Question

A wave travelling in the $+ve$ $x-direction$ having displacement along $y$-direction as $1m$, wavelength $2 \pi m$ and frequency of $ \frac{1}{\pi} Hz$ is represented by :

• A. $ y = \sin (2 \pi x + 2 \pi t)$
• B. $ y = \sin (x - 2 t)$
• C. $ y = \sin (2 \pi x - 2 \pi t)$
• D. $ y = \sin (10 \pi x - 20 \pi t)$

Answer 1

The Correct Option is B

To solve the given problem, we need to identify the correct wave equation that describes a wave traveling in the positive x-direction with a given displacement, wavelength, and frequency. Let's analyze the information provided:

Given Data:

• Displacement along the y-direction: 1 \, \text{m}
• Wavelength \lambda: 2 \pi \, \text{m}
• Frequency f: \frac{1}{\pi} \, \text{Hz}

Wave Equation:

The general form of a wave traveling in the positive x-direction is:

y = A \sin(kx - \omega t)

• A: Amplitude
• k: Wave number
• \omega: Angular frequency
• x, t: Position and time variables

Calculation of Wave Number k:

The wave number k is given by:

k = \frac{2 \pi}{\lambda} = \frac{2 \pi}{2 \pi} = 1 \, \text{m}^{-1}

Calculation of Angular Frequency \omega:

The angular frequency \omega is given by:

\omega = 2 \pi f = 2 \pi \times \frac{1}{\pi} = 2 \, \text{rad/s}

Constructing the Wave Equation:

The wave equation using the calculated values of k and \omega becomes:

y = A \sin(x - 2t)

Given that the amplitude is 1 m, the wave equation simplifies to:

y = \sin(x - 2t)

Conclusion:

The correct wave equation is therefore y = \sin(x - 2t), which corresponds to the given data.