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Concepts Of Physics MCQ Edition [Volume 1]PhysicsWave Motion and Waves on a String

A string fixed at both ends vibrates in a resonant mode, having a separation of 2.0 cm between consecutive nodes. In the next higher resonant frequency, the separation decreases to 1.6 cm . Calculate the length of the string.

Options

  1. A10.0 cm
  2. B4.0 cm
  3. C16.0 cm
  4. D8.0 cm

Correct answer

D. 8.0 cm

Step-by-step solution

Let the length of the string be L . For a string fixed at both ends vibrating in the n -th harmonic, the separation between consecutive nodes is 2 = L n . Given that for the n -th harmonic, the separation is 2.0 cm , we have: L = n 2.0 For the next higher resonant mode, the string vibrates in the (n+1) -th harmonic. The separation between consecutive nodes becomes 1.6 cm , so: L = (n+1) 1.6 Equating the two expressions for L : 2.0 n = 1.6 (n+1) 2.0 n = 1.6 n + 1.6 0.4 n = 1.6 n = 4 Substituting the value of n into

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