Question

A mechanical or electrical system is described by the differential equation \(y^{\prime \prime}+\omega^{2} y=f(t) .\) Find \(y\) if \(f(t)=\left\\{\begin{array}{ll}1, & 0a\) separately, remembering that \(f(t)=0\) for \(t>a .\) Show that \(y=\left\\{\begin{array}{ll}\frac{1}{\omega^{2}}(1-\cos \omega t), & ta.\end{array}\right.\) Sketch the motion if \(a=\frac{1}{3} T\) where \(T\) is the period for free vibrations of the system; if \(a=\frac{3}{2} T ;\) if \(a=\frac{1}{10} T\).

Short answer

For \( t < a, y(t) = \frac{1}{\omega^2}(1 - \cos \omega t) \)For \( t > a, y(t) = \frac{1}{\omega^2}(\cos(\omega(t - a)) - \cos(\omega t)) \)

Step-by-step solution

- Understand the given differential equation

The given differential equation is \( y'' + \omega^2 y = f(t) \). We need to find the solution for \( y(t) \) given the piecewise function \( f(t) \) and initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \).

- Use the convolution integral

The solution can be found by using the convolution integral because \( y(t) \) is expressed as \[ y(t) = \int_0^t h(t - \tau) f(\tau) d\tau \] where \( h(t) \) is the impulse response function for the homogeneous equation \( y'' + \omega^2 y = 0 \).

- Determine the impulse response function \( h(t) \)

Solve the homogeneous equation \( y'' + \omega^2 y = 0 \) assuming the initial conditions. This gives \( h(t) = \frac{1}{\omega} \sin(\omega t) \).

- Case: when \( t < a \)

For \( 0 < t < a \), the function \( f(t) = 1 \). Using the convolution integral, we compute \[ y(t) = \int_0^t h(t - \tau) d\tau \] Using \( h(t - \tau) = \frac{1}{\omega} \sin(\omega (t - \tau)) \), we get:\[ y(t) = \int_0^t \frac{1}{\omega} \sin(\omega (t - \tau)) d\tau = \frac{1}{\omega^2} (1 - \cos \omega t) \]

- Case: when \( t > a \)

For \( t > a\), the function \( f(t) = 0 \) and stops its effect. Thus, \( y(t) \) is only affected by the interval from \( 0 < \tau < a \). Using the convolution integral in this range, we have:\[ y(t) = \int_0^a \frac{1}{\omega} \sin(\omega (t - \tau)) d\tau = \frac{1}{\omega^2} [ \cos(\omega(t - a)) - \cos(\omega t)] \]

- Combine the cases

The final solution for \( y(t) \) is presented as a piecewise function by combining the results from steps 4 and 5:\[ y(t) = \left\{ \begin{array}{ll}\frac{1}{\omega^2}(1 - \cos \omega t), & t < a, \frac{1}{\omega^2}(\cos(\omega(t - a)) - \cos(\omega t)), & t > a \end{array}\right. \]

- Map the periods

Given that the period \( T = \frac{2 \pi}{\omega} \), substitute \( a = \frac{1}{3} T, \frac{3}{2} T, \frac{1}{10} T \) into the piecewise function to observe the motion.

Key concepts

Convolution Integral

The convolution integral is a powerful tool used to solve differential equations, especially in systems described by linear differential equations. It helps in finding the output of a system when the input and the system's response to a unit impulse (impulse response function) are known. The formula for convolution integral is given by:

• \[ y(t) = \int_0^t h(t - \tau) f(\tau) d\tau \]
where:
• y(t) is the output, or response of the system,
• h(t) is the impulse response function,
• f(t) is the input function,
• \( \tau \) is a dummy variable of integration.
In our problem, this integral helps us determine how the mechanical or electrical system responds over time, given the piecewise nature of \( f(t) \).

Impulse Response Function

The impulse response function, \( h(t) \), describes the reaction of a system to a single, instantaneous input (impulse). For the homogeneous differential equation \( y'' + \omega^2 y = 0 \), solving with the initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \) gives:

• \[ h(t) = \frac{1}{\omega} \sin(\omega t). \]

This function represents the free vibration behaviour of the system and plays a crucial role in the convolution integral for finding the overall response \( y(t) \). The impulse response function essentially captures the natural oscillating behaviour of the system, which is periodic in nature.

Piecewise Functions

A piecewise function is defined by different expressions for different intervals of its domain. In this exercise, the input function \( f(t) \) is a piecewise function defined as:

• \[ f(t) = \left\{ \begin{array}{ll} \1, & 0 < t < a, \ 0, & \text{otherwise}. \end{array} \right. \]

This means \( f(t) \) is '1' in the interval \( (0, a) \) and '0' elsewhere. Recognizing piecewise nature of \( f(t) \) is critical when applying the convolution integral since we need to handle different cases for \( t < a \) and \( t > a \). These cases lead to how \( y(t) \) is computed over different domains.

Homogeneous Differential Equations

A homogeneous differential equation is one where all terms involve the unknown function or its derivatives. In our problem, the homogeneous equation is:

• \[ y'' + \omega^2 y = 0. \]

To solve this, characteristic equations or methods such as undetermined coefficients are used. The solutions are generally of the form involving sine and cosine functions:

• \[ y(t) = C\cos(\omega t) + D\sin(\omega t), \]

where \( C \) and \( D \) are constants determined by initial conditions. For impulse response cases, specialized initial conditions are used to derive \( h(t) \), such as having \( y'(0) \) with a unit value.

Initial Conditions

Initial conditions specify the state of the system at the start of the observation. They are crucial in finding particular solutions to differential equations. Given the initial conditions:

• \[ y(0) = 0 \]
• \[ y'(0) = 0 \]

they imply that at time \( t = 0 \), there is no initial displacement or initial velocity in the system. Using these conditions, we can solve the differential equation specifically for our problem. These initial conditions are used in conjunction with the convolution integral and \( h(t) \) to find the complete response.