The equation of a wave travelling on a string is y=sin[20pi x+10pi t] where x and t are distance and time in SI units. The minimum distance between two points having the same oscillating speed is : [cite: 120, 121]

Solution & Explanation

### Related Formula k = frac2pilambda Delta x_textmin = fraclambda2 [cite: 715] ### Core Logic From the given wave equation, the wave number k is the coefficient of x [cite: 120]: k = 20pi\ textrad/m Now find the wavelength lambda: [cite: 717] lambda = frac2pik = frac2pi20pi = frac110\ textm = 10\ textcm [cite: 717] The minimum distance between any two points moving with identical speeds in a continuous wave cycle corresponds to a phase difference of pi, which translates spatially to a half-wavelength separation (fraclambda2) [cite: 715]: textDistance = fraclambda2 = frac102 = 5\ textcm [cite: 717] ### Pattern Recognition Oscillating speed configuration reaches identical value fields twice per spatial wavelength cycle[cite: 715]. Thus, minimum distance separation scales exactly to fraclambda2[cite: 715]. ### 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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