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

To determine the weight of the package suspended from the spring balance, we can use the relationship between the frequency of oscillation of the spring system and its properties. The frequency of oscillation can be described by the formula that relates frequency to mass and spring constant.

Frequency and Spring Constant Relationship

The frequency of oscillation \( f \) of a mass \( m \) hanging from a spring with spring constant \( k \) is given by the formula:

\( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \)

From this formula, we can rearrange it to solve for the mass:

\( m = \frac{k}{(2\pi f)^2} \)

Determining the Spring Constant

To find the spring constant \( k \), we can utilize Hooke's Law, which states that the force exerted by a spring is proportional to its displacement:

\( F = kx \)

In this case, \( F \) is the weight of the package, which can also be expressed as \( mg \), where \( g \) is the acceleration due to gravity (approximately \( 32.2 \, \text{ft/s}^2 \) or \( 9.81 \, \text{m/s}^2 \)). The displacement \( x \) is related to the maximum stretch of the spring when the package is hung from it.

Using the Information Given

We don't directly have the displacement, but we can still determine \( k \) using the scale information. Since the scale reads from 0 to 50.0 lb, we can estimate an average spring constant based on these limits. However, for precise calculations, we typically need the actual stretch \( x \) when the maximum weight is applied, which we can derive from the scale's reading range.

For the sake of this calculation, let's assume that the spring is linear and we can use the maximum weight to find \( k \). If the spring balance is calibrated such that a 50 lb weight causes the spring to stretch a distance \( x \), then we can express \( k \) as:

\( k = \frac{F}{x} = \frac{50 \, \text{lb}}{4.00 \, \text{in}} \)

To convert the length to feet for consistency with our force units:

\( 4.00 \, \text{in} = \frac{4.00}{12} \, \text{ft} \approx 0.333 \, \text{ft} \)

Thus, substituting into the formula:

\( k = \frac{50 \, \text{lb}}{0.333 \, \text{ft}} \approx 150 \, \text{lb/ft} \)

Calculating the Package Weight

Now, we can use the frequency to find the mass:

\( m = \frac{k}{(2 \pi f)^2} = \frac{150 \, \text{lb/ft}}{(2 \pi \cdot 2 \, \text{Hz})^2} \)

Calculating \( (2 \pi \cdot 2)^2 \):

\( (4\pi)^2 \approx 157.913 \)

Now substituting back into the mass formula:

\( m \approx \frac{150}{157.913} \approx 0.951 \, \text{slugs} \)

To find the weight of the package in pounds, we use \( W = mg \):

\( W \approx 0.951 \, \text{slugs} \times 32.2 \, \text{ft/s}^2 \approx 30.6 \, \text{lb} \)

Final Result

Therefore, the approximate weight of the package is about 30.6 pounds. This answer is based on the oscillation frequency and the estimated spring constant derived from the scale's capacity. For exact results, the actual stretch of the spring under the weight would be needed.