Question

A sound wave of frequency 210 Hz travels with a speed of 330 ms⁻¹ along the positive x-axis. Each particle of the wave moves a distance of 10 cm between the two extreme points. The equation of the displacement function (s) of this wave is (x in metre, t in second)

• A. \(s(x,t) = 0.10 \sin[4x - 1320t]\) m
• B. \(s(x,t) = 0.05 \sin[4x - 1320t]\) m
• C. \(s(x,t) = 0.05 \sin[1320x - 4t]\) m
• D. \(s(x,t) = 0.10 \sin[1320x - 4t]\) m

Hint:

In wave equation problems, determining the amplitude is often the first and easiest step. Here, recognizing that the "distance between extreme points" is \(2A\) quickly narrows down the choices. Then, check the relationship \(v = \omega/k\) for the remaining options to find the correct one.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept: The general equation of a traveling wave moving in the positive x-direction is: \[ s(x,t) = A \sin(kx - \omega t + \phi) \] Where: - \( A \) is the amplitude. - \( k \) is the angular wave number (\( k = \frac{2\pi}{\lambda} = \frac{\omega}{v} \)). - \( \omega \) is the angular frequency (\( \omega = 2\pi f \)). - \( v \) is the wave speed.
Step 2: Calculate Amplitude (A): The distance between the two extreme points is the total range of motion, which equals \( 2A \). Given \( 2A = 10 \, \text{cm} = 0.10 \, \text{m} \). Therefore, \( A = \frac{0.10}{2} = 0.05 \, \text{m} \).
Step 3: Calculate Angular Frequency (\( \omega \)): Given frequency \( f = 210 \, \text{Hz} \). \[ \omega = 2\pi f = 2 \times \frac{22}{7} \times 210 \] \[ \omega = 2 \times 22 \times 30 = 1320 \, \text{rad/s} \]
Step 4: Calculate Wave Number (k): Given wave speed \( v = 330 \, \text{ms}^{-1} \). \[ k = \frac{\omega}{v} = \frac{1320}{330} = 4 \, \text{m}^{-1} \]
Step 5: Formulate the Equation: Since the wave travels along the positive x-axis, the phase should be \( (kx - \omega t) \) or \( (\omega t - kx) \). Standard form: \( s(x,t) = A \sin(kx - \omega t) \) or \( s(x,t) = A \sin(\omega t - kx) \). The options use the form involving \( 4x \) and \( 1320t \). Let's check the options. Option (B) has \( A = 0.05 \), \( k = 4 \), \( \omega = 1320 \). \( s(x,t) = 0.05 \sin[4x - 1320t] \). This matches our calculated values.