In summary, the temperature indicated by the probe after 0.6 seconds is about 22.56°C, and the damping constant required to prevent overshoot in the second-order system is approximately 5.65 N·s/m. If you have any further questions or need clarification on any of these concepts, feel free to ask!
