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NVAEasyJEE 2023Superposition Principle & Standing Waves

JEE Physics 2023 Question with Solution

An organ pipe 40 cm40 \, \text{cm} long is open at both ends. The speed of sound in air is 360 m/s360 \, \text{m/s}. The frequency of the second harmonic is:

Answer

Correct answer:900

Step-by-step solution

Standard Method

Given: Length of the open organ pipe is L=0.4 mL = 0.4 \, \text{m} and speed of sound is v=360 m/sv = 360 \, \text{m/s}.

Find: The frequency of the second harmonic.

For an open organ pipe, the fundamental frequency is

f1=v2Lf_1 = \frac{v}{2L}

The second harmonic is

f2=2f1f_2 = 2f_1

Substituting the given values,

f1=3602×0.4=450 Hzf_1 = \frac{360}{2 \times 0.4} = 450 \, \text{Hz}

Therefore,

f2=2×450=900 Hzf_2 = 2 \times 450 = 900 \, \text{Hz}

Therefore, the frequency of the second harmonic is 900 Hz900 \, \text{Hz}.

Common mistakes

  • Using the closed pipe formula instead of the open pipe formula is incorrect because this pipe is open at both ends. Use f1=v2Lf_1 = \frac{v}{2L}, not the closed-pipe relation.

  • Taking the second harmonic equal to the fundamental frequency is wrong because harmonics are integer multiples of the fundamental. For an open pipe, the second harmonic is f2=2f1f_2 = 2f_1.

  • Forgetting to convert 40 cm40 \, \text{cm} into 0.4 m0.4 \, \text{m} leads to an incorrect numerical answer. Always convert length into SI units before substitution.

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