Problem 66 - University Physics with Modern Physics

Question

An object is undergoing SHM with period 0.300 s and amplitude 6.00 \(\mathrm{cm} .\) At \(t=0\) the object is instantaneously at rest at \(x=6.00 \mathrm{cm} .\) Calculate the time it takes the object to go from \(x=6.00 \mathrm{cm}\) to \(x=-1.50 \mathrm{cm} .\)

Step-by-Step Solution

Answer

The time is approximately 0.087 seconds.

Step 1: Understand Simple Harmonic Motion (SHM)

An object in SHM moves in a sinusoidal pattern described by its amplitude, period, and angular frequency. The motion occurs with a specific period, the time taken for a full cycle, and amplitude, the maximum displacement from equilibrium.

Step 2: Write the Displacement Equation

For SHM, the displacement from equilibrium can be modeled by the equation: \[ x(t) = A \cos(\omega t + \phi) \]where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant. Given that at \(t=0\), \(x = 6.00\) cm and \(\text{velocity} = 0\), the equation simplifies since \(\phi = 0\).

Step 3: Calculate Angular Frequency

The angular frequency \(\omega\) is related to the period \(T\) by:\[ \omega = \frac{2\pi}{T} \]Given the period \(T = 0.300\) s, calculate:\[ \omega = \frac{2\pi}{0.300} \approx 20.94\, \text{rad/s} \]

Step 4: Solve for the Time to Reach -1.50 cm

Set up the equation using the displacement formula:\[ -1.50 = 6.00 \cos(20.94 t) \]Solve for \(t\) by isolating the cosine function:\[ \cos(20.94 t) = -0.25 \]Find \(t\) by using the inverse cosine function to determine:\[ 20.94 t = \cos^{-1}(-0.25) \approx 1.823 \text{ radians} \] and then divide by \(\omega\) to find \(t\):\[ t \approx \frac{1.823}{20.94} \approx 0.087 \text{ seconds} \]

Key Concepts

Displacement Equation in Simple Harmonic Motion

The displacement equation is crucial to understanding Simple Harmonic Motion (SHM). This equation is used to describe the position of a particle at any given time during its motion. In SHM, the displacement from the equilibrium position can be represented as:

• \( x(t) = A \cos(\omega t + \phi) \)

Here:

• \( x(t) \) is the displacement at time \( t \).
• \( A \) is the amplitude, or the maximum displacement from the equilibrium position.
• \( \omega \) is the angular frequency.
• \( \phi \) is the phase constant, which adjusts the position of the wave at \( t = 0 \).

In situations where the object is at the amplitude position at \( t=0 \), like the problem where the object is at rest at \( x=6.00 \) cm, the phase constant \( \phi \) is generally zero. This simplifies our displacement equation to:

• \( x(t) = A \cos(\omega t) \)

This simplification is helpful as it makes the calculations straightforward when analyzing the motion of the object.

Understanding Angular Frequency

Angular frequency, denoted by \( \omega \), is an essential component of the displacement equation in Simple Harmonic Motion. It is a measure of how quickly the object oscillates back and forth:

• \( \omega = \frac{2\pi}{T} \)

where \( T \) is the period of the oscillation, which is the time it takes to complete one full cycle. In this exercise, the period \( T \) is given as 0.300 seconds, thus allowing us to calculate the angular frequency as:

• \( \omega = \frac{2\pi}{0.300} \approx 20.94 \) radians per second.

Understanding the angular frequency helps us predict the speed of oscillation. It is a cornerstone for solving problems related to how long it takes for an object in SHM to reach various points in its cycle.

The Role of Amplitude in SHM

Amplitude plays a critical role in Simple Harmonic Motion as it represents the maximum extent of the motion. In the equation \( x(t) = A \cos(\omega t + \phi) \), the amplitude \( A \) determines the furthest displacement from the central position:

• Amplitude \( A \) is the peak value of displacement and is measured from the equilibrium position to the highest or lowest point of the motion.
• It defines how "large" the oscillations are without influencing the frequency or period.

In the given problem, the amplitude is 6.00 cm, indicating that the object moves 6.00 cm away from the equilibrium to both its maximum and minimum points. This information is particularly useful when solving for how far the object moves at different times, such as understanding the trajectory from 6.00 cm to -1.50 cm during SHM.