Problem 20 - College Physics

Question

\(\bullet\) A 0.50-kg mass oscillates in simple harmonic motion on a spring with a spring constant of \(200 \mathrm{~N} / \mathrm{m}\). What are (a) the period and (b) the frequency of the oscillation?

Step-by-Step Solution

Answer

(a) Period \(T = 0.314\,\text{s}\), (b) Frequency \(f = 3.18\,\text{Hz}\).

Step 1: Understand the Concepts

Simple harmonic motion involves mass oscillating back and forth through equilibrium. Key parameters include mass \(m\), spring constant \(k\), period \(T\), and frequency \(f\).

Step 2: Write the Formula for Period

The period \(T\) of a mass-spring system is given by the formula: \[ T = 2\pi \sqrt{\frac{m}{k}} \] where \(m\) is the mass and \(k\) is the spring constant.

Step 3: Calculate the Period

Substitute the given values into the formula for period: \(m = 0.50\) kg and \(k = 200\, \text{N/m}\). Thus, \[ T = 2\pi \sqrt{\frac{0.50}{200}} \] Calculate the value inside the square root: \[ \frac{0.50}{200} = 0.0025 \] Then, \[ T = 2\pi \sqrt{0.0025} \] Finally, \(\sqrt{0.0025} = 0.05\), so \(T = 2\pi \times 0.05 = 0.314 \text{s}\).

Step 4: Understand Frequency Calculation

Frequency \(f\) is the reciprocal of the period: \[ f = \frac{1}{T} \] Use the period calculated to find the frequency.

Step 5: Calculate the Frequency

Using the period \(T = 0.314 \text{s}\), calculate the frequency: \[ f = \frac{1}{0.314} \approx 3.18 \text{Hz} \]

Step 6: Review and Conclude

We found that the period \(T\) is \(0.314\) seconds and the frequency \(f\) is \(3.18\) Hz. These values satisfy the conditions of the problem as the mass performs simple harmonic motion.

Key Concepts

Mass-Spring System

In simple harmonic motion, a mass-spring system is a classic example that helps illustrate the principle of oscillatory motion. Imagine a spring hanging vertically with a mass attached to its end. This system will naturally move back and forth when it's displaced from its rest position. The spring's restoring force pulls the mass back as it tries to return to equilibrium. This repetitive action is what causes oscillation.

• The mass in the system is the object that oscillates due to the spring's force.
• The spring, characterized by its spring constant, defines how stiff the spring is. A stiffer spring leads to faster oscillations.

This setup is governed by Hooke’s Law, which states that the force exerted by the spring is proportional to the displacement and opposite in direction. The force can be expressed mathematically as:
\[ F = -kx \]
Here, \(F\) is the force, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium.

Oscillation Period

The oscillation period \(T\) represents the time it takes for the mass-spring system to complete one full cycle of motion. It’s measured in seconds. This cycle includes moving from the equilibrium point to the maximum displacement, returning back, moving to the opposite side, and coming back to the starting position.
The period of a mass-spring system in simple harmonic motion can be calculated using the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
In this formula, \(m\) represents the mass, and \(k\) is the spring constant. The formula shows us that:

• A heavier mass \(m\) results in a longer period, meaning it takes more time for a complete cycle.
• A larger spring constant \(k\) leads to a shorter period, which means the oscillations occur more quickly.

Oscillation Frequency

Oscillation frequency, denoted \(f\), indicates how many cycles the mass-spring system completes per second. It is measured in Hertz (Hz), and it tells us about the speed of oscillations or how frequently they occur. If a system has a high frequency, it oscillates many times in a short period.
The frequency is the inverse of the period, determined by the relationship:
\[ f = \frac{1}{T} \]
In this relationship, when the period \(T\) reduces, the frequency \(f\) increases, and vice versa. This inverse relationship helps us understand that:

• A shorter period results in a higher frequency because the system cycles more rapidly.
• If the mass is lighter or the spring constant is larger, it leads to a higher frequency.

Understanding frequency is crucial in scenarios such as tuning musical instruments, where specific frequencies correspond to musical notes.

Spring Constant

The spring constant, denoted by \(k\), is a measure of a spring's stiffness. It tells us how hard it is to stretch or compress the spring. The metric unit for the spring constant is Newtons per meter (N/m).
In a mass-spring system, the spring constant plays a vital role in determining both the period \(T\) and frequency \(f\) of the oscillations.

• A larger spring constant means that the spring is very stiff. It requires more force to change the spring's position, resulting in faster oscillations.
• A smaller spring constant indicates a less stiff spring, making it easier to stretch but leading to slower oscillations.

Mathematically, the spring constant is usually included when calculating the force exerted by the spring using Hooke's Law:
\[ F = -kx \]
Where \(F\) is the force exerted by the spring, \(x\) is the displacement of the mass from its equilibrium position. Understanding the spring constant is fundamental in various applications, from designing shock absorbers in vehicles to ensuring accurate readings in precision scales.