Problem 6 - Fundamentals of Physics

Question

A \(0.50 \mathrm{~kg}\) body oscillates in SHM on a spring that, when extended \(2.0 \mathrm{~mm}\) from its equilibrium position, has an \(8.0 \mathrm{~N}\) restoring force. What are (a) the angular frequency of oscillation, (b) the period of oscillation, and (c) the capacitance of an \(L C\) circuit with the same period if \(L\) is \(5.0 \mathrm{H} ?\)

Step-by-Step Solution

Answer

(a) \( \approx 89.44 \text{ rad/s} \), (b) \( \approx 0.0702 \text{ s} \), (c) \( \approx 7.90 \times 10^{-5} \text{ F} \).

Step 1: Calculate Spring Constant

We start by applying Hooke's Law for springs, which states that the force exerted by the spring, \( F \), is directly proportional to the displacement \( x \) and can be expressed as \( F = kx \). Here \( F = 8.0 \text{ N} \) and \( x = 2.0 \text{ mm} = 0.002 \text{ m} \). Therefore, the spring constant \( k \) is given by:\[ k = \frac{F}{x} = \frac{8.0 \text{ N}}{0.002 \text{ m}} = 4000 \text{ N/m} \]

Step 2: Find Angular Frequency

The angular frequency \( \omega \) of a mass-spring system in simple harmonic motion is given by the formula \( \omega = \sqrt{\frac{k}{m}} \), where \( m \) is the mass of the body. With \( k = 4000 \text{ N/m} \) and \( m = 0.50 \text{ kg} \), we find:\[ \omega = \sqrt{\frac{4000 \text{ N/m}}{0.50 \text{ kg}}} = \sqrt{8000 \text{ s}^{-2}} = \sqrt{8.0 \times 10^3} \approx 89.44 \text{ rad/s} \]

Step 3: Calculate Period of Oscillation

The period \( T \) of oscillation is the reciprocal of the frequency, \( T = \frac{2\pi}{\omega} \). Using the angular frequency we found, \( \omega = 89.44 \text{ rad/s} \):\[ T = \frac{2\pi}{89.44} \approx 0.0702 \text{ seconds} \]

Step 4: Determine Capacitance in LC Circuit

For an LC circuit, the period \( T \) is given by the formula \( T = 2\pi \sqrt{LC} \). Rearranging for capacitance \( C \) and substituting \( T = 0.0702 \text{ s} \) and \( L = 5.0 \text{ H} \):\[ C = \left(\frac{T}{2\pi}\right)^2 \cdot \frac{1}{L} = \left(\frac{0.0702}{2\pi}\right)^2 \cdot \frac{1}{5.0} \approx 7.90 \times 10^{-5} \text{ F} \]

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle of physics that describes how springs work. It tells us that the force exerted by a spring is directly proportional to its displacement. Simply put, the further you stretch or compress a spring, the harder it pulls or pushes back to return to its original position. The formula for Hooke's Law is:

• Force (F) = spring constant (k) × displacement (x)

In this exercise, we have a force of 8.0 N when the spring is stretched 2.0 mm. To find the spring constant (k), we rearrange the formula to: \[k = \frac{F}{x}\]This constant tells us how stiff the spring is. A higher spring constant means a stiffer spring.

Spring Constant

The spring constant, represented by k , is a crucial factor in determining the behavior of a spring. It quantifies the stiffness of the spring. We've calculated this value in our example to be 4000 N/m. This means for every meter you stretch or compress the spring, it exerts a force of 4000 newtons in the opposite direction.

• Larger more force
• Smaller less force

Understanding the spring constant can help predict how much a spring will stretch or compress under a certain force, making it very useful in engineering and physics problems.

Angular Frequency

Angular frequency (\omega) is a measure of how quickly an object moves through one complete cycle of motion. In Simple Harmonic Motion (SHM), like that of a spring-mass system, it is calculated using the formula: \[\omega = \sqrt{\frac{k}{m}}\]where k is the spring constant and m is the mass. In this case, the angular frequency is approximately 89.44 rad/s. This shows how swiftly the mass oscillates back and forth through its equilibrium position. Angular frequency is important because it helps us understand the dynamics of oscillating systems.

LC Circuit

An LC circuit, composed of an inductor (L) and a capacitor (C), is another kind of oscillating system. It can resonate at a particular frequency, much like the mass-spring system. The period of oscillation of an LC circuit is given by:\[T = 2\pi \sqrt{LC}\]In this problem, by equating the period of the mechanical oscillator to that of the LC circuit, we can solve for the capacitance (C). Understanding the behavior of LC circuits is essential in electronics, particularly in radios and oscillators, because they can easily be tuned to different frequencies.

Period of Oscillation

The period (T) is the time it takes to complete one entire cycle of oscillation. It can be determined by finding the reciprocal of the angular frequency. Using the formula: \[T = \frac{2\pi}{\omega}\]The period reflects how long it takes for our spring-mass system to move from one extreme, back to the other, and return to its starting point. In our case, the period is approximately 0.0702 seconds. Knowing the period allows us to predict the behavior of oscillating systems, whether they involve springs, pendulums, or even electrical circuits.