Problem 18 - University Physics with Modern Physics

Question

\(\mathrm{A} 0.500\) -kg mass on a spring has velocity as a function of time given by \(v_{x}(t)=-(3.60 \mathrm{cm} / \mathrm{s}) \sin \left[\left(4.71 \mathrm{s}^{-1}\right) t-\pi / 2\right]\) What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

Step-by-Step Solution

Answer

(a) 1.334 s, (b) 0.764 cm, (c) 16.95 cm/s², (d) 11.10 N/m.

Step 1: Identify given function details

We have the velocity function \(v_{x}(t) = -(3.60 \ \mathrm{cm/s}) \sin\left((4.71 \ \mathrm{s^{-1}})t - \pi/2\right)\). This function is of the form \(v(t) = -v_{max} \sin(\omega t + \phi)\) where \(v_{max} = 3.60 \ \mathrm{cm/s}\), \(\omega = 4.71 \ \mathrm{s^{-1}}\), and \(\phi = -\pi/2\)."

Step 2: Calculate the period

The angular frequency \(\omega\) is related to the period \(T\) by the equation \(\omega = \frac{2\pi}{T}\). So, \(T = \frac{2\pi}{\omega} = \frac{2\pi}{4.71} \approx 1.334 \ \mathrm{s}\).

Step 3: Determine the amplitude

The maximum velocity \(v_{max} = \omega A\), where \(A\) is the amplitude. Therefore, \(A = \frac{v_{max}}{\omega} = \frac{3.60 \ \mathrm{cm/s}}{4.71 \ \mathrm{s^{-1}}} \approx 0.764 \ \mathrm{cm}\).

Step 4: Find the maximum acceleration

Maximum acceleration \(a_{max}\) is given by \(a_{max} = \omega^2 A\). Substituting \(\omega = 4.71 \ \mathrm{s^{-1}}\) and \(A = 0.764 \ \mathrm{cm}\), we have \(a_{max} = (4.71)^2 \times 0.764 \ \mathrm{cm/s^2} \approx 16.95 \ \mathrm{cm/s^2}\).

Step 5: Calculate the force constant of the spring

The force constant \(k\) is related to \(\omega\) and mass \(m\) by \(\omega = \sqrt{\frac{k}{m}}\). Thus, \(k = m\omega^2\). Substituting \(m = 0.500 \ \mathrm{kg}\) and \(\omega = 4.71 \ \mathrm{s^{-1}}\), we find \(k = 0.500 \times (4.71)^2 \approx 11.10 \ \mathrm{N/m}\).

Key Concepts

Angular Frequency

Let's delve into angular frequency, symbolized by \( \omega \). It is a measure of how quickly an object undergoing simple harmonic motion oscillates. Angular frequency is expressed in radians per second - Imagine the swinging motion of a pendulum. Angular frequency tells us how fast it goes back and forth.- Learn how to calculate angular frequency using the period \( T \) (the time it takes for one full cycle):\[ \omega = \frac{2\pi}{T} \]- It's important to note that a higher angular frequency means a faster oscillation. In our given problem, we determined that \( \omega = 4.71 \ \mathrm{s^{-1}} \). This tells us that the oscillation completes quickly, in roughly 1.334 seconds, per full cycle.

Amplitude

Amplitude, denoted as \( A \), is the peak value of an oscillation. It's the furthest distance the object travels from its resting position.- In simple terms, think of amplitude as the height of a wave in analogy to the water waves - For mechanical systems like a mass attached to a spring, amplitude measures how far the mass can stretch from its equilibrium point.- For velocity in harmonic motion, the maximum velocity \( v_{max} \) relates to amplitude via \( v_{max} = \omega A \).From our problem, we calculated the amplitude as \( A = 0.764 \ \mathrm{cm} \). This small amplitude indicates a gentle motion rather than a vigorous one, emphasizing subtle oscillations.

Force Constant

Consider the force constant \( k \), which is a measure of how stiff or strong a spring is. In the realm of oscillations, \( k \) defines the restoring force's effectiveness in Newton's law for springs, expressed by Hooke's law: \( F = -kx \).- If the force constant is high, the spring is stiff and hard to stretch.- It is directly related to angular frequency and mass: \[ \omega = \sqrt{\frac{k}{m}} \]- The relationship shows that with a greater force constant, there is a greater angular frequency for the same mass.For the given problem, \( k \) was found to be approximately \( 11.10 \ \mathrm{N/m} \), indicating the spring provides a moderately strong restoring force against displacement.

Maximum Acceleration

Maximum acceleration represents how fast an object can accelerate as it moves through its oscillation.- It's the peak acceleration experienced by the mass when it is at its two endpoints (maximum displacement) during the motion.- In simple harmonic motion, acceleration is directly tied to both angular frequency and amplitude: \[ a_{max} = \omega^2 A \]- A greater angular frequency or amplitude means more energy to accelerate quickly, resulting in a larger maximum acceleration.For our oscillating mass, the maximum acceleration was calculated as approximately \( 16.95 \ \mathrm{cm/s^2} \), showing how swiftly it can move from the peak of its motion back to the equilibrium.