Problem 29 - College Physics

Question

\(\bullet\bullet\) The equation of motion of a SHM oscillator is \(x=(0.50 \mathrm{~m}) \sin (2 \pi f) t,\) where \(x\) is in meters and \(t\) is in seconds. If the position of the oscillator is at \(x=0.25 \mathrm{~m}\) at \(t=0.25 \mathrm{~s},\) what is the frequency of the oscillator?

Step-by-Step Solution

Answer

The frequency of the oscillator is \( \frac{1}{3} \text{ Hz} \).

Step 1: Understand the Given Equation

The equation of motion for a Simple Harmonic Motion (SHM) oscillator is given as \( x = (0.50 \text{ m}) \sin(2\pi f t) \). This equation describes the position \( x \) in meters as a function of time \( t \) in seconds, and involves the sine of the product of the angular frequency \( 2\pi f \) and time \( t \). Here, the amplitude \( A \) is 0.50 m.

Step 2: Use Given Values in the Equation

We are given that the position \( x = 0.25 \text{ m} \) when \( t = 0.25 \text{ s} \). Substitute these values into the equation: \[ 0.25 = 0.50 \sin(2 \pi f \times 0.25) \]

Step 3: Solve for Sine Component

To isolate the sine term, divide both sides of the equation by the amplitude:\[ \sin(2 \pi f \times 0.25) = \frac{0.25}{0.50} = 0.50 \]

Step 4: Analyze the Sine Equation

We now need to find the angle \( \theta \) such that \( \sin(\theta) = 0.50 \). From trigonometry, \( \theta \) can be \( \frac{\pi}{6} \) or \( \frac{5\pi}{6} \) among other solutions due to periodicity.

Step 5: Calculate Frequency for Principal Solution

Consider the principal solution, \( \theta = \frac{\pi}{6} \). Since \( \theta = 2\pi f \times 0.25 \), solve for \( f \):\[ 2\pi f \times 0.25 = \frac{\pi}{6} \]\[ f = \frac{\frac{\pi}{6}}{2\pi \times 0.25} = \frac{1}{3} \text{ Hz} \]

Step 6: Verify Solution with Other Sine Solutions

The next possible angle \( \theta = \frac{5\pi}{6} \). Solve similarly:\[ 2\pi f \times 0.25 = \frac{5\pi}{6} \]\[ f = \frac{\frac{5\pi}{6}}{2\pi \times 0.25} = \frac{5}{3} \text{ Hz} \]Since this frequency edge is less practical due to periodicity handling (oscillations within one period), \( \frac{1}{3} \text{ Hz} \) is correct.

Key Concepts

Equation of Motion

The concept of an "equation of motion" in the context of Simple Harmonic Motion (SHM) is fundamental. For an SHM oscillator, the equation is usually written as \( x = A \sin(\omega t) \) or \( x = A \sin(2\pi ft) \), where:

• \(x\) is the displacement at any given time \(t\).
• \(A\) represents the amplitude, the maximum displacement from the equilibrium position.
• \(\omega\) is the angular frequency, and for SHM, it relates to normal frequency \(f\) as \(\omega = 2\pi f\).

In our exercise, the given equation is \( x = (0.50 \text{ m}) \sin(2\pi f t) \). This tells us that the oscillator swings back and forth symmetrically around its rest position. Here, the amplitude is 0.50 meters, meaning the farthest the oscillator travels from the center is 0.50 meters. The time \(t\) is in seconds, which is standard in physics for time-dependent equations. This equation perfectly describes how the position changes harmonically over time. Understanding this equation helps in predicting the behavior of oscillators like pendulums or springs.

Frequency Calculation

Frequency calculation is a key element when dealing with SHM and gives insights into how often a repeating event, like oscillation, occurs. To find the frequency \(f\), we manipulate the SHM equation provided. Given:

• Position \(x = 0.25 \text{ m}\)
• Time \(t = 0.25 \text{ s}\)

Substitute these into the equation, \(0.25 = 0.50 \sin(2\pi f \times 0.25)\). By solving for \(\sin(2\pi f \times 0.25)\), divide both sides by the amplitude 0.50, yielding \(\sin(2\pi f \times 0.25) = 0.50\). Knowing that \(\sin\) yields 0.50 at angles \(\theta = \frac{\pi}{6}\) or \(\frac{5\pi}{6}\), we solve \(2\pi f \times 0.25 = \frac{\pi}{6}\) for \(f\). This provides \(f = \frac{1}{3} \text{ Hz}\). This means the oscillator swings back and forth one third of a cycle per second. Ensuring the angle corresponds precisely is essential for physically realistic results.

Trigonometry in Physics

Utilizing trigonometry in physics allows us to work with oscillatory systems efficiently. Sine and cosine functions are especially useful to express periodic motions. For the SHM equation \( x = (0.50 \text{ m}) \sin(2\pi f t) \), the sine component determines how the displacement changes over time. Trigonometry tells us the angle where \(\sin\) becomes 0.50 can be found using basic trigonometric identity knowledge:

• \(\theta = \frac{\pi}{6}\)
• \(\theta = \frac{5\pi}{6}\)

These angles relate to the unit circle at specific positions where the sine value reaches 0.50. Employing this understanding helps identify valid angle values to find frequencies or phases in motion equations intuitively. In physics, this integration allows us to handle various conditions and changes within harmonic systems easily, such as phase differences or time shifts in oscillators, making the application of trigonometry invaluable.