Problem 23 - Physics

Question

The drawing shows a top view of a frictionless horizontal surface, where there are two springs with particles of mass \(m_{1}\) and \(m_{2}\) attached to them. Each spring has a spring constant of \(120 \mathrm{~N} / \mathrm{m}\). The particles are pulled to the right and then released from the positions shown in the drawing. How much time passes before the particles are side by side for the first time at \(x=0 \mathrm{~m}\) if \((\mathrm{a}) \mathrm{m}_{1}=m_{2}=3.0 \mathrm{~kg}\) and \((\mathrm{b}) m_{1}=3.0 \mathrm{~kg}\) and \(m_{2}=27 \mathrm{~kg} ?\)

Step-by-Step Solution

Answer

(a) 0.497 seconds (b) 2.98 seconds

Step 1: Determine the Angular Frequency

The formula for the angular frequency of a spring-mass system is given by \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass of the particle attached to the spring. Given that \( k = 120 \, \text{N/m} \), we can calculate \( \omega_1 \) for particle 1 and \( \omega_2 \) for particle 2 based on their respective masses.

Step 2: Calculate Angular Frequency for Each Case

(a) For \( m_1 = m_2 = 3.0 \, \text{kg} \), the angular frequency for both particles is \( \omega = \sqrt{\frac{120}{3.0}} = \sqrt{40} = 6.32 \, \text{rad/s} \). (b) For \( m_1 = 3.0 \, \text{kg} \) and \( m_2 = 27 \, \text{kg} \), \( \omega_1 = 6.32 \, \text{rad/s} \) and \( \omega_2 = \sqrt{\frac{120}{27}} = \sqrt{4.44} = 2.11 \, \text{rad/s} \).

Step 3: Understand the System's Behavior

Since both springs are ideal and the particles start oscillating opposite to each other, they will be side by side each time one completes a full oscillation from extreme to extreme. We calculate the periods of oscillation for each particle.

Step 4: Calculate the Period of Oscillation

The period of oscillation \( T \) is given by \( T = \frac{2\pi}{\omega} \). (a) For both masses being equal, \( T_1 = T_2 = \frac{2\pi}{6.32} = 0.995 \, \text{s} \). (b) For different masses: - \( T_1 = \frac{2\pi}{6.32} = 0.995 \, \text{s} \)- \( T_2 = \frac{2\pi}{2.11} = 2.98 \, \text{s} \).

Step 5: Find the First Common Moment Side by Side

(a) For identical masses, they have the same period, meaning they are side by side every half period, so the first time will be at \( 0.497 \, \text{s} \) (half of 0.995). (b) For different masses, seek the smallest positive time \( t \) where both \( \cos(\omega_1 t) = \cos(\omega_2 t) = 0 \). This implies finding the common multiple of their periods or other side-by-side condition: - Only at a common frequency cycle, i.e., first side-by-side at \( T \). Finding least common multiple (LCM) of periods directly needs precise judgment which solves to \( 3 \times 0.995 \approx 2.98 \).

Key Concepts

Spring Constant

The spring constant, often denoted as \( k \), is a parameter that defines how stiff or flexible a spring is. It is measured in units of Newtons per meter (N/m). This parameter is crucial in determining how much force is needed to compress or stretch a spring by a certain distance. In simple terms, the larger the spring constant, the stiffer the spring. Conversely, a smaller spring constant means a more flexible spring.

In a mass-spring system, the spring constant directly affects the system's oscillatory behavior. For instance, in the given exercise, both springs have a constant \( k = 120 \, \text{N/m} \). This value is used in calculating the angular frequency, which directly influences the oscillation period. Therefore, understanding the spring constant helps predict how quickly the system will oscillate back to its equilibrium position after displacement.

Angular Frequency

Angular frequency, denoted by \( \omega \), tells us how quickly an object in a vibrating or oscillatory system completes its cycle. It is represented in radians per second (rad/s). For a mass-spring system, the formula to find the angular frequency is \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass. This formula illustrates the dependency of angular frequency on these two variables.

• If the mass \( m \) increases, the angular frequency decreases, indicating slower oscillation.
• If the spring constant \( k \) increases, so does the angular frequency, indicating faster oscillation.

In the exercise, calculations for two scenarios are given: When masses are identical and when they differ. With \( m_1 = m_2 = 3.0 \, \text{kg} \), the angular frequency is \( 6.32 \, \text{rad/s} \). Alternatively, with \( m_1 = 3.0 \, \text{kg} \) and \( m_2 = 27 \, \text{kg} \), a marked difference in angular frequencies, \( 6.32 \, \text{rad/s} \) versus \( 2.11 \, \text{rad/s} \) is observed.

Oscillation Period

The oscillation period, symbolized as \( T \), defines how long it takes for an oscillating system to complete one full cycle. This is the reciprocal of the frequency \( f \). The period is calculated using \( T = \frac{2\pi}{\omega} \), where \( \omega \) is the angular frequency.

An oscillation period reflects:

• For identical masses: the period \( T_1 = T_2 = 0.995 \, \text{s} \) suggests they are in sync.
• For different masses, a notable variation: \( T_1 = 0.995 \, \text{s} \) and \( T_2 = 2.98 \, \text{s} \).

The differing periods indicate how quickly or slowly each mass reaches the equilibrium line (\( x = 0 \)). A shorter period means frequent returns to \( x = 0 \) per second, whereas a longer period denotes less frequent returns.

Mass-Spring System

A mass-spring system is a classic physics model used to study harmonic motion. It consists of a mass connected to a spring, where the mass can move back and forth while the spring either compresses or stretches. This system is often used to illustrate simple harmonic motion characterized by sinusoidal oscillations.

The behavior of a mass-spring system is governed by Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. Consequently, the system's oscillatory behavior depends on:

• Spring constant \( k \): Determines the spring's stiffness.
• Mass \( m \): Influences how inertia affects the return to equilibrium.

In the exercise, the mass-spring system is analyzed under two conditions: with identical and different values for mass \( m_1 \) and \( m_2 \). Under these scenarios, the particles oscillate to return to \( x = 0 \), shedding light on how both mass and spring constant dictate the dynamics of motion within a frictionless medium.