Question

Consider the initial value problem \(m y^{\prime \prime}+k y=20 \cos 8 \pi t, y(0)=0, y^{\prime}(0)=0\), modeling the response of a spring-mass system, initially at rest, to an applied force; assume that the unit of force is the newton. Suppose the motion shown in the figure is recorded and can be described mathematically by the formula \(y(t)=0.1 \sin (\pi t) \sin (7 \pi t) \mathrm{m}\). What are the values of mass \(m\) and spring constant \(k\) for this system? [Hint: Recall the identity \(\cos (\alpha \pm \beta)=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta .\).]

Short answer

Question: Find the mass and spring constant of a spring-mass system with an initial value problem given by: $$ my^{\prime\prime} + ky = 20 \cos 8\pi t \\ y(0)=0; y^{\prime}(0) = 0 $$ With motion described by the formula: $$ y(t) = 0.1 \sin(\pi t) \sin(7\pi t) $$ Answer: The mass, \(m\), for this system is approximately \(0.589\) kg and the spring constant, \(k\), is approximately \(5.792\) N/m.

Step-by-step solution

Rewrite the equation

We will first rewrite the given initial value problem in a more useful form. Dividing both sides of the equation by \(m\), we have: $$ y^{\prime \prime} + \frac{k}{m} y = \frac{20}{m} \cos 8 \pi t $$

Solve the homogeneous equation

Let's first find the homogeneous solution, which is the solution of the equation when the forcing term is zero. The homogeneous equation will be: $$ y^{\prime\prime} + \frac{k}{m} y = 0 $$ This is a second-order linear homogeneous differential equation with constant coefficients, which has the general solution: $$ y_h(t) = C_1 \cos\left(\sqrt{\frac{k}{m}} t\right) + C_2 \sin\left(\sqrt{\frac{k}{m}} t\right) $$

Find the particular solution

Now let's find a particular solution for the non-homogeneous equation. We can find a particular solution using the given hint: Recall the trigonometric identity \(\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta\). Assume that a particular solution has the form: $$ y_p(t) = A\cos(8\pi t) $$ Where \(A\) is a constant to be determined. Now, we need to determine its first and second derivatives: $$ y_p'(t) = -A(8\pi)\sin(8\pi t) $$ $$ y_p''(t) = -A(8\pi)^2\cos(8\pi t) $$ Plugging this into the non-homogeneous equation, we have: $$ -A(8\pi)^2\cos(8\pi t) + \frac{k}{m} A\cos(8\pi t) = \frac{20}{m} \cos 8 \pi t $$ Dividing both sides by \(\cos(8\pi t)\) and solving for \(A\), we obtain: $$ A = \frac{20}{m\left((8\pi)^2 - \frac{k}{m}\right)} $$ Thus, the particular solution is: $$ y_p(t) = \frac{20}{m\left((8\pi)^2 - \frac{k}{m}\right)} \cos(8\pi t) $$

Determine the general solution

The general solution can be obtained by combining the homogeneous and particular solutions: $$ y(t) = y_h(t) + y_p(t) = C_1 \cos\left(\sqrt{\frac{k}{m}} t\right) + C_2 \sin\left(\sqrt{\frac{k}{m}} t\right) + \frac{20}{m\left((8\pi)^2 - \frac{k}{m}\right)} \cos(8\pi t) $$

Apply the initial conditions

Now let's apply the initial conditions to find the coefficients \(C_1\) and \(C_2\): For \(y(0) = 0\): $$ 0 = C_1 + \frac{20}{m\left((8\pi)^2 - \frac{k}{m}\right)} $$ Thus, $$ C_1 = -\frac{20}{m\left((8\pi)^2 - \frac{k}{m}\right)} $$ For \(y'(0) = 0\): $$ 0 = -C_2 \sqrt{\frac{k}{m}} (8\pi) $$ Since \(\sqrt{\frac{k}{m}} \neq 0\), we get \(C_2 = 0\). Now the general solution becomes: $$ y(t) = -\frac{20}{m\left((8\pi)^2 - \frac{k}{m}\right)} \cos\left( \sqrt{\frac{k}{m}} t \right) + \frac{20}{m\left((8\pi)^2 - \frac{k}{m}\right)} \cos(8\pi t) $$ The given motion formula is: $$ y(t) = 0.1 \sin(\pi t) \sin(7\pi t) $$

Compare motion formula to general solution

Comparing the motion formula to our general solution, we can see that the terms with \(\pi t\) and \(7\pi t\) in the sine functions must match. Therefore, we have the following relationships: 1. \(\sqrt{\frac{k}{m}} = \pi\) 2. \(\frac{20}{m\left((8\pi)^2 - \frac{k}{m}\right)} = 0.1\) Solving the first equation for \(k\), we get \(k = \pi^2 m\).

Solve for mass and spring constant

Substituting the relationship from Step 6 into the second equation, we have: $$ \frac{20}{m\left((8\pi)^2 - \pi^2 m\right)} = 0.1 $$ Now, we can solve this equation for \(m\): $$ m = \frac{20}{0.1\left((8\pi)^2 - \pi^2 m\right)} $$ $$ m((8\pi)^2 - \pi^2 m) = 200 $$ $$ (64\pi^2 - \pi^2 m) m = 200 $$ Solving the quadratic equation, we get two solutions for \(m\): 1. \(m \approx 0.589\) kg 2. \(m \approx \frac{200}{(8\pi)^2} \approx 0.1003\) kg We will choose the positive solution for \(m\), which is \(\approx 0.589\) kg. Now, we can find \(k\) using the relationship \(k = \pi^2 m\): $$ k = \pi^2 (0.589) \approx 5.792 \, \text{N/m} $$ The mass, \(m\), for this system is approximately \(0.589\) kg and the spring constant, \(k\), is approximately \(5.792\) N/m.

Key concepts

Spring-Mass System

A spring-mass system is a classic physics problem that consists of a mass attached to a spring. The system can move in response to external forces or to the spring's attempt to return to its rest position. When the mass is displaced from its equilibrium position, it oscillates around that position due to the restoring force of the spring.

This behavior can be described by Hooke's Law, which states that the force exerted by the spring is proportional to the displacement, directed opposite to the displacement, and is given by the formula: \( F = -kx \). Here, \( k \) represents the spring constant, and \( x \) is the displacement.

If a mass is set into motion, with no other external forces (like friction or air resistance) acting on it, the motion will be a simple harmonic motion, which is periodic and oscillatory. This type of system is often modeled by a second-order linear homogeneous differential equation, as seen in the given exercise.

Second-Order Linear Homogeneous Differential Equation

A second-order linear homogeneous differential equation is an equation that involves the second derivative of a function and can be written in the general form \( ay'' + by' + cy = 0 \), where \( a \) , \( b \) , and \( c \) are constants, and the function \( y(t) \) is the solution we seek.

The term 'homogeneous' implies that the equation is equal to zero when the system is free from external forces. Equations of this kind are pivotal in modeling mechanical systems like the spring-mass system, electrical circuits, and other dynamic systems that can be described using second-order differential equations.

In our exercise, the homogeneous part of the equation \( y'' + \frac{k}{m} y = 0 \) determines the natural response of the system, or how the system behaves due to its internal properties, like mass and spring stiffness, without any external forcing.

Particular Solution

In the context of differential equations, the particular solution is a specific solution to a non-homogeneous differential equation, which includes terms that represent external forces or inputs. This solution by itself satisfies the entire non-homogeneous equation.

In our spring-mass example, we assume a particular solution of the form \( y_p(t) = A \cos(8\pi t) \) to account for the externally applied oscillating force. This assumed form helps in simplifying the process of finding the correct particular solution that will complement the homogeneous solution and thus, provide the general solution to the system's behavior.

After determining the particular solution, it can be combined with the homogeneous solution to form the general solution to the initial value problem. This general solution accounts for both the natural response of the system and the response due to external forces, effectively modeling the entire behavior of the spring-mass system under the given conditions.