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 equation for a one-dimensional sinusoidal wave traveling in the positive x-direction is given by \(s(x,t) = A \sin(kx - \omega t)\). We need to determine the amplitude (A), the wave number (k), and the angular frequency (\(\omega\)) from the given information.
Step 2: Key Formula or Approach:
- Amplitude (\(A\)): The maximum displacement from the equilibrium position.
- Angular frequency (\(\omega\)): \(\omega = 2\pi f\), where \(f\) is the frequency.
- Wave number (\(k\)): \(k = \frac{2\pi}{\lambda}\), where \(\lambda\) is the wavelength.
- Wave speed (\(v\)): \(v = f\lambda = \frac{\omega}{k}\).
Step 3: Detailed Explanation:
Determine the Amplitude (A):
The problem states that "each particle of the wave moves a distance of 10 cm between the two extreme points". The distance between the two extreme points of an oscillation is twice the amplitude (\(2A\)).
\[ 2A = 10 \text{ cm} = 0.10 \text{ m} \] \[ A = \frac{0.10 \text{ m}}{2} = 0.05 \text{ m} \] This immediately eliminates options (A) and (D).
Determine the Angular Frequency (\(\omega\)):
Given frequency \(f = 210\) Hz.
\[ \omega = 2\pi f = 2\pi (210) = 420\pi \text{ rad/s} \] Looking at the options, \(\pi\) is not present. This suggests an approximation. Let's calculate the numerical value. \(\omega \approx 420 \times 3.14 \approx 1319.5\) rad/s. This is very close to 1320 rad/s. Let's assume \(\omega = 1320\) rad/s.
Determine the Wave Number (k):
We are given the wave speed \(v = 330\) m/s. We can use the relation \(v = \omega/k\).
\[ k = \frac{\omega}{v} = \frac{1320 \text{ rad/s}}{330 \text{ m/s}} = 4 \text{ rad/m} \] Construct the Wave Equation:
Now we assemble the equation \(s(x,t) = A \sin(kx - \omega t)\).
\[ s(x,t) = 0.05 \sin(4x - 1320t) \] This matches option (B). The form is \((kx - \omega t)\) because the wave travels along the positive x-axis. Option (C) has the values for k and \(\omega\) swapped, which is incorrect.
Step 4: Final Answer:
The correct equation for the displacement function is \(s(x,t) = 0.05 \sin[4x - 1320t]\) m. Therefore, option (B) is correct.