Question

A plane progressive wave is given by \( y = 2 \cos 2\pi(330t - x) \, \text{m} \). The frequency of the wave is:

• A. 165 Hz
• B. 330 Hz
• C. 660 Hz
• D. 340 Hz

Answer 1

The Correct Option is B

To determine the frequency of the wave given by the equation \( y = 2 \cos 2\pi(330t - x) \, \text{m} \), we need to analyze the equation and extract the relevant parameters of a wave.

The standard form of a plane progressive wave is:

\(y = A \cos(2\pi f t - \frac{2\pi}{\lambda} x)\)

Where:

• \(A\) is the amplitude of the wave.
• \(f\) is the frequency of the wave.
• \(\lambda\) is the wavelength of the wave.

Given the wave equation:

\(y = 2 \cos 2\pi(330t - x) \, \text{m}\)

By comparing this with the standard form, we can identify the term \(2\pi f t\) in the given equation as \(2\pi \times 330 \times t\). Hence, the frequency \(f\) is 330 Hz.

Therefore, the correct frequency of the wave is 330 Hz.

This matches the given correct answer: 330 Hz.

Thus, the correct option is 330 Hz.

Answer 2

The general form of a plane progressive wave is:
\[y = A \cos(\omega t - kx).\]
Comparing with the given equation:
\[y = 2 \cos 2\pi (330 t - x),\]
we identify:
\[\omega = 2\pi \times 330.\]
The angular frequency \(\omega\) is related to the frequency \(f\) by:
\[\omega = 2\pi f \implies 2\pi f = 2\pi \times 330.\]
Thus:
\[f = 330 \, \text{Hz}.\]