In the resonance experiment, two air columns (closed at one end) of 100mathrm~cm and 120mathrm~cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is :

Solution & Explanation

### Related Formula For an air column closed at one end, the fundamental frequency f is given by: f = fracv4l where v is the velocity of sound and l is the length of the air column. ### Core Logic Given parameters: - l_1 = 100mathrm~cm = 1.0mathrm~m - l_2 = 120mathrm~cm = 1.2mathrm~m - Beats per second (f_1 - f_2) = 15 ### Step 1: Write the equation for beat frequency Since l_1 < l_2, the frequency f_1 > f_2. Hence: textBeat frequency = f_1 - f_2 = fracv4l_1 - fracv4l_2 15 = fracv4 left( frac1l_1 - frac1l_2 right) ### Step 2: Solve for velocity of sound (v) Substitute the lengths in meters: 15 = fracv4 left( frac11.0 - frac11.2 right) 15 = fracv4 left( 1 - frac56 right) 15 = fracv4 left( frac16 right) 15 = fracv24 v = 15 times 24 = 360mathrm~m/s ### Pattern Recognition Beat problems involving standing waves in organ pipes can be calculated faster by remembering that f propto frac1l. This allows setting up the proportion v = 4 cdot Delta f cdot fracl_1 l_2l_2 - l_1 directly as a short-cut. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

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Q45 jee_main_2024_31_jan_morning Organ Pipes
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60mathrm~cm, the length of the closed pipe will be:
  • A. 60mathrm~cm
  • B. 45mathrm~cm
  • C. 30mathrm~cm
  • D. 15mathrm~cm

Solution

### Related Formula f_textclosed, fundamental = fracv4L_c f_textopen, 1st overtone = frac2v2L_o ### Core Logic
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
For a closed organ pipe, the fundamental frequency (1st harmonic) is: f_1 = fracvlambda = fracv4L_1 where L_1 is the length of the closed pipe. For an open organ pipe, the first overtone (2nd harmonic) is: f_2 = frac2v2L_2 = fracvL_2 where L_2 is the length of the open pipe (L_2 = 60mathrm\,cm). ### Step 2: Equating Frequencies Given f_1 = f_2: fracv4L_1 = fracvL_2 L_2 = 4L_1 60 = 4 times L_1 L_1 = 15mathrm\,cm ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

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