Simple Harmonic Motion of Vertical Springs: Videos & Practice Problems

In vertical mass-spring systems, the equilibrium position is where the restoring force equals the weight of the mass, represented by the equation $k\mathrm{\Delta L}=\mathrm{mg}$. The distance the spring stretches to reach equilibrium is $\mathrm{\Delta L}$, while any additional displacement is the amplitude. Understanding these concepts is crucial for analyzing oscillatory motion and energy conservation in spring systems.

Simple Harmonic Motion of Vertical Springs

In vertical mass-spring systems, the behavior is similar to horizontal systems, with the primary distinction being the equilibrium position. When a spring is suspended from a ceiling and a mass is attached, the spring stretches due to the weight of the mass. This stretching creates a new equilibrium position where the restoring force of the spring balances the gravitational force acting on the mass. The original length of the spring is 0.5 meters, and when the mass is attached, it stretches down by a distance known as delta l, which in this case is 0.2 meters. This means the new equilibrium position is at 0.7 meters (0.5 m + 0.2 m).

The equilibrium condition can be expressed with the equation:

$$ k \cdot \Delta l = mg $$

where k is the spring constant, m is the mass, and g is the acceleration due to gravity (approximately 10 m/s²). In this scenario, if the mass is 5 kg, we can rearrange the equation to find the spring constant:

$$ k = \frac{mg}{\Delta l} = \frac{5 \cdot 10}{0.2} = 250 \, \text{N/m} $$

After reaching equilibrium, if the spring is pulled down an additional 0.3 meters, it will oscillate between two amplitudes. The amplitude represents the additional displacement from the equilibrium position, not to be confused with delta l. The total distance from the ceiling to the maximum height of the block can be calculated by considering the equilibrium position and the amplitude. The maximum height occurs when the spring is pulled down 0.3 meters and then released, oscillating back up.

To find the distance from the ceiling to the maximum height, we note that the maximum stretch downward is 0.7 m (equilibrium) + 0.3 m (amplitude) = 1.0 m. When the spring oscillates back up, the maximum height is at 0.7 m - 0.3 m = 0.4 m from the ceiling. Thus, the distance from the ceiling to the maximum height of the block is 0.4 meters.

Understanding these principles of vertical oscillations is crucial for analyzing systems involving springs and masses, as they illustrate the balance of forces and the dynamics of oscillatory motion.

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?

• A. .17 kg; .56 m/s
• B. .63 kg; .29 m/s
• C. .17 kg; 1.12 m/s
• D. .63 kg; .15 m/s

Answer: (A)

Example

In this problem, we analyze the behavior of an elastic cord, similar to a spring, under different weights. Initially, the cord is 65 centimeters long when a weight of 75 Newtons (derived from mass \( m_1 \) and gravitational acceleration \( g \)) is attached. This situation is described by the equilibrium condition:

\( k \cdot \Delta l_1 = m_1 g \)

Here, \( k \) is the spring constant, and \( \Delta l_1 \) represents the change in length of the cord due to the weight. The equilibrium condition indicates that the restoring force exerted by the cord equals the weight acting on it.

When a second weight of 180 Newtons (from mass \( m_2 \)) is added, the cord stretches further, reaching a new length of 85 centimeters. The equilibrium condition for this scenario is:

\( k \cdot \Delta l_2 = m_2 g \)

To clarify, \( \Delta l \) is defined as the final length minus the initial length of the cord. Thus, we can express the changes in length as:

\( \Delta l_1 = l_{f1} - l_i \quad \text{and} \quad \Delta l_2 = l_{f2} - l_i \

Given that \( l_{f1} = 0.65 \) m and \( l_{f2} = 0.85 \) m, we can set up the equations:

\( k \cdot (0.65 - l_i) = 75 \quad \text{(1)} \

and

\( k \cdot (0.85 - l_i) = 180 \quad \text{(2)} \

By distributing \( k \) in both equations, we have:

\( 0.65k - kl_i = 75 \quad \text{(3)} \

and

\( 0.85k - kl_i = 180 \quad \text{(4)} \

To eliminate the unknown \( l_i \), we can subtract equation (3) from equation (4):

\( (0.85k - kl_i) - (0.65k - kl_i) = 180 - 75 \

This simplifies to:

\( 0.20k = 105 \

Solving for \( k \), we find:

\( k = \frac{105}{0.20} = 525 \, \text{N/m} \)

This value represents the spring constant of the elastic cord, indicating its stiffness. Understanding these relationships between force, displacement, and the spring constant is crucial in mechanics, particularly in analyzing elastic materials.

A chair of mass 30 kg on top of a spring oscillates with a period of 2s. (a) Find the spring's force constant. You place an object on top of the chair, and it now oscillates with a period of 3s. (b) Find the object's mass.

• A. 296 N/m; 1286.9 kg
• B. 188 N/m; 42.9 kg
• C. 188 N/m; 817.4 kg
• D. 296 N/m; 38.1 kg

Answer: (D)

Here’s what students ask on this topic:

What is the equilibrium position in a vertical mass-spring system?

In a vertical mass-spring system, the equilibrium position is where the restoring force of the spring equals the weight of the attached mass. This is represented by the equation:

$k\mathrm{\Delta L}=mg$

Here, $k$ is the spring constant, $\mathrm{\Delta L}$ is the distance the spring stretches to reach equilibrium, $m$ is the mass, and $g$ is the acceleration due to gravity. At this position, the forces are balanced, and the system is in a state of rest.

How do you calculate the spring constant in a vertical mass-spring system?

To calculate the spring constant $k$ in a vertical mass-spring system, you can use the equilibrium condition where the restoring force equals the weight of the mass:

$k\mathrm{\Delta L}=mg$

Rearranging for $k$, we get:

$k=\frac{mg}{\mathrm{\Delta L}}$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $\mathrm{\Delta L}$ is the distance the spring stretches to reach equilibrium. By substituting the known values, you can solve for $k$.

What is the difference between the equilibrium position and amplitude in a vertical spring system?

The equilibrium position in a vertical spring system is the point where the restoring force of the spring equals the weight of the attached mass, represented by $k\mathrm{\Delta L}=mg$. This is the natural hanging distance of the spring.

Amplitude, on the other hand, is the additional displacement from the equilibrium position when the spring is pulled or pushed. It represents the maximum extent of oscillation on either side of the equilibrium position. Unlike $\mathrm{\Delta L}$, which is the static stretch due to the mass, amplitude is the dynamic displacement during oscillation.

How do you determine the maximum height of a block in a vertical spring system?

To determine the maximum height of a block in a vertical spring system, you need to consider the original length of the spring, the equilibrium position, and the amplitude of oscillation. The equilibrium position is given by:

$k\mathrm{\Delta L}=mg$

Let's denote the original length of the spring as $L$ and the equilibrium length as $L$_eq. The maximum height is reached when the block oscillates to its highest point, which is $L$_eq minus the amplitude $A$. Therefore, the distance from the ceiling to the block at maximum height is:

$d=L$_eq-A

What is the role of the spring constant in vertical oscillations?

The spring constant $k$ plays a crucial role in vertical oscillations as it determines the stiffness of the spring. A higher spring constant means a stiffer spring, which requires more force to stretch or compress. In the context of vertical oscillations, the spring constant is used to calculate the equilibrium position using the equation:

$k\mathrm{\Delta L}=mg$

It also affects the frequency of oscillation, with a stiffer spring resulting in a higher frequency. The spring constant is essential for understanding the dynamics of the system and predicting its behavior under various conditions.