Question

Displacement of a wave is expressed as $$ x(t) = 5 \cos \left( 628t + \frac{\pi}{2} \right) \, \text{m}. $$ The wavelength of the wave when its velocity is 300 m/s is:

• A. 5 m
• B. 0.5 m
• C. 0.33 m
• D. 0.33 m

Hint:

The wavelength of a wave can be calculated using the wave number, which is related to the angular frequency and velocity of the wave.

Answer 1

The Correct Option is B

The wave's displacement is described by the equation:

\(x(t) = 5 \cos \left( 628t + \frac{\pi}{2} \right) \, \text{m}\)

This equation aligns with the standard form:

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

Where:

• \(A\) represents the amplitude,
• \(\omega\) represents the angular frequency, and
• \(\phi\) represents the phase angle.

From the provided equation, the angular frequency is determined to be \(\omega = 628 \, \text{rad/s}\).

The relationship between angular frequency \(\omega\), wave velocity \(v\), and wavelength \(\lambda\) is defined by the formula:

\(\omega = \frac{2\pi v}{\lambda}\)

Rearranging this formula to solve for the wavelength yields:

\(\lambda = \frac{2\pi v}{\omega}\)

Given a wave velocity of \(v = 300 \, \text{m/s}\), substituting the known values results in:

\(\lambda = \frac{2\pi \times 300}{628}\)

Simplifying this expression gives:

\(\lambda = \frac{600\pi}{628}\)

Using the approximation \(\pi \approx 3.14\), the wavelength is calculated as:

\(\lambda \approx \frac{600 \times 3.14}{628} = 0.5 \, \text{m}\)

Therefore, the wavelength of the wave, given a velocity of 300 m/s, is 0.5 m.

The final answer is confirmed to be 0.5 m.