Problem 66 - University Physics with Modern Physics

Question

Two springs, each with \(k=125 \mathrm{~N} / \mathrm{m}\), are hung vertically, and \(1.00-\mathrm{kg}\) masses are attached to their ends. One spring is pulled down \(5.00 \mathrm{~cm}\) and released at \(t=0\); the othen is pulled down \(4.00 \mathrm{~cm}\) and released at \(t=0.300 \mathrm{~s}\). Find the phase difference, in degrees, between the oscillations of the two masses and the equations for the vertical displacements of the masses, taking upward to be the positive direction.

Step-by-Step Solution

Answer

Answer: The phase difference between the oscillations of the two masses is approximately 180.12 degrees. The equations for the vertical displacements of the masses are as follows: \(y_1(t) = 0.050\cdot \sin (\sqrt{125}\cdot t)\) \(y_2(t) = 0.040\cdot \sin (\sqrt{125}\cdot (t - 0.300))\)

Step 1: Determine the angular frequency of the oscillations

Since both springs have the same spring constant and mass, their angular frequency will be the same. To calculate the angular frequency, we can use the formula: \(\omega = \sqrt{\frac{k}{m}}\) where \(\omega\) is the angular frequency, \(k\) is the spring constant, and \(m\) is the mass. In this case, we have: \(\omega = \sqrt{\frac{125}{1}} = \sqrt{125}\)

Step 2: Write the equations for the vertical displacements of the masses

Now that we have the angular frequency, we can write the equations for the vertical displacements of the masses. The general equation for a simple harmonic oscillator is: \(y(t) = A\cdot \sin (\omega\cdot t + \phi)\) where \(y(t)\) is the vertical displacement at time \(t\), \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase angle. For the first mass, which is pulled down by \(5.00 \text{ cm}\) and released at \(t = 0\), the equation will be: \(y_1(t) = 0.050\cdot \sin (\sqrt{125}\cdot t)\) For the second mass, which is pulled down by \(4.00 \text{ cm}\) and released at \(t = 0.300 \text{ s}\), the equation will be: \(y_2(t) = 0.040\cdot \sin (\sqrt{125}\cdot (t - 0.300))\)

Step 3: Calculate the phase difference between the oscillations of the two masses

To find the phase difference, we need to compare the phase angles of the displacement equations for the two masses. The phase angle of \(y_1(t)\) is \(0\) and the phase angle of \(y_2(t)\) is \(-\sqrt{125}\cdot 0.300\). To find the phase difference in degrees, we can subtract the second phase angle from the first, and then multiply by the conversion factor \(\frac{180}{\pi}\): \(\Delta\phi = (0 - (-\sqrt{125}\cdot 0.300))\cdot \frac{180}{\pi} = 180.12\) Thus, the phase difference between the oscillations of the two masses is approximately \(180.12\) degrees, and the equations for the vertical displacements of the masses are: \(y_1(t) = 0.050\cdot \sin (\sqrt{125}\cdot t)\) \(y_2(t) = 0.040\cdot \sin (\sqrt{125}\cdot (t - 0.300))\)

Key Concepts

Angular Frequency

When studying simple harmonic motion (SHM), the term angular frequency represents the rate at which the object oscillates about the equilibrium position.

It's calculated using the formula \(\omega = \sqrt{\frac{k}{m}}\), where \(\omega\) is the angular frequency, \(k\) is the spring constant, and \(m\) is the mass of the object attached to the spring. In our example, with a spring constant \(k = 125 \text{ N/m}\) and mass \(m = 1.00 \text{ kg}\), the angular frequency is \(\omega = \sqrt{125}\) radians per second.

As an important aspect of oscillation, the angular frequency helps determine how fast the mass moves back and forth, which is crucial for predicting the position and velocity of the mass at any point in time. Understanding angular frequency is key in analyzing various systems in physics, such as pendulums, electrons in atoms, and even the vibration of molecules.

Spring Constant

The spring constant, denoted by \(k\), is a measure of a spring's stiffness. It indicates how much force is needed to extend or compress the spring by a certain distance. Higher values of \(k\) mean a stiffer spring that requires more force for the same displacement.

In our exercise, the spring constant is given as \(125 \text{ N/m}\), which means that it takes 125 newtons of force to stretch or compress the spring by one meter. This constant plays a critical role in determining angular frequency (as discussed previously) and is a fundamental parameter in Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement. Accurately knowing \(k\) is essential for solving any problem involving SHM.

Phase Difference

The phase difference in simple harmonic motion refers to the angular difference between the oscillatory positions of two bodies at any given time. It is a crucial concept when comparing the oscillations of multiple objects.

In the context of our exercise, we can see that the two masses are released at different times, leading to a phase difference in their oscillations. This phase difference can affect how the two objects interact if they are part of the same system, such as in waves or coupled oscillators. In this instance, the phase difference is found by comparing the phase angles of the displacement equations for both masses. Converting the result to degrees gives us an angle that represents how 'out of sync' the two oscillations are. A phase difference of \(180^\circ\) or \(\pi\) radians, for example, means that the two objects are oscillating completely in opposition to each other.

Vertical Displacement

The term vertical displacement in the context of simple harmonic motion describes the distance and direction an object moves from the equilibrium position along the vertical axis.

The equation for vertical displacement is \(y(t) = A\cdot \sin(\omega\cdot t + \phi)\), where \(A\) is the amplitude of the oscillation, \(\omega\) is the angular frequency, \(t\) is time, and \(\phi\) is the phase angle. In our example, one mass has a displacement equation of \(y_1(t) = 0.050\cdot \sin(\sqrt{125}\cdot t)\) and the other \(y_2(t) = 0.040\cdot \sin(\sqrt{125}\cdot (t - 0.300))\), reflecting the initial conditions of how far they are pulled down and the time they are released. Understanding this concept helps us visualize the movement of the mass in SHM and calculate its position at any time during the oscillation.