Question

These exercises deal with undamped vibrations of a spring-mass system, $$ m y^{\prime \prime}+k y=0, \quad y(0)=y_{0}, \quad y^{\prime}(0)=y_{0}^{\prime} . $$ Use a value of \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) or \(32 \mathrm{ft} / \mathrm{sec}^{2}\) for the acceleration due to gravity. (a) Derive an expression for the amplitude of the undamped vibrations modeled by equation (16). [Hint: From equations (10)-(12), the general solution of \(m y^{\prime \prime}+k y=0\) is \(y=c_{1} \cos \beta t+c_{2} \sin \beta t\). The amplitude \(R\) is given by \(R=\sqrt{c_{1}^{2}+c_{2}^{2}}\). Use the initial conditions in equation (16) to determine \(c_{1}\) and \(c_{2}\). Your expression for \(R\) will involve \(y_{0}, y_{0}^{\prime}\), and \(\beta=\sqrt{k / m}\).] (b) Two experiments are performed. A mass is given an initial downward displacement \(y_{0}\) and then released with a downward initial velocity \(y_{0}^{\prime} .\) Next, the mass is given the same downward displacement \(y_{0}\), but this time released with an upward initial velocity \(-y_{0}^{\prime}\). Which experiment (if any) would you expect to yield the larger amplitude? Using the result of part (a), compare the amplitudes of the resulting vibrations.

Short answer

Answer: No, there is no difference in the amplitude of undamped vibrations in both cases. The amplitude remains the same for both experiments.

Step-by-step solution

Write down the general solution

Given the equation of motion for undamped vibrations \(m y^{\prime \prime}+k y=0\), the general solution will be $$ y(t) = c_1 \cos(\beta t) + c_2 \sin(\beta t), $$ where \(\beta = \sqrt{\frac{k}{m}}\).

Determine \(c_1\) and \(c_2\) from the initial conditions

We have the initial conditions \(y(0)=y_0\) and \(y'(0)=y_0'\). We'll use these to find \(c_1\) and \(c_2\). Applying the first initial condition, we get: $$ y_0 = c_1 \cos(0) + c_2 \sin(0), $$ which simplifies to \(y_0 = c_1\). Applying the second initial condition, we need the derivative of the general solution: $$ y'(t) = -c_1 \beta \sin(\beta t) + c_2 \beta \cos(\beta t). $$ Then, we have: $$ y_0' = -c_1 \beta \sin(0) + c_2 \beta \cos(0), $$ which simplifies to \(y_0' = c_2 \beta\). Since \(c_1=y_0\), we can say that \(c_2 = \frac{y_0'}{\beta}\).

Calculate the amplitude of the undamped vibrations

According to the hint, the amplitude \(R\) can be found using the following equation: $$ R = \sqrt{c_1^2 + c_2^2}, $$ where \(c_1=y_0\) and \(c_2=\frac{y_0'}{\beta}\). Substituting these values, we get: $$ R = \sqrt{y_0^2 + \left(\frac{y_0'}{\beta}\right)^2}. $$

Compare the amplitudes for the two experiments

In the first experiment, the mass is given an initial downward displacement \(y_0\) and released with a downward initial velocity \(y_0'\). In this case, both \(y_0\) and \(y_0'\) have the same sign (both are positive or both are negative). In the second experiment, the mass is given the same downward displacement \(y_0\), but released with an upward initial velocity \(-y_0'\). In this case, \(y_0\) and \(y_0'\) have opposite signs. Consider the first case: the amplitude is $$ R_1 = \sqrt{y_0^2 + \left(\frac{y_0'}{\beta}\right)^2}. $$ For the second case: $$ R_2 = \sqrt{y_0^2 + \left(\frac{-y_0'}{\beta}\right)^2} = \sqrt{y_0^2 + \left(\frac{y_0'}{\beta}\right)^2} = R_1. $$ By comparing the amplitudes for both cases, we see that they are the same. Therefore, there is no difference in amplitude between the two experiments.

Key concepts

Spring-Mass System

The spring-mass system is a fundamental model used in classical mechanics to describe harmonic oscillators. It consists of a mass attached to a spring which, in turn, is secured to a fixed point. When this system is set into motion, typically by stretching or compressing the spring and then releasing it, the mass will oscillate around its equilibrium position. The back-and-forth motion is the result of the force exerted by the spring, which obeys Hooke's Law—this force is directly proportional to the displacement of the mass from its equilibrium position, and it acts in the opposite direction.

Mathematically, the spring-mass system can be modeled using the second-order linear differential equation mentioned in the exercise, where m represents the mass of the object, y the displacement from the equilibrium, and k the spring constant. Understanding the dynamics of this system is crucial for studying undamped vibrations, which occur in the absence of frictional forces that would otherwise dissipate energy and reduce the amplitude of oscillations over time.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In the context of the spring-mass system, the differential equation provided describes the relationship between the displacement of the mass and the restoring force exerted by the spring. The general form of this equation is a second-order differential equation because it involves the second derivative of the displacement, denoted as y'', which represents acceleration.

Solving such equations typically involves finding a function that satisfies the equation for all points in time. In our case, the solution represents the position of the mass as a function of time. Understanding how to solve these differential equations is essential for analyzing systems like our undamped spring-mass system and predicting their behavior over time.

Initial Conditions

Initial conditions in the context of differential equations are values that specify the state of the system at a starting point, usually at time t=0. For the spring-mass system, the initial conditions tell us the initial displacement, y(0)=y₀, and the initial velocity, y'(0)=y₀'. These values are necessary for determining the particular solution to the differential equation that corresponds to the physical situation we're examining.

In our exercise, the initial conditions allow us to solve for the constants c₁ and c₂ in the general solution of the differential equation. They are pivotal in defining the motion’s unique behavior, ensuring that the solution fits the physical events taking place in the system. Without initial conditions, we would have an infinite number of potential solutions, as opposed to the single trajectory that we observe in reality.

Amplitude of Vibrations

The amplitude of vibrations is a measure of how far the mass moves from its equilibrium position during oscillation. It's one of the most significant characteristics of a vibrating system. In undamped vibrations, where energy is conserved, the amplitude remains constant throughout the motion, as there is no energy loss due to friction or other resistive forces.

In our exercise, the amplitude is denoted by R and is calculated by taking the square root of the sum of the squares of the constants c₁ and c₂. These constants are determined by the initial conditions of the system. Amplitude is crucial for understanding the intensity of the vibrations, and it influences various practical applications of harmonic oscillators, like tuning musical instruments or designing suspension systems in vehicles.

Despite different initial velocities as stated in the experiment scenarios, the results showed that the amplitude remains the same regardless of the direction of the initial force. This outcome underscores the principle that amplitude is independent of the direction of motion and is solely dependent on the initial energy imparted in the system. Thus, it demonstrates a core feature of conservative forces where the vibration’s amplitude is a function of energy, not the direction of energy application.