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:

Show 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.

Updated On: Dec 31, 2025

5 m
0.5 m
0.33 m
0.33 m

Hide Solution

Verified By Collegedunia

The Correct Option is B

Approach Solution - 1

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.

Was this answer helpful?

Hide Solution

Verified By Collegedunia

Approach Solution -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).