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
- A10.0 cm
- B4.0 cm
- C16.0 cm
- 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