Question

A mechanical oscillator (such as a mass on a spring or a pendulum) subject to frictional forces satisfies the equation (called a differential equation) $$y^{\prime \prime}(t)+2 y^{\prime}(t)+5 y(t)=0,$$ where \(y\) is the displacement of the oscillator from its equilibrium position. Verify by substitution that the function \(y(t)=e^{-t}(\sin 2 t-2 \cos 2 t)\) satisfies this equation.

Short answer

Answer: Yes, the function \(y(t)=e^{-t}(\sin 2t-2\cos 2t)\) satisfies the given differential equation.

Step-by-step solution

Find the first derivative of the function

To find the first derivative, \(y'(t)\), of the given function, \(y(t)=e^{-t}(\sin 2t-2\cos 2t)\), apply the product rule: $$y'(t) = e^{-t}(-\sin 2t + 2\cos 2t) + e^{-t}(2\cos 2t + 4\sin 2t)$$ Combine the terms to get the first derivative: $$y'(t) = e^{-t}(-\sin 2t + 4\sin 2t + 2\cos 2t + 2\cos 2t)$$

Find the second derivative of the function

Now, find the second derivative, \(y''(t)\), of the function. Apply the product rule again to the first derivative, \(y'(t)\): $$y''(t) = e^{-t}(-2\cos 2t + 8\cos 2t - 4\sin 2t + 4\sin 2t)$$ Combine the terms to get the second derivative: $$y''(t) = e^{-t}(6\cos 2t)$$

Substitute the derivatives and function into the differential equation

Now, substitute \(y(t)\), \(y'(t)\), and \(y''(t)\) into the given differential equation to check if it holds true: $$y^{\prime \prime}(t)+2 y^{\prime}(t)+5 y(t) = e^{-t}(6\cos 2t) + 2e^{-t}(-\sin 2t + 4\sin 2t + 2\cos 2t + 2\cos 2t) + 5e^{-t}(\sin 2t-2\cos 2t)$$ Simplify the expression to see if it equals 0: $$e^{-t}(6\cos 2t - 2\sin 2t + 8\sin 2t + 4\cos 2t + 4\cos 2t + 5\sin 2t - 10\cos 2t)$$ Now, combine the terms with \(\cos 2t\) and \(\sin 2t\): $$e^{-t}(6 - 6 + 11\sin 2t - 11\sin 2t) = 0$$

Conclusion

Since the result after the substitution is 0, it means the function \(y(t)=e^{-t}(\sin 2t-2\cos 2t)\) satisfies the given differential equation \(y^{\prime \prime}(t)+2 y^{\prime}(t)+5 y(t)=0\).

Key concepts

Mechanical Oscillator

A mechanical oscillator is a system that exhibits periodic motion around an equilibrium position. Examples include a mass attached to a spring or a swinging pendulum. The motion of such systems can often be described mathematically by differential equations representing the forces acting on the system.

When considering the motion of mechanical oscillators, we typically use equations that account for various forces including restoring forces that seek to return the system to equilibrium, and in many cases, frictional forces that oppose the motion and cause the system to dissipate energy. In the given exercise, the equation reflects such forces and allows students to explore the dynamics of a damped oscillator.

Frictional Forces

In the context of mechanical oscillators, frictional forces play a critical role in the behavior of the system. These forces, which resist motion, lead to what is known as 'damping' in the oscillator. Damping is a type of friction that can be viscous, where the frictional force is proportional to the velocity of the object. The second term in the differential equation, with the first derivative of y, incorporates the damping effect. It demonstrates how the velocity of the oscillator contributes to the energy dissipation over time.

The coefficient of the first derivative in the differential equation is essentially a 'damping factor,' which quantifies the impact of these frictional forces on the system. Greater coefficients suggest stronger frictional forces, leading to quicker dissipation of the oscillator's kinetic energy.

Product Rule

The product rule is a fundamental technique of differentiation used when taking the derivative of a product of two functions. It states that the derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. Symbolically, if we have two functions f(t) and g(t), then their product f(t)g(t) has the derivative f'(t)g(t) + f(t)g'(t).

In our exercise, the product rule is used to find both the first derivative and the second derivative of the function y(t). This rule is crucial when dealing with variables that are functions of time, t, as it allows for the calculation of how their product changes over time.

First Derivative

The first derivative of a function represents the rate of change of the function concerning its variable. For motion problems like our oscillator, the first derivative of the displacement y(t) with respect to time t represents velocity. It indicates how fast the position of the oscillator is changing at any given moment and in which direction.

Applying the First Derivative to the Oscillator

When we calculate the first derivative of y(t), we're essentially seeking to understand how the oscillator's movement evolves over time. In the provided exercise, finding the first derivative is a step toward plugging it back into the differential equation to verify the solution.

Second Derivative

Similarly, the second derivative, which is the derivative of the first derivative, describes the rate at which the first derivative itself is changing — that is, it represents acceleration in the context of our mechanical oscillator problem. Specifically, since our given function y(t) denotes displacement, the second derivative describes how the velocity of the oscillator is changing over time.

A positive second derivative indicates that the velocity is increasing, while a negative value suggests it is decreasing. In the work through the exercise, we use the second derivative to contribute to verifying the validity of the provided solution to the differential equation.