Problem 36 - Precalculus Mathematics for Calculus

Question

Spring-Mass System The frequency of oscillation of an -object suspended on a spring depends on the stiffness \(k\) of the spring (called the spring constant) and the mass \(m\) of the object. If the spring is compressed a distance \(a\) and then allowed to oscillate, its displacement is given by $$ f(t)=a \cos \sqrt{k / m} t $$ a) A 10 -g mass is suspended from a spring with stiffness \(k=3 .\) If the spring is compressed a distance \(5 \mathrm{cm}\) and then released, find the equation that describes the oscillation of the spring. b) Find a general formula for the frequency (in terms of \(k\) and \(m\) ). c) How is the frequency affected if the mass is increased? Is the oscillation faster or slower? d) How is the frequency affected if a stiffer spring is used (larger \(k\) )? Is the oscillation faster or slower?

Step-by-Step Solution

Answer

a) The equation is \(f(t) = 5 \cos(10\sqrt{3} t)\). b) \(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\). c) Increasing mass results in slower oscillations. d) Increasing stiffness results in faster oscillations.

Step 1: Convert Units

Before solving the problem, we need to ensure all units are consistent. Convert the mass from grams to kilograms: \(10 \text{ g} = 0.01 \text{ kg}\). The spring compression is already given in centimeters, which is the same unit used in the function, so no conversion is needed.

Step 2: Identify Given Values

From the problem statement, we are given: \(m = 0.01 \text{ kg}\), \(k = 3\), and the amplitude of spring compression \(a = 5 \text{ cm}\).

Step 3: Setup the Displacement Equation

Use the displacement equation \(f(t) = a \cos \left( \sqrt{\frac{k}{m}} t \right)\). Substitute the known values: \(f(t) = 5 \cos \left( \sqrt{\frac{3}{0.01}} t \right)\).

Step 4: Simplify the Expression inside the Cosine Function

Calculate the inside expression: \(\sqrt{\frac{3}{0.01}} = \sqrt{300} = 10\sqrt{3}\). Thus, the simplified oscillation equation is \(f(t) = 5 \cos \left(10\sqrt{3} t \right)\).

Step 5: Derive Frequency Formula

In general, the oscillation frequency \(\omega\) can be derived as \(\omega = \sqrt{\frac{k}{m}}\). The frequency \(f\) is given by \(f = \frac{\omega}{2\pi}= \frac{1}{2\pi} \sqrt{\frac{k}{m}} \).

Step 6: Analyze Mass Increase Effect

If the mass \(m\) is increased, the value under the square root decreases, leading to a decrease in frequency. Therefore, the oscillation becomes slower.

Step 7: Analyze Spring Stiffness Effect

If the spring stiffness \(k\) is increased, the value under the square root increases, leading to an increase in frequency. Therefore, the oscillation becomes faster.

Key Concepts

Frequency of Oscillation

In a spring-mass system, the frequency of oscillation represents how often the system completes a full cycle of movement in a second. It dictates how quick or slow the object attached to the spring moves back and forth. The frequency is an essential aspect of oscillatory motion, particularly in systems involving springs.
For any given spring and mass, the natural frequency determines the pace of oscillation. It can be mathematically represented using the formula for angular frequency \[\omega = \sqrt{\frac{k}{m}}\] where:

• \( \omega \) is the angular frequency.
• \( k \) is the spring constant, a measure of the spring's stiffness.
• \( m \) is the mass of the object.

The actual frequency \( f \) in hertz (Hz) is given by the equation:
\[f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\]This formula shows that frequency is inversely proportional to the mass and directly proportional to the spring constant. Understanding this allows students to predict and control the behavior of oscillating springs by adjusting either the mass or the stiffness of the spring.

Spring Constant

The spring constant, noted as \( k \), measures a spring's stiffness or rigidity—the force required to displace the spring by a unit length. It plays a crucial role in defining the characteristics of a spring-mass system.
Springs with higher \( k \) values are stiffer and less compressible, meaning they require more force for the same displacement compared to springs with lower \( k \). This stiffness directly influences the frequency of oscillation. A stiffer spring, or a higher \( k \), results in a faster oscillation frequency.
The relationship between the spring constant and oscillation frequency is depicted in the frequency equation:\[f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\]From this, it's clear that increasing \( k \) will increase \( f \), making the system oscillate faster. Conversely, using a spring with a lower \( k \) will slow down the oscillation frequency. This concept is vital when designing systems requiring specific oscillation rates, such as mechanical clocks or vibration isolation systems.

Mass Effect on Oscillation

In a spring-mass system, the mass \( m \) attached to the spring significantly affects the oscillation frequency. The relationship is inversely proportional, meaning as the mass increases, the frequency decreases.
To understand why, consider the frequency equation:\[f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\]Here, as \( m \) increases, the value inside the square root becomes smaller because the numerator \( k \) remains constant, leading to a reduction in the overall value of the frequency \( f \).
A larger mass leads to a slower oscillation as the system takes more time to complete one cycle. This principle is important in many real-world applications where altering the mass can control the speed of oscillations. For example, adding mass to a pendulum results in slower swings, which is useful in regulating the timing of a clock.

Displacement Equation

In the context of spring-mass systems, the displacement equation describes the position of the mass over time as the system oscillates. The equation given in the exercise, \[f(t) = a \cos\left(\sqrt{\frac{k}{m}} t\right)\]provides a way to calculate where the mass will be at any given time \( t \).

• \( a \) represents the amplitude, which is the maximum displacement from the equilibrium position.
• The square root expression \( \sqrt{\frac{k}{m}} \) corresponds to the angular frequency \( \omega \).
• The function \( \cos\left(\right) \) signifies the oscillatory nature of the system.

Using this equation, one can predict the displacement, velocity, and acceleration of the mass at any time, aiding in the design and analysis of systems like car suspensions and building structures. The equation underscores the harmonic nature of oscillatory systems, characterized by predictable and repeatable motion patterns.