Question:medium
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 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?
Show 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 \( kx \).
Updated On: Nov 28, 2025
- A node occurs at \( x = 0.15 \, \text{m} \).
- An antinode occurs at \( x = 0.3 \, \text{m} \).
- The speed of the wave is 4 m/sec.
- The wavelength of the wave is 0.2 m.
Hide Solution
The Correct Option is A
Solution and Explanation
The provided wave function is:
\[
y(x,t) = 0.02 \cos\left(50 \pi t + \frac{\pi}{2}\right) \cos(10 \pi x)
\]
This aligns with the standard wave equation:
\[
y(x,t) = A \cos(\omega t + \phi) \cos(kx)
\]
where:
- \( A = 0.02 \) represents the wave's amplitude,
- \( \omega = 50 \pi \) is the angular frequency,
- \( k = 10 \pi \) is the wave number,
- \( \phi = \frac{\pi}{2} \) denotes the phase constant.
Step 1: Analyzing the Wave Number and Wavelength The wave number \( k \) and wavelength \( \lambda \) are related by: \[ k = \frac{2 \pi}{\lambda} \] Substituting the value of \( k \): \[ 10 \pi = \frac{2 \pi}{\lambda} \] Solving for \( \lambda \): \[ \lambda = \frac{2 \pi}{10 \pi} = 0.2 \, \text{m} \] Therefore, the wavelength is \( 0.2 \, \text{m} \), validating Option (D). Step 2: Analyzing Node and Antinode Positions Nodes, where displacement is always zero, occur when \( \cos(kx) = 0 \), which happens at: \[ kx = (2n + 1)\frac{\pi}{2}, \quad n = 0, 1, 2, \dots \] Substituting \( k = 10 \pi \): \[ 10 \pi x = (2n + 1)\frac{\pi}{2} \] Simplifying: \[ x = \frac{(2n + 1)}{20} \] For \( n = 0 \), the first node is at: \[ x = \frac{1}{20} = 0.05 \, \text{m} \] For \( n = 1 \), the second node is at: \[ x = \frac{3}{20} = 0.15 \, \text{m} \] Thus, a node exists at \( x = 0.15 \, \text{m} \), confirming Option (A). Antinodes, where displacement is maximum, occur when \( \cos(kx) = \pm 1 \), 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 first antinode is at: \[ x = \frac{1}{10} = 0.1 \, \text{m} \] 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 exists at \( x = 0.3 \, \text{m} \), confirming Option (B). Step 3: Speed of the Wave The wave's speed is related to wave number and angular frequency by: \[ v = \frac{\omega}{k} \] Substituting \( \omega = 50 \pi \) and \( k = 10 \pi \): \[ v = \frac{50 \pi}{10 \pi} = 5 \, \text{m/s} \] Therefore, the wave's speed is 5 m/s, which means Option (C) is incorrect. The actual 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 wave speed is 5 m/s, not 4 m/s. - Option (D): Correct, the wavelength is \( 0.2 \, \text{m} \). Therefore, the correct options are (A), (B), and (D).
- \( \omega = 50 \pi \) is the angular frequency,
- \( k = 10 \pi \) is the wave number,
- \( \phi = \frac{\pi}{2} \) denotes the phase constant.
Step 1: Analyzing the Wave Number and Wavelength The wave number \( k \) and wavelength \( \lambda \) are related by: \[ k = \frac{2 \pi}{\lambda} \] Substituting the value of \( k \): \[ 10 \pi = \frac{2 \pi}{\lambda} \] Solving for \( \lambda \): \[ \lambda = \frac{2 \pi}{10 \pi} = 0.2 \, \text{m} \] Therefore, the wavelength is \( 0.2 \, \text{m} \), validating Option (D). Step 2: Analyzing Node and Antinode Positions Nodes, where displacement is always zero, occur when \( \cos(kx) = 0 \), which happens at: \[ kx = (2n + 1)\frac{\pi}{2}, \quad n = 0, 1, 2, \dots \] Substituting \( k = 10 \pi \): \[ 10 \pi x = (2n + 1)\frac{\pi}{2} \] Simplifying: \[ x = \frac{(2n + 1)}{20} \] For \( n = 0 \), the first node is at: \[ x = \frac{1}{20} = 0.05 \, \text{m} \] For \( n = 1 \), the second node is at: \[ x = \frac{3}{20} = 0.15 \, \text{m} \] Thus, a node exists at \( x = 0.15 \, \text{m} \), confirming Option (A). Antinodes, where displacement is maximum, occur when \( \cos(kx) = \pm 1 \), 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 first antinode is at: \[ x = \frac{1}{10} = 0.1 \, \text{m} \] 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 exists at \( x = 0.3 \, \text{m} \), confirming Option (B). Step 3: Speed of the Wave The wave's speed is related to wave number and angular frequency by: \[ v = \frac{\omega}{k} \] Substituting \( \omega = 50 \pi \) and \( k = 10 \pi \): \[ v = \frac{50 \pi}{10 \pi} = 5 \, \text{m/s} \] Therefore, the wave's speed is 5 m/s, which means Option (C) is incorrect. The actual 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 wave speed is 5 m/s, not 4 m/s. - Option (D): Correct, the wavelength is \( 0.2 \, \text{m} \). Therefore, the correct options are (A), (B), and (D).
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