NEET Physics Question - Waves - Standing Wave in String and Organ Pipe

Assertion (A): For a closed organ resonating pipe, the first resonance length is 60 cm. The second resonating length will be 180 cm.

Reason (R): For a particular closed pipe \(n_2 = 3n_1\).

Both (A) & (R) are true and the (R) is the correct explanation of the (A)

Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

(A) is true but (R) is false

Both (A) and (R) are false

Solution:

For a closed organ pipe, the resonance lengths are in the ratio \(L_1 : L_2 : L_3 = 1 : 3 : 5\). If \(L_1 = 60 text{ cm}\), then \(L_2 = 3 times 60 \text{ cm} = 180 \text{ cm}\). So (A) is true. The resonant frequencies for a closed pipe are \(f_n = (2n-1)f_1\), thus the second resonance (third harmonic) is \(f_2 = 3f_1\). (R) is true and correctly explains (A).

Rankers Physics

Closed Organ Pipe Resonance

Solution:

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