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:

Question

A steel wire with mass per unit length $ 7.0 \times 10^{-3} \, \text{kg/m} $ is under a tension of $ 70 \, \text{N} $. The speed of transverse waves in the wire will be:

Answer 1

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.

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:

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The Correct Option is B

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