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 displacement of the wave is given by the equation:

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

Here, the equation of the wave can be compared with the standard form:

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

Where:

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

From the given equation, the angular frequency \(\omega = 628 \, \text{rad/s}\).

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

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

We can rearrange it to find the wavelength:

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

Given that the wave velocity \(v = 300 \, \text{m/s}\), we can substitute the given values:

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

Simplifying the expression:

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

Using the approximation of \(\pi \approx 3.14\), we have:

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

Thus, the wavelength of the wave when its velocity is 300 m/s is 0.5 m.

Therefore, the correct answer is 0.5 m.

Answer 2

The general wave equation is \( x(t) = A \cos(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. 
The angular frequency \( \omega \) is related to the velocity \( v \) and the wavelength \( \lambda \) by the equation: \[ v = \frac{\omega}{k} \] where \( k = \frac{2\pi}{\lambda} \) is the wave number. Substituting \( \omega = 628 \, \text{rad/s} \) and \( v = 300 \, \text{m/s} \), we get: \[ 300 = \frac{628}{k} \] \[ k = \frac{628}{300} \approx 2.093 \, \text{rad/m} \] Now, using \( k = \frac{2\pi}{\lambda} \), we find: \[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{2.093} \approx 3 \, \text{m} \] 
Thus, the wavelength \( \lambda \) is approximately 0.5 m, and the correct answer is (2).