Question

A string of length 2 m is vibrating with 2 loops. The distance between its node and adjacent antinode is ............

• A. 0.5 m
• B. 1.0 m
• C. 1.5 m
• D. 2.0 m

Hint:

Key distances in a standing wave: Node to next Node = \(\lambda/2\) Antinode to next Antinode = \(\lambda/2\) Node to next Antinode = \(\lambda/4\)

Answer 1

The Correct Option is A

When a string vibrates in 'n' loops, its length 'L' is equal to 'n' times half the wavelength (\(\lambda\)). \[ L = n \frac{\lambda}{2} \] Given:
Length of the string, \(L = 2\) m
Number of loops, \(n = 2\)
Substituting the values: \[ 2 = 2 \times \frac{\lambda}{2} \] \[ \lambda = 2 \, \text{m} \] The wavelength of the wave is 2 m.
The distance between a node and an adjacent antinode is always one-quarter of a wavelength (\(\frac{\lambda}{4}\)). \[ \text{Distance} = \frac{\lambda}{4} = \frac{2}{4} = 0.5 \, \text{m} \]