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 \( x \) and \( y \) are in meters and \( t \) is in seconds. Which statement(s) is/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 m/sec.
• D. The wavelength of the wave is 0.2 m.

Hint:

For wave equations of the form \( y(x,t) = A \cos(\omega t + \phi) \cos(kx) \), remember that the wavelength \( \lambda = \frac{2\pi}{k} \) and the speed of the wave \( v = \frac{\omega}{k} \). Nodes and antinodes are determined by the value of \( kx \).

Answer 1

The Correct Option is A

The given wave function is: \[ y(x,t) = 0.02 \cos\left(50 \pi t + \frac{\pi}{2}\right) \cos(10 \pi x) \] This is a standard wave equation of the form: \[ y(x,t) = A \cos(\omega t + \phi) \cos(kx) \] where: - \( A = 0.02 \) is the amplitude of the wave,
- \( \omega = 50 \pi \) is the angular frequency,
- \( k = 10 \pi \) is the wave number,
- \( \phi = \frac{\pi}{2} \) is the phase constant.
Step 1: Analyzing the Wave Number and Wavelength
The wave number \( k \) is related to the wavelength \( \lambda \) by the equation: \[ k = \frac{2 \pi}{\lambda} \] Substitute the value of \( k \): \[ 10 \pi = \frac{2 \pi}{\lambda} \] Solving for \( \lambda \): \[ \lambda = \frac{2 \pi}{10 \pi} = 0.2 \, \text{m} \] Thus, the wavelength of the wave is \( 0.2 \, \text{m} \). This confirms that Option (D) is correct. Step 2: Analyzing the Node and Antinode Positions Nodes occur where the displacement of the wave is always zero, and antinodes occur where the displacement is maximum. The displacement is zero when \( \cos(kx) = 0 \), which happens at: \[ kx = (2n + 1)\frac{\pi}{2}, \quad n = 0, 1, 2, \dots \] Substitute \( k = 10 \pi \): \[ 10 \pi x = (2n + 1)\frac{\pi}{2} \] Simplifying: \[ x = \frac{(2n + 1)}{20} \] For \( n = 0 \), the position of the first node is: \[ x = \frac{1}{20} = 0.05 \, \text{m} \] For \( n = 1 \), the second node is at: \[ x = \frac{3}{20} = 0.15 \, \text{m} \] Thus, a node occurs at \( x = 0.15 \, \text{m} \), which means Option (A) is correct. Antinodes occur where \( \cos(kx) = \pm 1 \), which happens at: \[ kx = n\pi, \quad n = 0, 1, 2, \dots \] For \( k = 10 \pi \), we get: \[ 10 \pi x = n\pi \] Thus, \( x = \frac{n}{10} \). For \( n = 1 \), the position of the first antinode is: \[ x = \frac{1}{10} = 0.1 \, \text{m} \] For \( n = 2 \), the second antinode is at: \[ x = \frac{2}{10} = 0.2 \, \text{m} \] For \( n = 3 \), the third antinode is at: \[ x = \frac{3}{10} = 0.3 \, \text{m} \] Thus, an antinode occurs at \( x = 0.3 \, \text{m} \), confirming Option (B) is correct. Step 3: Speed of the Wave The speed of the wave is related to the wave number and the angular frequency by the equation: \[ v = \frac{\omega}{k} \] Substitute \( \omega = 50 \pi \) and \( k = 10 \pi \): \[ v = \frac{50 \pi}{10 \pi} = 5 \, \text{m/s} \] Thus, the speed of the wave is 5 m/s, which confirms that Option (C) is incorrect. The correct speed is 5 m/s, not 4 m/s. Conclusion - Option (A): Correct, a node occurs at \( x = 0.15 \, \text{m} \). - Option (B): Correct, an antinode occurs at \( x = 0.3 \, \text{m} \). - Option (C): Incorrect, the speed of the wave is 5 m/s, not 4 m/s. - Option (D): Correct, the wavelength of the wave is \( 0.2 \, \text{m} \). Thus, the correct options are (A), (B), and (D).