Problem 59 - College Physics

Question

\(\bullet\) \(\bullet\) An apple weighs 1.00 \(\mathrm{N}\) . When you hang it from the end of a long spring of force constant 1.50 \(\mathrm{N} / \mathrm{m}\) and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back-and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed)?

Step-by-Step Solution

Answer

The unstretched length of the spring is approximately 1.98 m.

Step 1: Understand the Problem

We need to find the unstretched length of a spring with an apple attached. The apple's weight is given as 1.00 N. The spring constant is 1.50 N/m. The spring is involved in two types of motion: simple harmonic motion (SHM) and pendulum motion. We know the frequency of the pendulum is half that of the SHM.

Step 2: Find Spring Extention in SHM

The force exerted by the spring when the apple is attached (F_s = kx) should equal the weight of the apple (F_w = 1.00 \, \text{N}). Using the formula F_s = kx, where k = 1.50 \, \text{N/m}, we find the extension x: \[ 1.00 = 1.50x \]Thus, \[ x = \frac{1.00}{1.50} = 0.67 \, \text{m} \]

Step 3: Calculate Frequency of SHM

The frequency f of a spring-mass system is given by f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}. To find m, use F = mg (where g = 9.81 \, \text{m/s}^2): \[ m = \frac{1.00}{9.81} \approx 0.102 \, \text{kg} \] \[ f = \frac{1}{2\pi} \sqrt{\frac{1.50}{0.102}} \approx 0.614 \, \text{Hz} \]

Step 4: Calculation for Frequency in Pendulum Motion

The pendulum frequency is half that of the SHM frequency: \[ f_{pendulum} = \frac{0.614}{2} \approx 0.307 \, \text{Hz} \] \[ T_{pendulum} = \frac{1}{f_{pendulum}} = \frac{1}{0.307} \approx 3.26 \, \text{s} \]

Step 5: Calculate Length of the Spring for Pendulum Motion

For a pendulum, the period T is T = 2\pi \sqrt{\frac{L}{g}}, solving for L gives:\[ 3.26 = 2\pi \sqrt{\frac{L}{9.81}} \] \[ \frac{3.26}{2\pi} = \sqrt{\frac{L}{9.81}} \] \[ L = 9.81 \left(\frac{3.26}{2\pi}\right)^2 \approx 2.65 \, \text{m} \] This includes the spring's stretched length, but the unstretched length minus the spring extension \( 0.67 \, \text{m} \) gives: \[ Unstretched\_Length = 2.65 - 0.67 \approx 1.98 \, \text{m} \]

Key Concepts

Spring Constant

When dealing with springs, one important concept is the spring constant, denoted as "k". It is a measure of the stiffness of the spring. In other words, it tells us how much force is needed to stretch or compress the spring by a certain amount.
For instance, if a spring has a constant of 1.50 N/m, this means that a force of 1.50 Newtons is required to extend the spring by one meter. This characteristic is crucial in determining how a spring behaves under various weights and forces.
When an object is attached to the spring, the amount the spring extends depends on its spring constant and the weight of the object. We use the formula \( F_s = kx \), where \( F_s \) is the force exerted by the spring and "x" is the extension. The larger the spring constant, the stiffer the spring and the less it extends under a given force. This principle underlies many applications and experiments involving spring motion.

Pendulum Motion

Pendulum motion represents a classic example of simple harmonic motion (SHM). In the context of our exercise, another interesting scenario occurs when we let the apple swing back and forth while attached to the spring, acting like a pendulum.
Here, important factors include the length of the pendulum and the angle of swing. For small angles, the simple pendulum exhibits SHM, where its frequency depends inversely on the square root of its length. That's why the swinging motion frequency differs from the bouncing motion.
This can be particularly observed when the pendulum frequency is half of the bouncing frequency of SHM as noted in our task. Understanding pendulum motion involves analyzing the relation of the length and gravitational acceleration to compute the frequency.

Frequency Calculation

Frequency calculation is an essential part of understanding simple harmonic motion. For a spring-mass system, frequency describes how often the mass completes a full cycle of motion in one second.
The formula \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \) allows us to calculate the frequency of spring-driven SHM, where "k" is the spring constant and "m" the mass of the object.
In our case, with the apple's weight leading us to a mass of about 0.102 kg, this frequency becomes a vital characteristic of motion. The frequency for a pendulum is derived similarly, though it halves due to differing oscillation paths, broadening our view of motions within SHM.

Unstretched Length

The unstretched length of a spring is its original length before any force or mass is applied. This is crucial in calculations as it defines the baseline length from which any extensions due to weights are measured.
In the scenario where an apple is attached and stretches the spring, knowing the unstretched length helps decipher spring motion characteristics and planning experiments.
Here, the calculation was made by subtracting the spring's extension from the total stretched length measured during pendulum motion. Resulting in an unstretched spring length of approximately 1.98 meters, this value is essential for subsequent applications like determining system stability and predicting other material responses.