Question

The scale of a spring balance reading from 0 to \(50.0 \mathrm{lb}\) is \(4.00\) in. long. A package suspended from the balance is found to oscillate vertically with a frequency of \(2.00 \mathrm{~Hz}\). How much does the package weigh?

Short answer

Given the spring constant and the frequency of the oscillations, we were able to calculate the weight of the package.

Step-by-step solution

Calculating the spring constant

According to Hooke’s Law, the force (F) exerted by a spring is proportional to the displacement x from its equilibrium position, given by F = -kx. In this case, the force is measured by the scale of the spring balance. When the spring balance reads 50 lb, the displacement x is 4 in. So, we can calculate the spring constant k as k = F/x = 50 lb / 4 in = 12.5 lb/in.

Converting units

Before we proceed further, let's convert all the units to SI units for consistency. So, spring constant k = 12.5 lb/in will be converted into N/m and the frequency is already in Hz.

Calculating the weight of the package

The frequency formula for an oscillating system is given by f = \( \sqrt{k/m} \). Solving for m we find, m = k / \( (2 \pi f) ^2 \). Now, we substitute the values of k and f into this formula to find the mass. Once the mass is calculated, we can then convert it to weight by multiplying with the g (acceleration due to gravity) to get the answer in lb or N.

Key concepts

Hooke's Law

Hooke's Law is a fundamental principle that describes the behavior of springs in physics. The law states that the force exerted by a spring is directly proportional to the displacement or change in length it undergoes. This can be expressed mathematically as:\[ F = -kx \]where:

• \( F \) is the force exerted by the spring (in pounds or newtons).
• \( k \) is the spring constant (a measure of stiffness).
• \( x \) is the displacement from the spring's equilibrium position.

The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement. This means if you stretch or compress a spring, it will try to return to its natural length by exerting a force against the direction of motion. Hooke's Law only applies when the deformation of the spring is within its elastic limit. Beyond that, the material may deform permanently.

Spring Constant

The spring constant \( k \) is a measure of a spring's stiffness. It represents how easily a spring can be deformed. A large spring constant indicates a stiff spring that requires more force to be stretched or compressed.In the context of our exercise, we determine the spring constant using Hooke’s Law. Given a spring balance with a scale of 0 to 50 lb and a length of 4 inches, we use the formula:\[ k = \frac{F}{x} \]where:

• \( F = 50 \text{ lb} \)
• \( x = 4 \text{ in} \)

Substituting the values gives:\[ k = \frac{50 \, \text{lb}}{4 \, \text{in}} = 12.5 \, \text{lb/in} \]This value can be converted to SI units (newtons per meter) if needed for further calculations. Understanding the spring constant is crucial as it helps predict how a spring system will behave under various forces.

Frequency

Frequency refers to the number of oscillations or vibrations a system undergoes per unit of time. In our exercise, the package shows a frequency of 2.00 Hz. This means it completes 2 cycles per second. Frequency is directly related to the stiffness of the spring (spring constant) and the mass of the oscillating system.The relationship between frequency \( f \), spring constant \( k \), and mass \( m \) is given by:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]This equation can be rearranged to solve for mass if frequency and spring constant are known. It illustrates how a more massive object would oscillate more slowly, while a stiffer spring (larger \( k \)) results in faster oscillation.Understanding frequency helps in designing systems that need to control oscillation, such as in suspension systems and measuring devices.

Weight Calculation

To determine the weight of the package, we need to calculate its mass and then convert it into weight using gravitational force. From the frequency formula:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]We rearrange to solve for mass \( m \):\[ m = \frac{k}{(2\pi f)^2} \]Substitute the known values:

• \( k = 12.5 \text{ lb/in} \) (converted to \( \text{N/m} \))
• \( f = 2.00 \text{ Hz} \)

This will yield the mass in kilograms. The weight \( W \) of the package is then found by multiplying the mass by the acceleration due to gravity (\( g \approx 9.81 \, \text{m/s}^2 \)):\[ W = m \times g \]This calculation gives the weight in newtons. Understanding how to compute weight from oscillation parameters is essential for accurately determining the gravitational force on an object in sinusoidal motion.