Problem 102 - Precalculus

Question

If a sound wave is represented by \(y=0.006 \cos (1000 \pi t) \mathrm{cm},\) what are its amplitude and frequency? See Exercise 99.

Step-by-Step Solution

Answer

The amplitude is 0.006 cm and the frequency is 500 Hz.

Step 1: Identify the Amplitude from the Equation

The general equation for a cosine wave is given by \(y = A \cos(\omega t)\), where \(A\) is the amplitude. In the equation \(y = 0.006 \cos (1000 \pi t)\), we can see that the amplitude \(A\) is 0.006 cm.

Step 2: Identify the Angular Frequency

The term inside the cosine function is given as \(1000 \pi t\). This means the angular frequency \(\omega\) is \(1000 \pi\). Angular frequency \(\omega = 2 \pi f\), where \(f\) is the frequency.

Step 3: Calculate the Frequency

Using the relationship \(\omega = 2 \pi f\), substitute \(\omega = 1000 \pi\) and solve for \(f\). This gives:\[1000 \pi = 2 \pi f\]. Divide both sides by \(2 \pi\) to find \(f\):\[f = \frac{1000 \pi}{2 \pi} = 500\]. So, the frequency \(f\) is 500 Hz.

Key Concepts

Amplitude

In trigonometry, the amplitude of a wave or a periodic function signifies its maximum absolute displacement from the equilibrium position. Think of amplitude as how "tall" the wave is—the higher the amplitude, the stronger or louder the sound, in the case of a sound wave. For a cosine wave expressed by the equation \(y = A \cos(\omega t)\), the amplitude is the coefficient \(A\).

In our given exercise, the wave is represented by \(y = 0.006 \cos(1000 \pi t)\). Therefore, the amplitude \(A\) is 0.006 cm. This tells us that at its maximum, the wave reaches 0.006 cm above and below its middle, or equilibrium, position. The amplitude does not affect the frequency or angular frequency of the wave, but it significantly influences the wave's intensity or energy, especially in sound waves.

Frequency

Frequency refers to how often the wave cycles occur in one second and is measured in Hertz (Hz). If we think about a wave traveling through time, the frequency tells us how many full wave cycles are completed in one second.

For a wave described by \(y = A \cos(\omega t)\), we can determine the frequency from the angular frequency \(\omega\) using the formula \(\omega = 2 \pi f\). Here, \(f\) denotes the frequency. The relationship captures how angular frequency, which pertains to the rate of rotation in circular motion, translates into the linear frequency in wave motion.

• Given the wave \(y = 0.006 \cos(1000 \pi t)\), the angular frequency \(\omega\) is \(1000 \pi\).
• Using \(\omega = 2 \pi f\), we solve for \(f\): \(f = \frac{1000 \pi}{2 \pi} = 500\).
• This means the frequency \(f\) is 500 Hz, which implies the wave completes 500 cycles per second.

Such frequency indicates how fast the oscillations occur, affecting how we perceive the wave, especially in scenarios like sound, where higher frequencies correspond to higher-pitched sounds.

Angular Frequency

Angular frequency, often denoted by the symbol \(\omega\), is a measure of how quickly the wave oscillates in terms of angles. Unlike regular frequency—which counts full cycles in a second—angular frequency measures how many radians (an angular measurement) the wave covers per second.

In the equation \(y = A \cos(\omega t)\), \(\omega\) represents this angular speed. Angular frequency is linked to the normal frequency \(f\) using the relationship \(\omega = 2 \pi f\). This formula shows how changes in angular frequency can affect the standard frequency of the wave motion.

• For the given wave \(y = 0.006 \cos(1000 \pi t)\), \(\omega = 1000 \pi\).
• Knowing \(\omega\), you can calculate the normal frequency, as we did in the previous section.

Understanding angular frequency is vital in fields where phase relationships matter, such as electrical engineering or acoustics, since it provides a more precise understanding of wave motion in radial terms.