Problem 200 - Calculus Volume 1

Question

A mass on a spring bounces up and down in simple harmonic motion, modeled by the function \(s(t)=-6 \cos t\) where \(s\) is measured in inches and \(t\) is measured in seconds. Find the rate at which the spring is oscillating at \(t=5\) s.

Step-by-Step Solution

Answer

The rate of oscillation at \( t=5 \) seconds is approximately \(-5.7534\) inches per second.

Step 1: Understand the Function

The function given is \( s(t) = -6 \cos t \), which describes the position of the mass on the spring over time \( t \). This is a simple harmonic motion scenario, where \( s(t) \) represents the displacement at time \( t \).

Step 2: Determine the Rate of Change

To find the rate at which the spring is oscillating, find the first derivative \( s'(t) \) of the function \( s(t) \). This derivative represents the velocity, or rate of change of position.

Step 3: Calculate the Derivative

Calculate \( s'(t) \) by differentiating \( s(t) = -6 \cos t \) with respect to \( t \). The derivative of \( \cos t \) is \( -\sin t \), so \( s'(t) = 6 \sin t \).

Step 4: Evaluate the Derivative at the Given Time

Substitute \( t = 5 \) seconds into the derivative \( s'(t) = 6 \sin t \) to find the rate of change at this specific time. So, \( s'(5) = 6 \sin(5) \).

Step 5: Compute the Result

Using a calculator, evaluate \( 6 \sin(5) \). The value of \( \sin(5) \) is approximately \(-0.9589\). Multiply this by 6 to get \( s'(5) \approx 6 \times (-0.9589) \approx -5.7534 \). Thus, the rate of oscillation at \( t=5 \) seconds is approximately \(-5.7534\) inches per second.

Key Concepts

Understanding Derivatives

The concept of derivatives is integral to understanding rates of change in calculus. In simple terms, the derivative of a function at any given point tells us the slope of the tangent to the function at that point. It provides insight into how the function behaves as the input values change. For the function \( s(t) = -6 \cos t \), the derivative, \( s'(t) \), represents the velocity of the mass on the spring. This velocity indicates how fast and in what direction the spring is moving at any moment in time.

To find the derivative, you differentiate the function with respect to \( t \). In this case, differentiating \( -6 \cos t \), gives \( s'(t) = 6 \sin t \). Here’s why it's important:

• It tells us the rate at which the mass's position changes.
• The sign and value of \( s'(t) \) show the direction and speed of this change.

At \( t = 5 \) seconds, plugging into \( s'(t) \) provides \( s'(5) = 6 \sin(5) \) which is approximately \(-5.7534\). This negative value indicates that the mass is moving downward at that moment.

Grasping the Oscillation Rate

In simple harmonic motion, the oscillation rate refers to how frequently the motion repeats itself and how quickly the position changes over time. This is fundamentally tied to the velocity of the object. The velocity, determined by the derivative \( s'(t) \), accounts for not just the speed but also the direction of movement.

Understanding the oscillation rate is crucial because it tells us:

• How quickly the mass attached to the spring is moving away from or towards the equilibrium position.
• Whether the motion is speeding up or slowing down depending on the forces acting on it.

At \( t = 5 \) seconds, when we calculated \( 6 \sin(5) \) resulting in approximately \(-5.7534\) inches per second, this value provides the oscillation rate. The negative sign reflects a downward movement in simple harmonic motion.

The Role of Trigonometric Functions

Trigonometric functions like sine and cosine are essential when describing periodic phenomena, such as simple harmonic motion. These functions encapsulate the repetitive nature of oscillatory motion, fully capturing the characteristics of the system's movement. In the function \( s(t) = -6 \cos t \), the cosine component describes the displacement of the spring at any time \( t \).

The use of trigonometric functions in this context offers numerous advantages:

• Cosine and sine functions exhibit periodic behavior, perfectly modeling oscillations.
• The amplitude of \(-6\) indicates the maximum displacement from the equilibrium position.

When differentiating, the transition from cosine to sine in \( s'(t) = 6 \sin t \) describes how the function shifts between velocity and position. This ties into how energy is transferred between kinetic and potential forms in simple harmonic motion. Thus, trigonometric functions provide a comprehensive framework for understanding the behavior of oscillating systems.