Question

A wave disturbance in a medium is described by \[ y(x,t)=0.02\cos\left(50\pi t + \frac{\pi}{2}\right)\cos(10\pi x), \] where \( x,y \) are in meters and \( t \) in seconds. Which statements are correct?

• A. A node occurs at \( x=0.15\,\text{m} \)
• B. An antinode occurs at \( x=0.3\,\text{m} \)
• C. The speed of the wave is \( 4\,\text{m/s} \)
• D. The wavelength of the wave is \( 0.2\,\text{m} \)

Hint:

The wave is a standing wave of the form \( \cos\omega t \cos kx \). Nodes are where \( \cos kx = 0 \) and antinodes where \( \cos kx = \pm 1 \). Speed is \( \omega / k \).

Answer 1

The Correct Option is A, B, D

Step 1: Understanding the Question:
The disturbance has a time part and a position part multiplied together, so it is a standing wave. Four statements are given and we check each one.

Step 2: Read off the constants:
Compare with \( y = A\cos(\omega t + \phi)\cos kx \). This gives \( \omega = 50\pi \) rad/s and \( k = 10\pi \) rad/m.

Step 3: Check statement (D), wavelength:
We use \( k = 2\pi / \lambda \).
\[ \lambda = \frac{2\pi}{10\pi} = 0.2\ \text{m} \]
So (D) is TRUE.

Step 4: Check statement (C), speed:
Wave speed is \( v = \omega / k \).
\[ v = \frac{50\pi}{10\pi} = 5\ \text{m/s} \]
The statement says 4 m/s, so (C) is FALSE.

Step 5: Check statement (A), node at 0.15 m:
A node is a point where \( \cos(10\pi x) = 0 \), so \( 10\pi x = (2n+1)\pi/2 \).
\[ x = \frac{2n+1}{20}\ \text{m} = 0.05,\ 0.15,\ 0.25,\ \ldots \]
Put \( n = 1 \) and we get 0.15 m. So (A) is TRUE.

Step 6: Check statement (B), antinode at 0.3 m:
An antinode is a point where \( |\cos(10\pi x)| = 1 \), so \( 10\pi x = n\pi \).
\[ x = \frac{n}{10}\ \text{m} = 0,\ 0.1,\ 0.2,\ 0.3,\ \ldots \]
Put \( n = 3 \) and we get 0.3 m. So (B) is TRUE.

Final Answer:
Statements (A), (B) and (D) are correct. \[ \boxed{\text{(A), (B) and (D)}} \]