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A steel rod of length 1 m is clamped at its middle. If the fundamental frequency of vibrations is 3 kHz, then the speed of sound in steel is
A steel rod of length 1 m is clamped at its middle. If the fundamental frequency of vibrations is 3 kHz, then the speed of sound in steel is
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For a rod fixed at both ends, the fundamental mode has \(L = \lambda/2\). For a rod free at both ends, the fundamental mode also has \(L = \lambda/2\). For a rod clamped at the middle, it behaves like two rods fixed at one end (the clamp) and free at the other, so the fundamental has \(L = \lambda/4\) for each half? Wait, the full rod length is from one free end to the other. The clamp at the center is a node. The distance from node to antinode is \(\lambda/4\). The rod has two such segments (node to antinode on left, and node to antinode on right). So total length \(L = \lambda/4 + \lambda/4 = \lambda/2\). Correct.
Updated On: May 10, 2026
- \(2000 ms^{-1}\)
- \(6000 ms^{-1}\)
- \(8000 ms^{-1}\)
- \(4000 ms^{-1}\)
- \(9000 ms^{-1}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
The question describes a standing wave in a steel rod. When a rod is clamped at its middle, this point becomes a node (a point of zero displacement). The ends of the rod are free to vibrate, so they become antinodes (points of maximum displacement). The fundamental frequency is the lowest frequency at which the rod can vibrate in this configuration.
Step 2: Key Formula or Approach:
The relationship between wave speed (\(v\)), frequency (\(f\)), and wavelength (\(\lambda\)) is given by:
\[ v = f \lambda \] We need to determine the wavelength (\(\lambda\)) for the fundamental mode of vibration based on the boundary conditions.
Step 3: Detailed Explanation:
For the fundamental mode of vibration in a rod clamped at the middle:
- The center is a Node (N).
- The two ends are Antinodes (A).
The simplest standing wave pattern that fits this condition is A-N-A.
The distance between an antinode and the next node is always \(\frac{\lambda}{4}\).
The total length of the rod, L, covers the distance from one antinode (end) to the node (middle) and then to the other antinode (other end).
So, \(L = \frac{\lambda}{4} + \frac{\lambda}{4} = \frac{\lambda}{2}\).
This means the wavelength of the fundamental mode is twice the length of the rod:
\[ \lambda = 2L \] Given:
- Length of the rod, \(L = 1\) m.
- Fundamental frequency, \(f = 3\) kHz = \(3 \times 10^3\) Hz.
First, calculate the wavelength:
\[ \lambda = 2 \times 1 \, \text{m} = 2 \, \text{m} \] Now, calculate the speed of sound (\(v\)) using the wave equation:
\[ v = f \lambda \] \[ v = (3 \times 10^3 \, \text{Hz}) \times (2 \, \text{m}) \] \[ v = 6000 \, \text{m/s} \] Step 4: Final Answer:
The speed of sound in the steel rod is 6000 ms\(^{-1}\), which corresponds to option (B).
The question describes a standing wave in a steel rod. When a rod is clamped at its middle, this point becomes a node (a point of zero displacement). The ends of the rod are free to vibrate, so they become antinodes (points of maximum displacement). The fundamental frequency is the lowest frequency at which the rod can vibrate in this configuration.
Step 2: Key Formula or Approach:
The relationship between wave speed (\(v\)), frequency (\(f\)), and wavelength (\(\lambda\)) is given by:
\[ v = f \lambda \] We need to determine the wavelength (\(\lambda\)) for the fundamental mode of vibration based on the boundary conditions.
Step 3: Detailed Explanation:
For the fundamental mode of vibration in a rod clamped at the middle:
- The center is a Node (N).
- The two ends are Antinodes (A).
The simplest standing wave pattern that fits this condition is A-N-A.
The distance between an antinode and the next node is always \(\frac{\lambda}{4}\).
The total length of the rod, L, covers the distance from one antinode (end) to the node (middle) and then to the other antinode (other end).
So, \(L = \frac{\lambda}{4} + \frac{\lambda}{4} = \frac{\lambda}{2}\).
This means the wavelength of the fundamental mode is twice the length of the rod:
\[ \lambda = 2L \] Given:
- Length of the rod, \(L = 1\) m.
- Fundamental frequency, \(f = 3\) kHz = \(3 \times 10^3\) Hz.
First, calculate the wavelength:
\[ \lambda = 2 \times 1 \, \text{m} = 2 \, \text{m} \] Now, calculate the speed of sound (\(v\)) using the wave equation:
\[ v = f \lambda \] \[ v = (3 \times 10^3 \, \text{Hz}) \times (2 \, \text{m}) \] \[ v = 6000 \, \text{m/s} \] Step 4: Final Answer:
The speed of sound in the steel rod is 6000 ms\(^{-1}\), which corresponds to option (B).
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