Question

For a travelling harmonic wave $y(x, t) = 2.0 \cos 2\pi(10t – 0.0080 x + 0.35)$, where $x$ and $y$ are in cm and $t$ in s. The phase difference between oscillatory motion of two points separated by a distance of 0.5 m is: ____.

• A. 0.8 $\pi$ rad
• B. 8 $\pi$ rad
• C. 0.008 $\pi$ rad
• D. 0.08 $\pi$ rad

Hint:

Always ensure units are consistent. In wave problems, the most common mistake is mixing meters (distance) with centimeters (from the wave equation).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The phase difference ($\Delta \phi$) between two points in a travelling wave is directly proportional to the distance (path difference $\Delta x$) between them.

Step 2: Key Formula or Approach:
1. General wave equation: $y = A \cos(\omega t - kx + \phi_0)$ 2. Comparing with given equation: $y = 2.0 \cos [2\pi(10t) - 2\pi(0.0080x) + 2\pi(0.35)]$ 3. Phase difference: $\Delta \phi = k \cdot \Delta x$

Step 3: Detailed Explanation:
1. Identify wave number ($k$): From the equation, $k = 2\pi(0.0080) \text{ rad/cm}$. 2. Calculate path difference ($\Delta x$): Given distance is 0.5 m. Since $x$ is in cm, we must convert: \[ \Delta x = 0.5 \text{ m} = 50 \text{ cm} \] 3. Calculate Phase Difference: \[ \Delta \phi = [2\pi(0.0080)] \times 50 \] \[ \Delta \phi = 2\pi \times 0.40 \] \[ \Delta \phi = 0.8\pi \text{ rad} \]

Step 4: Final Answer:
The phase difference is 0.8 $\pi$ rad.