Question

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. 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

Question: Verify that the function \(y(t) = e^{-t}(\sin 2t - 2\cos 2t)\) satisfies the differential equation \(y^{\prime \prime}(t) + 2y^{\prime}(t) + 5y(t) = 0\). Answer: We found the first and second derivatives of the given function and substituted them into the differential equation. After simplifying the expression, we found that it equals zero. Therefore, 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 y(t)

To find the first derivative (\(y'(t)\)), we apply the product rule and the chain rule: $$y'(t) = \frac{d(e^{-t}(\sin 2t - 2\cos 2t))}{dt}$$ $$y'(t) = e^{-t} \frac{d(\sin 2t - 2\cos 2t)}{dt} + (\sin 2t - 2\cos 2t) \frac{d(e^{-t})}{dt}$$ Now, we find the derivatives: $$\frac{d(\sin 2t - 2\cos 2t)}{dt} = 2\cos 2t + 4\sin 2t$$ $$\frac{d(e^{-t})}{dt} = -e^{-t}$$ Then substitute the derivatives back: $$y'(t) = e^{-t}(2\cos 2t + 4\sin 2t) - e^{-t}(\sin 2t - 2\cos 2t)$$

Find the second derivative of y(t)

To find the second derivative (\(y''(t)\)), we once again apply the product rule and the chain rule: $$y''(t) = \frac{d(e^{-t}(2\cos 2t + 4\sin 2t - \sin 2t + 2\cos 2t))}{dt}$$ $$y''(t) = e^{-t} \frac{d(2\cos 2t + 4\sin 2t - \sin 2t + 2\cos 2t)}{dt} + (2\cos 2t + 4\sin 2t - \sin 2t + 2\cos 2t) \frac{d(e^{-t})}{dt}$$ Now, we find the derivatives: $$\frac{d(2\cos 2t + 4\sin 2t - \sin 2t + 2\cos 2t)}{dt} = -4\sin 2t + 8\cos 2t + 2\cos 2t - 4\sin 2t$$ $$\frac{d(e^{-t})}{dt} = -e^{-t}$$ Then substitute the derivatives back: $$y''(t) = e^{-t}(-4\sin 2t + 8\cos 2t + 2\cos 2t - 4\sin 2t) - (2\cos 2t + 4\sin 2t - \sin 2t + 2\cos 2t)e^{-t}$$

Substitute y(t), y'(t), and y''(t) into the differential equation

Now we need to plug our expressions for \(y(t)\), \(y'(t)\), and \(y''(t)\) into the given differential equation to verify the solution: $$y''(t) + 2y'(t) + 5y(t)$$ $$= e^{-t}(-4\sin 2t + 8\cos 2t + 2\cos 2t - 4\sin 2t) - (2\cos 2t + 4\sin 2t - \sin 2t + 2\cos 2t)e^{-t}$$ $$+ 2(e^{-t}(2\cos 2t + 4\sin 2t) - e^{-t}(\sin 2t - 2\cos 2t))$$ $$+ 5e^{-t}(\sin 2t - 2\cos 2t)$$

Simplify the expression and see if it equals zero

Now we simplify the expression and check if it is equal to zero: $$= e^{-t}(-4\sin 2t + 10\cos 2t - 8\sin 2t + 4\cos 2t - 5\sin 2t + 10\cos 2t)$$ $$= e^{-t}(20\cos 2t - 17\sin 2t) - e^{-t}(20\cos 2t - 17\sin 2t)$$ $$= 0$$ Since the simplified expression is equal to zero, we can verify that the given function \(y(t) = e^{-t}(\sin 2t - 2\cos 2t)\) satisfies the differential equation \(y^{\prime \prime}(t) + 2y^{\prime}(t) + 5y(t) = 0\).

Key concepts

Mechanical Oscillator

A mechanical oscillator is a system that experiences repeated back and forth movement about a stable equilibrium position. One common example is a mass attached to a spring. When displaced from its equilibrium, the system will move back and forth. This is due to the restoring force exerted by the spring and the inertia of the mass.

The mechanical oscillator can serve different purposes in real-world applications:

• It can measure time in clocks and watches.
• It is used in musical instruments to produce sound vibrations.
• It is utilized in sensors to detect motion or force.

In an idealized system, oscillations occur without loss of energy. However, in most practical systems, some energy is lost, often due to friction or air resistance.

Mass-Spring System

A mass-spring system consists of an object with mass attached to a spring. This setup is crucial for studying mechanical oscillations. The behavior of the mass-spring system depends on various factors:

• The mass of the object, represented as \( m \).
• The stiffness of the spring, represented by the spring constant \( k \).

The spring constant \( k \) determines how much force is needed to stretch or compress the spring by a unit length. A larger \( k \) indicates a stiffer spring.

In a mass-spring system, if we pull down on the mass and then release it, the spring exerts a force to bring the mass back to the equilibrium position. This force is described by Hooke's Law, \( F = -kx \), where \( x \) is the displacement from the equilibrium.

In absence of friction, and when the system is displaced from equilibrium, it will undergo simple harmonic motion, swapping potential and kinetic energy back and forth.

Harmonic Oscillation

Harmonic oscillation refers to oscillatory motion that follows a consistent and repeating pattern. The motion is typically sinusoidal in nature, characterized by a sine or cosine function. This is observed in many natural systems and engineered devices.

• For the mass-spring system, the motion can be represented as \( y = y_0 \cos (\omega t) \), where \( \omega = \sqrt{\frac{k}{m}} \) is the angular frequency of the oscillation.
• In our specific exercise, the displacement \( y \) can be calculated using the expression \( y = y_0 \cos(t \sqrt{\frac{k}{m}}) \).

Harmonic oscillation is particularly important because it serves as the basis for understanding more complex vibrations and wave phenomena in physics and engineering.

Under ideal conditions without damping or external forces, harmonic oscillations continue indefinitely. These oscillations also help in analyzing vibrations, resonance, and stability in various systems.