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 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).