Problem 53 - Algebra and Trigonometry

Question

Loudspeaker A loudspeaker diaphragm is oscillating in simple harmonic motion described by the function \(d(t)=a \cos (\omega t)\) with a frequency of 520 hertz (cycles per second) and a maximum displacement of 0.80 millimeter. Find \(\omega\) and then find a function that describes the movement of the diaphragm.

Step-by-Step Solution

Answer

\(\omega = 3267.24\,\text{radians per second}\), \(d(t) = 0.80 \cos(3267.24 t)\)

Step 1: Understand the Given Information

The loudspeaker diaphragm is oscillating in simple harmonic motion described by the function \(d(t) = a \cos(\omega t)\). Given the frequency \(f = 520\,\text{Hz}\) and the maximum displacement \(a = 0.80\,\text{mm}\).

Step 2: Relate Frequency to Angular Frequency

Angular frequency \(\omega\) is related to the frequency \(f\) by the formula \(\omega = 2\pi f\).

Step 3: Calculate Angular Frequency \(\omega\)

Using the formula \[\omega = 2\pi \times 520\,\text{Hz}\], \[\omega = 2\pi \times 520 \approx 3267.24\,\text{radians per second}\].

Step 4: Write the Final Function

Substitute \(a\) and \(\omega\) into the given function. Therefore, the function describing the movement is \[d(t) = 0.80 \cos(3267.24 t)\].

Key Concepts

angular frequency

Angular frequency, often denoted by \( \omega \), is a fundamental concept in simple harmonic motion (SHM). To understand angular frequency, think of it as a way of expressing how fast something oscillates in radians per second. This contrasts with the regular frequency, which is in cycles per second, or Hertz (Hz).
In the problem provided, we know the frequency \( f \) is 520 Hz. By knowing this, we can find the angular frequency using the formula: \[ \omega = 2 \pi f \]
This formula implies that we multiply the frequency by \( 2\pi \) to convert it from cycles per second (Hz) to radians per second. Following the provided step-by-step solution:
\[ \omega = 2 \pi \times 520 = 2 \pi\times 520 \approx 3267.24 \ \text{radians per second} \]
This result serves as the angular speed of the diaphragm's oscillation in the loudspeaker, giving us a more precise understanding of its motion.

displacement function

The displacement function in SHM describes how the position of an oscillating object changes over time. For our loudspeaker problem, the displacement function is given as:
\[ d(t) = a \cos(\omega t) \]
Here, \( d(t) \) represents the displacement at time \( t \), \( a \) is the maximum displacement (or amplitude), and \( \omega \) is the angular frequency.
Plugging in the details from our problem, we have \( a = 0.80 \text{mm} \) as the maximum displacement. From the previous calculation, we found \( \omega \approx 3267.24 \). Thus, the displacement function becomes:
\[ d(t) = 0.80 \cos(3267.24 t) \]
This function describes the movement of the diaphragm over time, oscillating back and forth as a cosine wave with an amplitude of 0.80 mm and a cycle determined by the angular frequency.

frequency calculation

Calculating frequency is crucial to understanding any system in SHM. Frequency (denoted as \( f \)) tells us how many cycles of oscillation occur per second.
The frequency was given directly in the problem: \[ f = 520 \text{Hz} \]
This value tells us the diaphragm completes 520 oscillations every second. However, if we were not given this value, we could still find it based on other information, such as the period (\( T \)), which is the time for one complete cycle. The relationship between period and frequency is:
\[ f = \frac{1}{T} \]
In this loudspeaker problem, we can also translate this frequency to angular frequency using the formula noted earlier: \[ \omega = 2 \pi f \]
Worked this way, understanding frequency calculation helps one easily translate between SHM representations and comprehend the oscillation dynamics better.