Problem 7 - Advanced Engineering Mathematics

Question

Answer Problems \(1-8\) without referring back to the text. Fill in the blank or answer true/false. The differential equation describing the motion of a mass attached to a spring is \(x^{\prime \prime}+16 x=0\). If the mass is released at \(t=0\) from 1 meter above the equilibrium position with a downward velocity of \(3 \mathrm{~m} / \mathrm{s}\), the amplitude of vibrations is _________ meters.

Step-by-Step Solution

Answer

The amplitude of vibrations is \( \frac{5}{4} \) meters.

Step 1: Identify the Type of Differential Equation

The given differential equation is \( x'' + 16x = 0 \). This is a second-order linear homogeneous differential equation with constant coefficients.

Step 2: Determine the General Solution

For a second-order linear homogeneous differential equation of the form \( x'' + ext{constant} \times x = 0 \), we can use the characteristic equation \( r^2 + 16 = 0 \) to find the general solution. Solving for \(r\), we get \( r = ±4i \). Thus, the general solution is \( x(t) = C_1 \cos(4t) + C_2 \sin(4t) \).

Step 3: Apply Initial Conditions

We know that at \( t = 0 \), \( x(0) = 1 \) and \( x'(0) = -3 \). First, plug these conditions into the general solution:For \( x(0) = 1 \), \( C_1 \cdot 1 + C_2 \cdot 0 = 1 \), hence, \( C_1 = 1 \).For \( x'(t) = -C_1 \, 4 \sin(4t) + C_2 \, 4 \cos(4t) \) and \( x'(0) = -3 \), \(-C_1 \, 4 \cdot 0 + C_2 \, 4 \cdot 1 = -3 \), therefore, \( C_2 = -\frac{3}{4} \).

Step 4: Determine the Amplitude

The amplitude of a sinusoidal function \( C_1 \cos(4t) + C_2 \sin(4t) \) is given by \( \sqrt{C_1^2 + C_2^2} \). Substitute the values \( C_1 = 1 \) and \( C_2 = -\frac{3}{4} \) to find the amplitude:\[ \text{Amplitude} = \sqrt{1^2 + \left(-\frac{3}{4}\right)^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \].

Key Concepts

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is a mathematical expression that involves derivatives of a function and equals zero. The term 'linear' suggests that each term can be added up or subtracted from one another, and 'homogeneous' indicates all the terms require this summation to be zero. In the context of differential equations, "linear" refers to the dependent variable (often represented as \(x\) or \(y\)) and its derivatives having only constant multiples.

The equation \(x'' + 16x = 0\) is a perfect example of a second-order linear homogeneous differential equation with constant coefficients. The "second-order" specifies the highest derivative, which here, is \(x''\) (or the second derivative of \(x\)).

The process of solving such equations often involves using a characteristic equation, in this case, \( r^2 + 16 = 0 \), to determine an expression for the dependent variable. The resulting general solution, \(x(t) = C_1 \cos(4t) + C_2 \sin(4t)\), is based on solving for \(r\) and includes constants \(C_1\) and \(C_2\), which are determined from initial or boundary conditions.

Initial Conditions

Initial conditions are crucial in the process of solving differential equations. They provide specific values for the function and its derivatives at a certain point, typically at the beginning of the observation period (\(t = 0\)). This information helps to determine the specific solution to the differential equation out of the potentially infinite number of solutions to the general differential equation.

In our exercise, the mass-spring system's initial conditions are given by: the mass is released from 1 meter above the equilibrium (\(x(0) = 1\)) and has a downward velocity of \(3 \text{ m/s}\) at time \(t = 0\) (\(x'(0) = -3\)).

• \(x(0) = 1\) applies the condition that at \(t = 0\), the position \(x\) equals 1 meter.
• \(x'(0) = -3\): the velocity, represented as the derivative of \(x\) with respect to time, is -3 m/s, indicating the initial movement is downwards.

These initial conditions are substituted into the general solution \(x(t) = C_1 \cos(4t) + C_2 \sin(4t)\) to find values for the constants \(C_1\) and \(C_2\). By substituting and calculating, we find \(C_1 = 1\) and \(C_2 = -\frac{3}{4}\).

Amplitude of Vibrations

The amplitude of vibrations in a differential equation related to motion describes how far the motion extends from the equilibrium position. In sinusoidal functions derived from differential equations, the amplitude measures the peak value of the sine or cosine functions. For the function \(x(t) = C_1 \cos(4t) + C_2 \sin(4t)\), the amplitude can be calculated using the formula \(\sqrt{C_1^2 + C_2^2}\).

This formula derives from the Pythagorean theorem, where the coefficients of \(\cos\) and \(\sin\) correspond to the two legs of a right triangle, and the amplitude represents the hypotenuse.

• Given \(C_1 = 1\) and \(C_2 = -\frac{3}{4}\), the amplitude is \(\sqrt{1^2 + \left(-\frac{3}{4}\right)^2} = \sqrt{\frac{25}{16}} = \frac{5}{4}\).

This result tells us that the maximum displacement from the equilibrium position is \(\frac{5}{4}\) meters or 1.25 meters. Understanding the amplitude is essential for analyzing how a system vibrates, helping in efficiently designing and controlling mechanical systems like springs and oscillators.