91Ó°ÊÓ

21\. Massless Spring A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position \(y_{1}\) such that the spring is at its rest length. The object is then released from \(y_{1}\) and oscillates up and down, with its lowest position being \(10 \mathrm{~cm}\) below \(y_{1}\). (a) What is the frequency of the oscillation? (b) What is the speed of the object when it is \(8.0 \mathrm{~cm}\) below the initial position? (c) An object of mass \(300 \mathrm{~g}\) is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object? (d) Relative to \(y_{1}\) where is the new equilibrium (rest) position with both objects attached to the spring?

Step by step solution

01

- Understanding the Problem

Identify the key points of the problem: The object at y1 is at the spring's rest length, the object oscillates below y1, and we need to find frequency, speed at a certain position, the mass of the first object, and the new equilibrium position with added mass.

02

- Calculate the Frequency of Oscillation (a)

Use the information that the object's lowest position is 10 cm (0.1 m) below y1. This distance is the amplitude (A) of the oscillation. The formula for frequency is given by \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \] and the relationship for equilibrium force and spring constant is \[ mg = k \Delta y \]. Use these two equations to find the frequency.

03

- Determine the Speed at a Specific Position (b)

At y = -0.08 m (8 cm below y1), use the conservation of energy principle. Total mechanical energy E in oscillation is:\[ E = \frac{1}{2} k A^2 = \frac{1}{2} k y^2 + \frac{1}{2} m v^2 \]Solve for velocity (v) given k from the force equation and previous step results.

04

- Find the Mass of the First Object (c)

After adding the new mass (300 g or 0.3 kg) and given that this new system oscillates with half the original frequency, use the equation \[ f' =\frac{f}{2} = \frac{1}{2\pi} \sqrt{\frac{k}{m_1 + 0.3}} \]Match it to original frequency to solve for the mass of the first object (m1).

05

- New Equilibrium Position (d)

Determine where the new equilibrium position is by finding where the gravitational force is balanced by the spring force:\[ m_{total}g = k \Delta y' \]where m_total = m1 + 0.3 kg. Solve for \(\Delta y'\) relative to y1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

spring constant

The spring constant, represented by the symbol 'k', is a measure of a spring's stiffness. It tells us how much force is needed to stretch or compress the spring by a specific distance. The larger the spring constant, the stiffer the spring.
In our exercise, the spring constant can be found using the gravitational force and the displacement from the equilibrium position. We use the formula:
\[ mg = k \, \Delta y \]
Where 'm' is the mass of the object, 'g' is the acceleration due to gravity, and 'Δy' is the displacement of the spring from its natural length. With this relationship, we can link the spring's behavior to the object's oscillations.
Understanding the spring constant is key to determining other characteristics of the spring system, such as the frequency of oscillation.

oscillation frequency

The oscillation frequency, denoted as 'f', is the number of oscillations the spring performs per unit time. It is influenced by the mass attached to the spring and the spring constant.
The formula to find the frequency of a spring-mass system is:
\[ f = \frac{1}{2\pi} \, \sqrt{\frac{k}{m}} \]
Where 'k' is the spring constant and 'm' is the mass of the object attached to the spring. This formula tells us that a higher spring constant or a smaller mass will result in a higher frequency of oscillation.
In the exercise, we determined the initial frequency and then how the frequency changes when additional mass is attached to the system. Since the frequency is inversely proportional to the square root of the mass, doubling the mass would result in halving the frequency.

conservation of energy

The conservation of energy principle states that in an isolated system, total mechanical energy remains constant if there are no non-conservative forces (like friction) acting.
In the spring system, the two main forms of energy are kinetic energy (KE) and potential energy (PE). As the object oscillates:

• At the highest or lowest points, the speed is zero, so KE is zero, and PE is maximized.
• At the equilibrium position, speed is maximized, so KE is at its maximum, and PE is minimized.

The total energy E of the system at any point is given by:
\[ E = \frac{1}{2} k A^2 \]
where 'A' is the amplitude of the oscillation. This energy is constant and is the sum of KE and PE at any point in the oscillation:
\[ \frac{1}{2} k A^2 = \frac{1}{2} k y^2 + \frac{1}{2} m v^2 \]
This equation helps us to find the speed of the object at a specific position during its oscillation.

equilibrium position

The equilibrium position is where the net force on the object is zero. For a spring system, it is where the spring force balances the gravitational force.
In our exercise, with only one object attached, the equilibrium position is at the spring's natural length when the object is at rest. When additional mass is added, the new equilibrium position is determined by:
\[ m_{total}g = k \, \Delta y' \]
Here, 'm_total' is the combined mass of the original object and the added mass. This position will be lower than the original equilibrium position due to the increased weight. Finding the new equilibrium position involves balancing the total weight with the restoring force of the spring.

mass and weight

Mass is a measure of the amount of matter in an object and is usually measured in kilograms or grams. Weight, on the other hand, is the force exerted by gravity on an object.
The relationship between mass (m) and weight (W) is given by:
\[ W = mg \]
where 'g' is the acceleration due to gravity (approximately 9.8 m/s² on Earth). In the spring oscillation problem, the mass affects both the gravitational force and the resulting oscillations.
We saw that adding a mass of 300 grams to an existing system required recalculating the frequency and equilibrium position. Understanding mass and weight is crucial as they directly influence how the spring stretches and the system's dynamics.