Problem 15 - Differential Equations and Linear Algebra

Question

Consider the general solution for an underdamped spring-mass system. (a) Show that the time between successive maxima (or minima) of \(y(t)\) is $$T=\frac{2 \pi}{\mu}=\frac{4 \pi m}{\sqrt{4 k m-c^{2}}}$$ (b) Show that if \(\frac{c^{2}}{4 k m}<<1\), then $$T=2 \pi \sqrt{\frac{m}{k}}$$ Is this result reasonable? Explain.

Step-by-Step Solution

Answer

The time between successive maxima (or minima) of an underdamped spring-mass system is \[ T=\frac{2 \pi}{\mu}=\frac{4 \pi m}{\sqrt{4 k m-c^{2}}} \] For the specific condition \(\frac{c^{2}}{4 k m}<<1\), the time period can be approximated as \[ T=2 \pi \sqrt{\frac{m}{k}} \] This result is reasonable because it shows that when the damping effects are negligible, the time period of the system approaches that of an undamped system.

Step 1: Part (a) - Time period derivation

The general solution for an underdamped spring-mass system can be written as: \[y(t) = e^{-\frac{c}{2m} t}(A \cos{(\mu t)} + B \sin{(\mu t)})\] where \(A\) and \(B\) are constants, \(\mu = \frac{\sqrt{4km - c^2}}{2m}\), and the exponential term represents the damping effect. To find the successive maxima (or minima) of \(y(t)\), we need to find the time when the velocity is zero. To do this, differentiate \(y(t)\) with respect to time: \[y'(t) = \frac{d y(t)}{dt}\] After differentiating, we get: \[y'(t) = e^{-\frac{c}{2m} t}[-\frac{c}{2m}(A \cos{(\mu t)} + B \sin{(\mu t)}) + \mu(-A \sin{(\mu t)} + B \cos{(\mu t)})]\] Now, we need to find the time \(t_n\) when \(y'(t_n) = 0\). To simplify the equation, we can divide both sides by \(e^{-\frac{c}{2m} t}\) as it is never zero. Thus, we obtain: \[-\frac{c}{2m}(A \cos{(\mu t_n)} + B \sin{(\mu t_n)}) + \mu(-A \sin{(\mu t_n)} + B \cos{(\mu t_n)}) = 0\] Let's solve for the time between successive maxima (or minima), i.e., \(T = t_{n+1} - t_n\). We have: \[\mu (t_{n+1} - t_n) = 2\pi\] Next, we can solve for \(T\): \[T = \frac{2\pi}{\mu} = \frac{4\pi m}{\sqrt{4km - c^2}}\] This is the required expression for the time period \(T\).

Step 2: Part (b) - Time period for specific condition

Now, let's consider the case \(\frac{c^2}{4km} \ll 1\). Since this term is very small, we can approximate the square root in the denominator as follows: \[\sqrt{4km - c^2} \approx \sqrt{4km}\] Then, we can write the expression for the time period \(T\) as: \[T = \frac{4\pi m}{\sqrt{4km}}\] \[T = 2\pi \sqrt{\frac{m}{k}}\] Now, let's address if this result is reasonable: The value of \(T\) we derived is the time period for an undamped spring-mass system. It means that when the damping constant \(c\) is very small, the system would behave almost like an undamped system, with a time period of \(2\pi \sqrt{\frac{m}{k}}\). This result is reasonable because it agrees with the fact that when damping effects are negligible (\(\frac{c^2}{4km} \ll 1\)), the time period of the system should approach that of an undamped system.

Key Concepts

General Solution

An underdamped spring-mass system can be expressed using a general solution that characterizes the oscillations of the system. The key form of this solution is:\[y(t) = e^{-\frac{c}{2m} t}(A \cos{(\mu t)} + B \sin{(\mu t)})\]Here's what each term represents:

• The factor \(e^{-\frac{c}{2m} t}\) captures the damping effect, which gradually reduces the amplitude of the oscillation over time.
• Constants \(A\) and \(B\) are determined by initial conditions and define the oscillation's starting characteristics.
• \(\mu = \frac{\sqrt{4km - c^2}}{2m}\) is related to the system's frequency and determines how fast the system oscillates.

This expression is pivotal because it defines how the system behaves over time under the influence of damping.

Time Period Derivation

In underdamped systems, determining the time period between maxima (or minima) is a critical element. To find this, we analyze when the velocity, \(y'(t)\), is zero. This occurs at the points of maxima or minima of the oscillation. From the provided general solution, the derivative \(y'(t)\) guides us in finding these points. Through calculation, by setting \(y'(t) = 0\), we derive:\[\mu (t_{n+1} - t_n) = 2\pi\]Solving this equation for \(T\), the period between successive maxima, we obtain:\[T = \frac{2\pi}{\mu} = \frac{4\pi m}{\sqrt{4km - c^2}}\]This formula shows how the time period is influenced by mass \(m\), stiffness \(k\), and damping coefficient \(c\).

Damping Effect

The damping effect in a spring-mass system is represented by the term \(e^{-\frac{c}{2m} t}\) in the general solution. This exponential function models how the oscillations decay over time. Let's break down its components:

• **\(c\)**: This is the damping constant, a measure of how significant the damping is. Larger \(c\) values lead to faster decay of the oscillations.
• **\(m\)**: The mass of the object. It affects how much the system is impacted by damping.

The overall influence of damping is to reduce the amplitude of oscillations, causing the system to settle into equilibrium more quickly than it would in an undamped scenario. When the damping is weak (small \(c\)), the system retains the ability to oscillate visibly for longer periods.

Successive Maxima

To understand successive maxima in the context of an underdamped system, consider what happens over time. Maxima (and minima) refer to the highest (and lowest) points reached by the system in one complete oscillation cycle. The timing of these points is crucial for predicting the system's behavior. To determine these points, we look for the values of \(t\) where \(y'(t) = 0\), which involves solving:\[-\frac{c}{2m}(A \cos{(\mu t_n)} + B \sin{(\mu t_n)}) + \mu(-A \sin{(\mu t_n)} + B \cos{(\mu t_n)}) = 0\]When you solve this, you find that the time interval between successive maxima, \(T\), is consistent, as derived previously:\[T = \frac{2\pi}{\mu}\]. This understanding helps in analyzing how quickly the peaks in the oscillation are occurring and thus gives a complete view of the system’s behavior over time.