Question

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 damped oscillator The displacement of a mass on a spring suspended from the ceiling is given by \(y=10 e^{-t / 2} \cos \frac{\pi t}{8}\) a. Graph the displacement function. b. Compute and graph the velocity of the mass, \(v(t)=y^{\prime}(t)\) c. Verify that the velocity is zero when the mass reaches the high and low points of its oscillation.

Short answer

Answer: From the graph of the displacement function of the damped oscillator, we can observe a decaying cosine function which results in a sinusoidal oscillation with decreasing amplitude.

Step-by-step solution

Graph the displacement function

To graph the displacement function \(y(t) = 10e^{-t/2}\cos\frac{\pi t}{8}\), you need a graphing tool or software. Plug the function into the graphing tool and observe the shape of the curve. The result should be a decaying cosine function which gives a sinusoidal oscillation with decreasing amplitude.

Compute the velocity function

To compute the velocity function, we need to differentiate the displacement function \(y(t)\) with respect to time \(t\). Here's how to differentiate the given function: $$v(t) = y'(t) = \frac{d}{dt}(10e^{-t/2}\cos\frac{\pi t}{8})$$ To differentiate this function, we can use the product rule \((u\cdot v)' = u' \cdot v + u \cdot v'\). Let \(u = 10e^{-t/2}\) and \(v = \cos\frac{\pi t}{8}\). First, find the derivative of \(u\) $$u'(t) = \frac{d}{dt}(10e^{-t/2}) = -5e^{-t/2}$$ Second, find the derivative of \(v\) $$v'(t) = \frac{d}{dt}(\cos\frac{\pi t}{8}) = -\frac{\pi}{8}\sin\frac{\pi t}{8}$$ Now apply the product rule, $$v(t) = -5e^{-t/2}\cos\frac{\pi t}{8} - \frac{\pi}{8} \cdot 10e^{-t/2}\sin\frac{\pi t}{8}$$ So the velocity function is $$v(t) = -5e^{-t/2}\cos\frac{\pi t}{8} - \frac{5\pi}{4} e^{-t/2}\sin\frac{\pi t}{8}$$

Graph the velocity function

To graph the velocity function \(v(t) = -5e^{-t/2}\cos\frac{\pi t}{8} - \frac{5\pi}{4} e^{-t/2}\sin\frac{\pi t}{8}\), use a graphing tool or software. Plug the function into the graphing tool and observe the shape of the curve. The result should be a damped oscillation, similar to the displacement function, but with a phase shift.

Verify that the velocity is zero at high and low points

To verify that the velocity is zero when the mass reaches the high and low points of its oscillation, observe the graph of the displacement function. At the high and low points, the displacement changes direction. At these instances, the slope of the displacement curve should be zero, which means the velocity should be zero as well. Now, look at the graph of the velocity function and check if it is zero at the same points where the displacement function has its high and low points. If they coincide, then the velocity is indeed zero at the high and low points of the oscillation, which verifies the given condition.

Key concepts

Mass-Spring System

A mass-spring system is a classical model in physics used to describe how an object attached to a spring behaves when it is displaced from its equilibrium position. This system is fundamental in understanding harmonic motion. Here, the object of mass \(m\) is connected to a spring with a stiffness constant \(k\). This constant \(k\), often called the spring constant, determines how stiff the spring is:

• The larger the \(k\), the stiffer the spring, meaning it requires more force to stretch or compress the spring.
• The basic principle is governed by Hooke's Law, which states that the force needed to extend or compress a spring is directly proportional to the distance it is stretched.

When displaced, the mass will oscillate around the equilibrium position. In its simplest form, without the influence of external forces like friction, the oscillation follows a sinusoidal pattern. The position function \(y = y_0 \cos (t \sqrt{\frac{k}{m}})\) describes this motion where the cosine part reflects the periodic nature of oscillations, and \(y_0\) indicates the maximum displacement from the equilibrium.

Damped Oscillator

A damped oscillator is an extension of the mass-spring system that includes damping effects like friction or air resistance, which gradually reduce the amplitude of oscillations over time. This damping typically follows an exponential decay model. The position function for a damped oscillator can be expressed as \(y = 10 e^{-t / 2} \cos \frac{\pi t}{8}\). In this equation:

• \(10 e^{-t / 2}\) represents the amplitude decay factor.
• The exponential term \(e^{-t / 2}\) reflects how quickly the oscillations are damping out.
• The cosine function gives the remaining oscillatory behavior with decreasing amplitude.

As time passes (\(t\rightarrow \infty \)), the exponential term tends towards zero, causing the oscillation to eventually stop, which is observed as the motion slowing and settling.

Differentiation

Differentiation is the process of finding the derivative of a function. In the context of harmonic motion, differentiation helps us find velocity or acceleration from the displacement function. Here, we're differentiating \(y(t) = 10 e^{-t/2} \cos \frac{\pi t}{8}\) with respect to time \(t\), which requires the application of calculus rules.

• The derivative of the position function \(y(t)\) gives us the velocity function \(v(t)\).
• Finding a derivative involves evaluating how the function changes as the variable changes, which implies finding the slope of the tangent at any point on the function curve.

In harmonic oscillation problems, the derivative plays a crucial role because it describes how quickly and in what direction the object is moving at any given point in time.

Product Rule

The product rule in calculus is a fundamental tool for finding the derivative of a product of two functions. If you have two functions \(u(t)\) and \(v(t)\), their product's derivative is given by \((uv)' = u'v + uv'\). In the exercise, this rule helps differentiate the displacement function \(y(t) = 10e^{-t/2}\cos\frac{\pi t}{8}\). Here's how it works:

• Identify \(u = 10 e^{-t/2}\) and \(v = \cos \frac{\pi t}{8}\).
• Compute the derivative \(u' = -5 e^{-t/2}\), using the chain rule for exponential decay.
• Compute \(v' = -\frac{\pi}{8} \sin \frac{\pi t}{8}\) with the chain rule for trigonometric derivatives.
• Substitute into the product rule: \(v(t) = -5e^{-t/2} \cos \frac{\pi t}{8} - \frac{5\pi}{4} e^{-t/2} \sin \frac{\pi t}{8}\).

This expression gives us the velocity function, essential for analyzing the system's dynamics. Understanding and applying the product rule is crucial for working with compounded functions like those in harmonic oscillators.