Question:
A string of length \( 1 \text{ m} \) fixed at both ends vibrates in second harmonic. Wavelength is:
A string of length \( 1 \text{ m} \) fixed at both ends vibrates in second harmonic. Wavelength is:
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Visualizing standing waves is simple: the harmonic number \( n \) always matches the total number of loops (segments) formed on the string. Since one full loop is equal to half a wavelength (\(\lambda / 2\)), a string vibrating in the second harmonic (\(n=2\)) forms two loops, which equals exactly one full wavelength (\(\lambda\))!
Updated On: May 20, 2026
- \( 1 \text{ m} \)
- \( 2 \text{ m} \)
- \( 0.5 \text{ m} \)
- \( 0.25 \text{ m} \)
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The Correct Option is A
Solution and Explanation
Concept:
When a stretched string of length \( L \) is fixed at both ends and set into vibration, progressive transverse waves travel along the string and reflect back from the boundaries. The superposition of these incident and reflected waves forms stationary (standing) waves.
Because the ends are firmly anchored, the air/string particles cannot move at those boundaries, producing a displacement node at both endpoints. The different possible stable modes of vibration are called harmonics:
• First Harmonic (Fundamental Mode): Stretches across a single loop with nodes at the ends and one antinode in the middle. The length equals a half wavelength: \( L = \frac{\lambda_1}{2} \).
• Second Harmonic (First Overtone): The string vibrates in exactly two complete loops, adding a central node. The length matches one full wavelength: \( L = 2 \cdot \left(\frac{\lambda_2}{2}\right) = \lambda_2 \). In general, for the \( n \)-th harmonic, the relationship between the length and wavelength is: \[ L = n \cdot \frac{\lambda_n}{2} \implies \lambda_n = \frac{2L}{n} \]
Step 1: Identifying the given string parameters and mode configuration.
From the question statement, we have:
• Length of the fixed string, \( L = 1 \text{ m} \)
• Order of the harmonic mode, \( n = 2 \)
Step 2: Calculating the wavelength for the second harmonic.
Substitute \( L = 1 \) and \( n = 2 \) into the general harmonic wavelength relation: \[ \lambda_2 = \frac{2(1)}{2} \] Simplifying the fraction: \[ \lambda_2 = 1 \text{ m} \] This calculated value corresponds directly to option (A).
• First Harmonic (Fundamental Mode): Stretches across a single loop with nodes at the ends and one antinode in the middle. The length equals a half wavelength: \( L = \frac{\lambda_1}{2} \).
• Second Harmonic (First Overtone): The string vibrates in exactly two complete loops, adding a central node. The length matches one full wavelength: \( L = 2 \cdot \left(\frac{\lambda_2}{2}\right) = \lambda_2 \). In general, for the \( n \)-th harmonic, the relationship between the length and wavelength is: \[ L = n \cdot \frac{\lambda_n}{2} \implies \lambda_n = \frac{2L}{n} \]
Step 1: Identifying the given string parameters and mode configuration.
From the question statement, we have:
• Length of the fixed string, \( L = 1 \text{ m} \)
• Order of the harmonic mode, \( n = 2 \)
Step 2: Calculating the wavelength for the second harmonic.
Substitute \( L = 1 \) and \( n = 2 \) into the general harmonic wavelength relation: \[ \lambda_2 = \frac{2(1)}{2} \] Simplifying the fraction: \[ \lambda_2 = 1 \text{ m} \] This calculated value corresponds directly to option (A).
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