Problem 4 - AP Physics 1 with 2 Practice Test 2nd

Question

A mass of \(0.5 \mathrm{~kg}\) is connected to a massless spring with a force constant \(k\) of \(50 \mathrm{~N} / \mathrm{m}\). The system is oscillating on a frictionless, horizontal surface. If the amplitude of the oscillations is \(2 \mathrm{~cm},\) the total energy of the system is (A) \(0.01 \mathrm{~J}\) (B) \(0.1 \mathrm{~J}\) (C) \(0.5 \mathrm{~J}\) (D) \(0.3 \mathrm{~J}\)

Step-by-Step Solution

Answer

Option (A) 0.01 J

Step 1 - Understanding the Problem

We need to find the total energy of a mass-spring system oscillating on a frictionless, horizontal surface given the mass, spring constant, and amplitude of oscillation.

Step 2 - Identify Key Formulas

The total energy in a simple harmonic oscillator is given by the formula: \[ E = \frac{1}{2} k A^2 \]where \(k\) is the spring constant and \(A\) is the amplitude.

Step 3 - Convert Amplitude Units

The amplitude is given as 2 cm. First, convert this to meters: 2 cm = 0.02 m

Step 4 - Substitute Values into the Formula

Now, substitute the given values into the formula: \[ E = \frac{1}{2} \times 50 \times (0.02)^2 \]

Step 5 - Calculation

Compute the value: \[ E = \frac{1}{2} \times 50 \times 0.0004 \] \[ E = 0.01 \, \text{J} \]

Step 6 - Choose the Correct Answer

The total energy of the system is 0.01 J, which corresponds to option (A).

Key Concepts

harmonic oscillator

A harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force proportional to the displacement. In a mass-spring system, this means that if you pull the mass to one side and let go, it will oscillate back and forth around its equilibrium position. The oscillations are due to the restoring force exerted by the spring, which follows Hooke's law. This can be observed in many everyday situations, such as a playground swing or a vibrating guitar string. Each time the object oscillates, it moves through its cycle with consistent behavior, assuming no friction or energy loss occurs.
The energy in a harmonic oscillator is spread between kinetic energy and potential energy. As the mass moves through its equilibrium point, its speed is highest (max kinetic energy) and the potential energy is at its lowest. Conversely, at the extreme points of its motion (maximum displacement), its speed is zero, and the potential energy is at its highest due to the compressed or stretched spring.

spring constant

The spring constant, denoted by the symbol \(k\), is a measure of the stiffness of a spring. It defines the relationship between the force applied to a spring and the displacement caused by this force. According to Hooke's Law, the force \(F\) needed to stretch or compress a spring by a distance \(x\) is given by: \[ F = kx \] where: \(F\) is the force exerted by or on the spring, \(k\) is the spring constant (in N/m), and \(x\) is the distance the spring is stretched or compressed from its equilibrium position.
In our mass-spring example, the given spring constant is 50 N/m. This means that for every meter you stretch or compress the spring, it exerts a force of 50 Newtons. Therefore, a higher spring constant denotes a stiffer spring that is harder to stretch or compress, whereas a lower spring constant indicates a more flexible spring.

amplitude of oscillation

The amplitude of oscillation is the maximum distance the mass moves from its equilibrium position. It's a crucial variable in measuring the extent of the motion in a harmonic oscillator. In our example, the amplitude is given as 2 cm, which we converted to 0.02 meters for the energy calculation. When calculating energy, it's essential to ensure all units are consistent. The amplitude affects how much energy is stored in the system, as the energy is directly proportional to the square of the amplitude: \[ E \propto A^2 \] This means that if you were to double the amplitude, the system's total energy would increase by a factor of four. Therefore, even small changes in amplitude can significantly impact the energy within the system.

energy calculation

To find the total energy of a mass-spring system, we use the formula for the potential energy stored in a harmonic oscillator: \[ E = \frac{1}{2} k A^2 \] Here, \(E\) is the total energy, \(k\) is the spring constant, and \(A\) is the amplitude. In the given exercise: \1. The spring constant \(k\) is 50 N/m.
2. The amplitude \(A\) is 0.02 meters (after converting from 2 cm).
Plug these values into the formula: \[E = \frac{1}{2} \times 50 \times (0.02)^2 \] Calculate the value inside the parenthesis first: \[ (0.02)^2 = 0.0004 \] Now, multiply this by the spring constant and the coefficient: \[ E = \frac{1}{2} \times 50 \times 0.0004 = 0.01 \] Therefore, the total energy of the system is 0.01 Joules. This calculation helps us understand how different factors like the spring constant and amplitude affect the energy of an oscillating system.