Question

A 0.500-kg glider, attached to the end of an ideal spring with force constant \(k =\) 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x = -\)0.015 m; (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x = -\)0.015 m; (e) the total mechanical energy of the glider at any point in its motion.

Short answer

(a) Maximum speed ≈ 1.80 m/s; (b) Speed at x = -0.015 m ≈ 1.69 m/s; (c) Maximum acceleration ≈ 36 m/s²; (d) Acceleration at x = -0.015 m ≈ -13.5 m/s²; (e) Total mechanical energy ≈ 0.36 J.

Step-by-step solution

Identify Known Values and Formulas

We have a glider with mass \( m = 0.500 \) kg, a spring constant \( k = 450 \) N/m, amplitude \( A = 0.040 \) m, and position \( x = -0.015 \) m. In SHM, the total mechanical energy \( E \) is \( \frac{1}{2} k A^2 \). The maximum speed \( v_{max} \) is \( \omega A \), where \( \omega = \sqrt{\frac{k}{m}} \). Acceleration \( a \) is \( \omega^2 x \).

Calculate Maximum Speed

The angular frequency \( \omega \) is \( \sqrt{\frac{k}{m}} = \sqrt{\frac{450}{0.500}} \). Calculate \( \omega \) and use it to find \( v_{max} = \omega A \).

Calculate Speed at Given Position

Use the formula \( v = \omega \sqrt{A^2 - x^2} \) with \( \omega \) known and \( x = -0.015 \) m to find the speed of the glider at this position.

Calculate Maximum Acceleration

Maximum acceleration \( a_{max} = \omega^2 A \). Use the known values of \( \omega \) and \( A \) to calculate \( a_{max} \).

Calculate Acceleration at Given Position

Use the formula \( a = \omega^2 x \) with \( x = -0.015 \) m and \( \omega \) calculated earlier to find the acceleration at this position.

Calculate Total Mechanical Energy

The total mechanical energy is constant at any point during the motion and is given by \( E = \frac{1}{2} k A^2 \). Substitute the values of \( k \) and \( A \) to find \( E \).

Key concepts

Spring Constant

The spring constant, often denoted as \(k\), is a measure of a spring's stiffness. In simple harmonic motion (SHM), the spring constant determines how strong the spring force is. The spring force is direct evidence of Hooke's law, which states that the force exerted by the spring is proportional to the displacement from its equilibrium position. In mathematical terms, the force \( F \) can be expressed as:\[F = -k x\]where \(x\) is the displacement from the equilibrium position. The negative sign indicates that the force exerted by the spring is always directed opposite to the direction of displacement, seeking to restore the object to equilibrium.

For example, a spring constant \( k = 450 \text{ N/m} \) in our exercise means that for every meter the spring is compressed or extended, it applies a force of 450 Newtons. A larger spring constant indicates a stiffer spring, which is harder to stretch or compress.

Mechanical Energy

In simple harmonic motion, the total mechanical energy is conserved, meaning it remains constant throughout the motion. It is the sum of potential energy stored in the spring and the kinetic energy of the moving mass. The formula for total mechanical energy \(E\) in SHM is:\[E = \frac{1}{2} k A^2\]where \(A\) is the amplitude and \(k\) is the spring constant.
This equation shows that mechanical energy depends directly on the spring constant and the square of the amplitude.

In our example, the spring's potential energy is highest when the glider is at the extremes of its motion (maximum displacement), and kinetic energy is highest when passing through the equilibrium position. Understanding this energy interchange is key to grasping SHM's predictability:

• At maximum displacement (either extreme), velocity is zero, and energy is all potential.
• At equilibrium (amplitude is zero), all energy is kinetic, hence, maximum speed.

Angular Frequency

Angular frequency, denoted \(\omega\), is a crucial concept in simple harmonic motion, describing how quickly the system oscillates. It reflects the rate of change of the phase of the oscillation relative to time. The angular frequency in simple harmonic motion is derived from the equation:\[\omega = \sqrt{\frac{k}{m}}\]where \(k\) is the spring constant and \(m\) is the mass of the glider.
This equation highlights the inverse relationship between mass and frequency.

In essence, the lighter the object or the stiffer the spring, the higher the frequency of oscillation. Angular frequency is measured in radians per second, depicting how many complete oscillations can occur in that time.
In our specific problem, the angular frequency calculated using a mass of 0.500 kg and \(k = 450 \text{ N/m}\) provides the foundation for further calculations, such as maximum speed and acceleration. This concept is vital, as it connects the physical properties of the system with its motion behavior.

Amplitude

Amplitude is one of the defining characteristics of simple harmonic motion. It represents the maximum extent of the oscillation measured from the system's equilibrium position. The amplitude \(A\) in our exercise is 0.040 m, and it directly measures how far the glider moves from its central position.
The larger the amplitude, the further the object moves during each oscillation.

Amplitude is essential because it affects the total mechanical energy within the system, as seen in the formula:\[E = \frac{1}{2} k A^2\]Additionally, it influences calculations for maximum speed and acceleration:

• Maximum Speed: The maximum speed \(v_{max}\) happens as the object passes through the equilibrium, where energy is all kinetic.
• Maximum Acceleration: Acceleration is highest at the maximum amplitude, where the restoring force is the greatest.

Amplitude not only indicates the scale of motion but also enhances understanding of the energy dynamics involved in SHM. The easy recovery of these values from amplitude makes it a convenient and insightful parameter in analyzing harmonic systems.