Question

A 2.0 -kg mass attached to a spring is displaced \(8.0 \mathrm{~cm}\) from the equilibrium position. It is released and then oscillates with a frequency of \(4.0 \mathrm{~Hz}\) a) What is the energy of the motion when the mass passes through the equilibrium position? b) What is the speed of the mass when it is \(20 \mathrm{~cm}\) from the equilibrium position?

Short answer

Answer: The energy of the motion when the mass passes through the equilibrium position is approximately 252.9 J. We cannot calculate the speed of the mass when it is 20 cm away from the equilibrium position because the maximum displacement is only 8 cm. It is not possible for the mass to reach a displacement of 20 cm, as shown by the negative kinetic energy result in our calculations.

Step-by-step solution

Calculate the angular frequency (ω)

The angular frequency (ω) of the oscillating mass can be calculated using the given frequency (f) with the formula: \(ω = 2πf\). Using the given frequency (4.0 Hz), we can calculate the angular frequency as follows: $$ω = 2π (4.0) = 8π\,\mathrm{rad/s}$$

Calculate the spring constant (k)

To find the spring constant (k), we can use the formula: \(ω^2 = \frac{k}{m}\). We know the mass (m) and have found the angular frequency (ω). Rearranging the formula to solve for k: $$k = ω^2 m = (8π)^2 (2.0\,\mathrm{kg}) = 128π^2\,\mathrm{N/m}$$

Calculate the energy of the motion (E) when the mass passes through the equilibrium position.

At the equilibrium position, all the energy in the system is kinetic energy (KE). We can determine the total energy (E) of the motion at this point, knowing that the maximum potential energy (PE) occurs at the maximum displacement (x_max): $$PE_{max} = \frac{1}{2} k x_{max}^2$$ Therefore, the energy of the motion when the mass passes through the equilibrium position is: $$E = PE_{max}$$ Plugging in our values, we have: $$E = \frac{1}{2} (128π^2\,\mathrm{N/m}) (0.08\, \mathrm{m})^2 = 25.60π^2\,\mathrm{J} ≈ 252.9\,\mathrm{J}$$

Calculate the potential energy (PE) when the mass is 20 cm from the equilibrium position.

We can calculate the potential energy (PE) at a displacement of 20 cm (0.20 m) using the following equation: $$PE = \frac{1}{2}kx^2$$ Plugging in our values, we get: $$PE = \frac{1}{2} (128π^2\,\mathrm{N/m}) (0.20\, \mathrm{m})^2 = 128π^2\, \mathrm{J}$$

Determine the kinetic energy (KE) when the mass is 20 cm from the equilibrium position

At any given point during the oscillation, the sum of the kinetic energy (KE) and potential energy (PE) will always be equal to the total energy (E). Therefore, we can determine the KE when the mass is 20 cm from the equilibrium position by subtracting the potential energy (PE) at that position from the total energy (E): $$KE = E - PE = 25.60π^2 \,\mathrm{J} - 128π^2 \,\mathrm{J} = -102.40π^2 \,\mathrm{J}$$ However, we got a negative value for KE, indicating an error. This happened because we incorrectly assumed that the displacement of 20 cm is possible. The maximum displacement (x_max) is only 8 cm, which means the mass can never reach a displacement of 20 cm. We must provide an explanation rather than calculating the speed in this case.

Explain the peculiar calculation at the displacement of 20 cm

The mass can never be 20 cm away from the equilibrium position because the maximum displacement (x_max) is only 8 cm. Our calculations in Step 5 resulted in a negative kinetic energy, which is not possible. This confirms that the given displacement of 20 cm is not achievable in this situation.

Key concepts

Angular Frequency

Angular frequency, often represented by the symbol \( \omega \), determines how fast an object oscillates in terms of radians per second (rad/s). It is a crucial parameter for any harmonic oscillator, one that follows repetitive movement, such as a mass attached to a spring. To find the angular frequency, you can use the relation \( \omega = 2\pi f \), where \( f \) is the frequency of the oscillation. For instance, a mass oscillating with a frequency of 4.0 Hz has an angular frequency \( \omega \) calculated as \( 8\pi \mathrm{rad/s} \). Understanding this concept is essential because it links the temporal aspect of oscillatory motion with spatial displacement and force.

Spring Constant

The spring constant \( k \) is a measure of the stiffness of a spring in the context of Hooke's law. It's expressed in Newtons per meter (N/m) and directly affects the system's angular frequency. The stiffer the spring, the larger the spring constant, and the faster the system oscillates when displaced from equilibrium. Using the formula \( \omega^2 = \frac{k}{m} \) and rearranging it to solve for \( k \) gives us a spring constant for a given mass and angular frequency. For a 2.0-kg mass oscillating with an angular frequency of \( 8\pi \mathrm{rad/s} \) the spring constant is worked out to be \( 128\pi^2 \mathrm{N/m} \).

Harmonic Oscillator

A harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force proportional to the displacement, as described by Hooke's law. The mass attached to a spring behaves as a harmonic oscillator. Such systems show periodic motion, which repeats at a constant frequency. The total energy in an ideal harmonic oscillator remains constant, composed of kinetic and potential energy, which interchange as the mass oscillates. At the equilibrium position, the energy is purely kinetic, and energy is primarily potential at maximum displacement.

Kinetic Energy

Kinetic energy (KE) in oscillatory motion is the energy due to the motion of the oscillator. At the equilibrium position, the velocity of the mass is greatest, and thus the kinetic energy is at a maximum. Mathematically, this is given by the formula \( KE = \frac{1}{2} mv^2 \) where \( m \) is the mass of the object and \( v \) is its velocity. It's important to note that during oscillation, kinetic energy converts to potential energy and vice versa, with their sum equating to the total mechanical energy of the system.

Potential Energy

The potential energy (PE) in the context of oscillatory motion is the energy stored due to the position of the mass within the elastic limits of the spring. This energy is highest at maximum displacement, calculated as \( PE_{max} = \frac{1}{2} k x_{max}^2 \). If the mass attached to the spring is displaced 8.0 cm from the equilibrium, we use this formula to determine the potential energy at this position. However, it's crucial to remember that the potential energy cannot be calculated for a displacement greater than the maximum, such as 20 cm in the exercise, since it exceeds the spring's elastic limits, leading to an incorrect negative value for kinetic energy.