Question

Consider the initial value problem \(m y^{\prime \prime}+\gamma y^{\prime}+k y=\bar{F} \cos \sqrt{k / m} t, y(0)=0\), \(y^{\prime}(0)=0 .\) If we set \(\gamma / m=2 \delta, \omega_{0}^{2}=k / m\), and \(\bar{F} / m=F\), we obtain initial value prob\(\operatorname{lem}(11 \mathrm{a})\). Assume that \(\omega_{0}^{2}>2 \delta\). Note that the radian frequency of the applied force is \(\omega_{0}\); this is the resonant radian frequency of the corresponding undamped system. (a) Derive equation (11b), showing that the solution of this initial value problem is $$ y(t)=\frac{F}{2 \delta}\left[\frac{\sin \left(\omega_{0} t\right)}{\omega_{0}}-\frac{e^{-\delta t} \sin \left(\sqrt{\omega_{0}^{2}-\delta^{2}} t\right)}{\sqrt{\omega_{0}^{2}-\delta^{2}}}\right] $$ (b) Show, for any fixed values \(t>0\) and \(\omega_{0}>0\), that $$ \lim _{\delta \rightarrow 0^{+}}\left\\{\frac{F}{2 \delta}\left[\frac{\sin \left(\omega_{0} t\right)}{\omega_{0}}-\frac{e^{-\delta t} \sin \left(\sqrt{\omega_{0}^{2}-\delta^{2}} t\right)}{\sqrt{\omega_{0}^{2}-\delta^{2}}}\right]\right\\}=\frac{F}{2 \omega_{0}} t \sin \left(\omega_{0} t\right) $$ This limit is the response of the undamped spring-mass system to resonant frequency excitation. (c) Suppose that we know the values of mass \(m\) and spring constant \(k\) (and \(\bar{F}\), the amplitude of the applied force). Explain how we might use our knowledge of the solution in part (a) (observed over a long time interval) to estimate the damping constant \(\delta\).

Short answer

Question: Compute the limit of the response function as the damping constant approaches 0. Answer: The limit of the response function as the damping constant approaches 0 is \(\frac{F}{2\omega_0} t \sin(\omega_0 t)\).

Step-by-step solution

Rewrite the given equation

Substitute the given expressions for \(\gamma/m\), \(\omega_0^2\), and \(\bar{F}/m\) into the initial value problem equation: $m y^{\prime\prime}+2\delta m y^{\prime}+\omega_0^2 m y = F \cos(\omega_0 t), y(0) = 0, y^{\prime}(0)=0$ Divide the equation by \(m\): \(y^{\prime\prime}+2\delta y^{\prime}+\omega_0^2 y = F \cos(\omega_0 t)\)

Find the general solution for the given differential equation

Observe that this differential equation has a homogeneous part and a non-homogeneous part. The homogeneous part is: \(y^{\prime\prime}+2\delta y^{\prime}+\omega_0^2 y = 0\) The general solution is \(y_h(t) = c_1 e^{-\delta t} \cos(\sqrt{\omega_0^2 - \delta^2} t)+c_2 e^{-\delta t} \sin(\sqrt{\omega_0^2 - \delta^2} t)\) Now for the non-homogeneous part, we can assume a particular solution of the form: \(y_p(t) = A \cos(\omega_0 t) + B \sin(\omega_0 t)\) Differentiate \(y_p(t)\) and \(y_p'(t)\) with respect to \(t\): \(y_p'(t) = -A\omega_0 \sin(\omega_0 t) + B\omega_0 \cos(\omega_0 t)\) \(y_p''(t) = -A\omega_0^2 \cos(\omega_0 t) - B\omega_0^2 \sin(\omega_0 t)\) Substitute \(y_p(t)\), \(y_p'(t)\), and \(y_p''(t)\) into the non-homogeneous part of the differential equation: \(-A\omega_0^2 \cos(\omega_0 t) - B\omega_0^2 \sin(\omega_0 t) + 2\delta (-A\omega_0 \sin(\omega_0 t) + B\omega_0 \cos(\omega_0 t))+\omega_0^2 (A \cos(\omega_0 t) + B \sin(\omega_0 t)) = F \cos(\omega_0 t)\) Comparing the coefficients, we find: \(A = \frac{F}{2\delta}\) \(B = 0\) Therefore, the general solution is: \(y(t) = y_h(t) + y_p(t) = c_1 e^{-\delta t} \cos(\sqrt{\omega_0^2 - \delta^2} t)+c_2 e^{-\delta t} \sin(\sqrt{\omega_0^2 - \delta^2} t) + \frac{F}{2\delta} \cos(\omega_0 t)\)

Find the particular solution using the initial conditions

Using the initial conditions \(y(0) = 0\) and \(y^{\prime}(0) = 0\), we have: \(0 = c_1 + \frac{F}{2\delta}\) \(c_1 = -\frac{F}{2\delta}\) \(0 = -\delta c_1 + \sqrt{\omega_0^2 - \delta^2} c_2\) Plug in \(c_1\) into the equation above and solve for \(c_2\): \(c_2 = \frac{F}{2\delta} \cdot \frac{\delta}{\sqrt{\omega_0^2 - \delta^2}}\) Thus the particular solution is: \(y(t) = \frac{F}{2\delta}\left[\frac{\sin(\omega_0 t)}{\omega_0} - \frac{e^{-\delta t}\sin(\sqrt{\omega_0^2 - \delta^2} t)}{\sqrt{\omega_0^2 - \delta^2}}\right]\) (Equation 11b)

Find the limit for the solution as the damping constant approaches 0

Now we need to compute the limit: \(\lim_{\delta \rightarrow 0^+} \left\{\frac{F}{2\delta}\left[\frac{\sin(\omega_0 t)}{\omega_0} - \frac{e^{-\delta t}\sin(\sqrt{\omega_0^2 - \delta^2} t)}{\sqrt{\omega_0^2 - \delta^2}}\right]\right\}\) As \(\delta \rightarrow 0\), \(\sqrt{\omega_0^2 - \delta^2} \rightarrow \omega_0\) and \(e^{-\delta t} \rightarrow 1\). Therefore, we have: \(\lim_{\delta \rightarrow 0^+} \left\{\frac{F}{2\delta}\left[\frac{\sin(\omega_0 t)}{\omega_0} - \frac{\sin(\omega_0 t)}{\omega_0}\right]\right\} = \frac{F}{2\omega_0} t \sin(\omega_0 t)\) This limit represents the response of the undamped spring-mass system to resonant frequency excitation.

Discuss a method to estimate the damping constant

To estimate the damping constant given the observed solution \(y(t)\) in part (a) over a long time interval, we can follow these steps: 1. Inspect the observed data to identify the amplitude of oscillations after several periods. 2. Determine the observed damped frequency by taking the ratio of the time interval between two successive zero crossing points and the oscillation period. 3. Use the observed damped frequency and amplitude of oscillations to estimate the damping constant, \(\delta\), by comparing them with the computed solution from part (a). 4. Tune the theoretical model using the estimated damping constant to obtain better agreement with the observed data, if necessary.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation that comes with accompanying conditions that specify the state of the system at a starting point. For example, the condition might dictate the initial position and velocity of a particle. These conditions are crucial because they allow us to determine a unique solution to the differential equation. In our spring-mass system problem, we start with two initial conditions: when time (\(t\)) is zero, the displacement (\(y\)) and its derivative (velocity, \(y'\)) are both zero.

• The initial value problem concerns both a differential equation and specific initial conditions.
• This problem provides a specially tailored solution that fits the scenario, starting from a defined point in time.

Understanding and setting the initial value accurately is critical for modeling physical situations, such as objects at rest before they start moving.

Spring-Mass System

A Spring-Mass System is a classic example in physics used to model motion dynamics. It consists of a mass attached to a spring, which moves back and forth when displaced. The basic idea here is that the spring applies a force proportional to its displacement from the equilibrium position based on Hooke's Law. This force causes the mass to accelerate back towards its original position, potentially leading to oscillations.

• A simple spring-mass system oscillates repeatedly, exhibiting harmonic motion unless affected by external forces.
• Such systems can be described by second-order differential equations.

The behavior of this system under different conditions aids in understanding concepts like resonance and damping.

Damping Constant

The Damping Constant, denoted here as \(\delta\), represents how quickly the oscillations in a system decay over time. It measures the resistance the system experiences as it moves, such as friction-due resistance. The greater the damping constant, the faster the oscillations diminish.

• It affects how energy is lost in the system, causing the amplitude of oscillations to decrease.
• Without damping (\(\delta = 0\)), the system would continue oscillating forever.

Understanding the damping constant is crucial for predicting how systems behave in realistic scenarios, where resistance and energy loss cannot be ignored.

Resonant Frequency

The Resonant Frequency is a critical parameter in oscillating systems. It is the natural frequency at which a system tends to oscillate in the absence of any driving or damping force. When an external frequency matches the resonant frequency, even a small force can produce large amplitude oscillations because the system is moving in harmony with its natural rhythm.

• Resonant frequency is vital because it helps in designing systems to avoid catastrophic failure due to resonance.
• It is calculated based on factors like the stiffness and mass of the system.

In our exercise, resonant frequency is key because it represents the frequency of the applied force, leading to significant system responses when matched.