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A 50 -g mass hangs at the end of a Hookean spring. When \(20 \mathrm{~g}\) more are added to the end of the spring, it stretches \(7.0 \mathrm{~cm}\) more. ( \(a\) ) Find the spring constant. (b) If the \(20-\mathrm{g}\) mass is now removed, what will be the period of the motion? (a) Under the weight of the \(50-\mathrm{g}\) mass, \(F_{\text {ext } 1}=k x_{1}\), where \(x_{1}\) is the original stretching of the spring. When \(20 \mathrm{~g}\) more are added, the force becomes \(F_{\text {ext } 1}+F_{\text {ext } 2}=k\left(x_{1}+x_{2}\right)\), where \(F_{\text {ext } 2}\) is the weight of \(20 \mathrm{~g}\) and \(x_{2}\) is the stretching it causes. Subtracting the two force equations leads to $$ F_{\text {ext } 2}=k x_{2} $$ (Note that this is the same as \(F_{\text {eut }}=k x\), where \(F_{\text {et }}\) is the additional stretching force and \(x\) is the amount of stretch due to it. Hence, we could have ignored the fact that the spring had the \(50-\mathrm{g}\) mass at its end to begin with.) Solving for \(k\). $$ k=\frac{F_{\text {ext } 2}}{x_{2}}=\frac{(0.020 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}{0.070 \mathrm{~m}}=2.8 \mathrm{~N} / \mathrm{m} $$ (b) \(T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{0.050 \mathrm{~kg}}{2.8 \mathrm{~N} / \mathrm{m}}}=0.84 \mathrm{~s}\)

Step by step solution

01

Identify the additional force

When an additional 20 g is added to the spring, it causes a stretch of 7.0 cm. First, convert 20 g to kg (since standard SI units are needed): 20 g = 0.020 kg. The force exerted by this mass due to gravity is given by \( F = mg \), where \( m \) is the added mass and \( g \approx 9.81 \text{ m/s}^2 \) is the acceleration due to gravity. This gives \( F = (0.020 \text{ kg})(9.81 \text{ m/s}^2) \approx 0.196 \text{ N} \).

02

Calculate the spring constant

Use Hooke's Law, which states \( F = kx \), where \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the extension caused by the force. We know \( x = 7.0 \text{ cm} = 0.070 \text{ m} \). Therefore, solve for \( k \): \[ k = \frac{F}{x} = \frac{0.196 \text{ N}}{0.070 \text{ m}} = 2.8 \text{ N/m} \]

03

Determine the period of oscillation

Using the spring constant, find the period of oscillation for the original 50 g mass. The period \( T \) of a mass-spring system is given by: \[ T = 2\pi\sqrt{\frac{m}{k}} \] where \( m = 0.050 \text{ kg} \) is the mass and \( k = 2.8 \text{ N/m} \) is the spring constant. Therefore, \[ T = 2\pi\sqrt{\frac{0.050 \text{ kg}}{2.8 \text{ N/m}}} \approx 0.84 \text{ seconds} \]

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Key Concepts

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

Hooke's Law

In physics, Hooke's Law is fundamental to understanding how springs behave when forces are applied. It expresses the relationship between the force applied to a spring and the displacement it causes. Specifically, Hooke's Law is stated as \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant that characterizes the spring's stiffness, and \( x \) is the displacement or stretch of the spring.

If you imagine pulling on a spring, the force you apply changes its length. Hooke's Law tells us that the more force you apply, the more the spring stretches, as long as you don't exceed the spring's elastic limit. The spring constant \( k \) is crucial because it measures how resistant the spring is to being stretched. Higher values of \( k \) mean a stiffer spring.

• The spring constant \( k \) is determined in units of Newtons per meter (N/m).
• It depends on the material and dimensions of the spring.

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of regular, repeating motion that springs often exhibit. It's characterized by a restoring force that is directly proportional to the displacement, meaning the further away the object is from its equilibrium position, the stronger the force pulling it back.

In the context of springs, we often see SHM when a mass is attached to a spring and then displaced. It moves back and forth around an equilibrium point in a predictable pattern. This periodic motion is defined by

• A constant frequency, which is independent of the amplitude, the maximum extent of the oscillation.
• An equilibrium position that the spring-mass system always returns to.

Such motion is not only interesting but also foundational in physics, providing a simple model for more complex oscillations found in nature.

Mass-Spring System

A mass-spring system is a common physical model used to illustrate SHM and Hooke's Law. It consists of a mass attached to a spring. When displaced from its equilibrium state, the system will undergo oscillations.

This setup is useful because it clearly shows how energy is exchanged between potential and kinetic forms:

• Potential energy is stored in the spring when it is compressed or stretched.
• Kinetic energy is at maximum when the mass passes through the equilibrium point at a high speed.

The behavior of this system can be described by the differential equation derived from Hooke's Law and Newton's second law of motion, which leads to the formula for the period of oscillation. This system is often an idealization, assuming no air resistance or other forms of damping.

Oscillation Period

The oscillation period \( T \) of a mass-spring system refers to the time it takes for the system to complete one full cycle of motion. Determining the period is important for various applications, from designing springs in mechanical watches to tuning the suspension systems in vehicles.

For a simple mass-spring system, the period depends on two factors: the mass (\( m \)) and the spring constant (\( k \)). The formula for the period is given by \[T = 2\pi\sqrt{\frac{m}{k}}\]where

• \( m \): mass attached to the spring, measured in kilograms (kg).
• \( k \): spring constant, measured in Newtons per meter (N/m).

This formula shows that the period is independent of the amplitude, meaning that large and small oscillations of the same mass-spring system have the same period, as long as the spring's linear range is not exceeded.