91Ó°ÊÓ

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 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

01

Convert Mass to Kilograms

First, we need to convert the mass from grams to kilograms. Since 1 gram equals 0.001 kilograms, the mass is \( m = 10 \times 0.001 = 0.01 \) kg.

02

Determine the square root of k/m

The spring stiffness is given as \( k = 3 \). Therefore, the value of \( \sqrt{\frac{k}{m}} \) becomes \( \sqrt{\frac{3}{0.01}} = \sqrt{300} \). Simplifying gives \( \sqrt{300} = 10\sqrt{3} \).

03

Formulate the Oscillation Equation

Using these values in the given displacement equation \( f(t) = a \cos \sqrt{\frac{k}{m}} t \), with compression distance \( a = 0.05 \), the equation becomes \( f(t) = 0.05 \cos (10\sqrt{3}t) \).

04

Formulate the Frequency Formula

The angular frequency is given by \( \omega = \sqrt{\frac{k}{m}} \). The frequency \( f \) is related to the angular frequency by \( f = \frac{\omega}{2\pi} \), giving us \( f = \frac{\sqrt{k/m}}{2\pi} \).

05

Frequency Effect with Increased Mass

When the mass \( m \) is increased, the term \( \sqrt{\frac{k}{m}} \) becomes smaller, reducing the frequency and leading to slower oscillations.

06

Frequency Effect with Increased Stiffness

Increasing the spring constant \( k \) results in an increase in \( \sqrt{\frac{k}{m}} \), which raises the frequency and leads to faster oscillations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Constant

The spring constant, denoted by the symbol \( k \), measures how stiff or flexible a spring is. It is an essential parameter in understanding the behavior of a harmonic oscillator. The spring constant has the units of force per unit length (typically Newtons per meter in the metric system), and it dictates how much force is needed to compress or extend the spring by a certain amount.

• A larger spring constant means a stiffer spring, which requires more force to compress or expand.
• A smaller spring constant implies a more flexible spring, needing less force for the same displacement.

Understanding the spring constant helps in predicting the motion of springs in various applications, from simple toys to complex mechanical systems.

Angular Frequency

Angular frequency, represented by the Greek letter \( \omega \), is crucial in studying oscillatory systems, such as springs and pendulums. It defines how fast an object revolves in a circular path or oscillates in simple harmonic motion. Angular frequency is related to the spring constant and the mass of the object via the formula:\[\omega = \sqrt{\frac{k}{m}}\]Where:

• \( \omega \) is the angular frequency in radians per second.
• \( k \) is the spring constant.
• \( m \) is the mass of the object.

This relationship shows that angular frequency increases with a stiffer spring and decreases with a larger mass. In harmonic oscillation, angular frequency is linked to how often the oscillations occur in a set period.

Mass Effect

The mass of an object, \( m \), plays a significant role in determining the frequency of oscillation. As the mass increases, the frequency of oscillation decreases, causing the system to oscillate more slowly. Mathematically, this is evident from the angular frequency equation \( \omega = \sqrt{\frac{k}{m}} \), as the mass \( m \) is in the denominator:

• A larger mass results in a smaller value of \( \sqrt{\frac{k}{m}} \), reducing the oscillation frequency.
• Conversely, a smaller mass increases \( \sqrt{\frac{k}{m}} \), leading to a higher frequency.

This mass effect is crucial for designing systems that rely on specific frequencies, such as musical instruments or engineering structures.

Stiffness Impact

Stiffness, characterized by the spring constant \( k \), influences how quickly a system can oscillate. The effect of stiffness is directly proportional to the frequency of oscillation. If the stiffness increases (i.e., a larger \( k \)), the oscillation frequency becomes higher, as shown by the equation \( \omega = \sqrt{\frac{k}{m}} \).

• Increasing the spring's stiffness causes the oscillation to be faster, as the system returns to its equilibrium position more quickly after being displaced.
• A stiffer spring results in larger forces acting against the displacement, enhancing the speed of oscillation cycles.

This understanding helps optimize the design and function of systems where precise oscillatory responses are needed, like in shock absorbers or seismic isolation systems.