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A closed organ pipe is vibrating in its second overtone. The length of the pipe is 10cm and maximum amplitude of vibration of particles of the air in the pipe is 2mm. Then the amplitude of S.H.M. of the particles at 9cm from the open end is:
- a)√3 mm
- b)√2 mm
- c)√3/2 mm
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
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A closed organ pipe is vibrating in its second overtone. The length of...
4L/5 = λ ⇒ λ = 8cm
hus 2 cm corresponds to Δϕ = z/2
1 cm corresponds to Δϕ = z/4
hus 2 cm corresponds to Δϕ = z/2
1 cm corresponds to Δϕ = z/4
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A closed organ pipe is vibrating in its second overtone. The length of...
To find the amplitude of the SHM of the particles at 9cm from the open end, we can use the concept of standing waves in closed organ pipes.
In a closed organ pipe, the length of the pipe is equal to an odd multiple of a quarter wavelength for the fundamental frequency. In this case, the length of the pipe is 10cm, which is equal to 1/4 of the wavelength of the second overtone.
The second overtone is the third harmonic, so the wavelength of the second overtone is equal to 3 times the wavelength of the fundamental frequency.
Therefore, the wavelength of the second overtone is (10cm)(3) = 30cm.
To find the amplitude of the SHM of the particles at 9cm from the open end, we need to determine the position of the node at 9cm from the open end.
In a standing wave, the nodes are the points where the particles do not move. The distance between adjacent nodes is equal to half the wavelength.
Since the wavelength of the second overtone is 30cm, the distance between adjacent nodes is 30cm/2 = 15cm.
So, the node at 9cm from the open end is located 15cm - 9cm = 6cm from the closed end.
At the node, the amplitude of the SHM of the particles is zero.
Thus, the amplitude of the SHM of the particles at 9cm from the open end is zero.
In a closed organ pipe, the length of the pipe is equal to an odd multiple of a quarter wavelength for the fundamental frequency. In this case, the length of the pipe is 10cm, which is equal to 1/4 of the wavelength of the second overtone.
The second overtone is the third harmonic, so the wavelength of the second overtone is equal to 3 times the wavelength of the fundamental frequency.
Therefore, the wavelength of the second overtone is (10cm)(3) = 30cm.
To find the amplitude of the SHM of the particles at 9cm from the open end, we need to determine the position of the node at 9cm from the open end.
In a standing wave, the nodes are the points where the particles do not move. The distance between adjacent nodes is equal to half the wavelength.
Since the wavelength of the second overtone is 30cm, the distance between adjacent nodes is 30cm/2 = 15cm.
So, the node at 9cm from the open end is located 15cm - 9cm = 6cm from the closed end.
At the node, the amplitude of the SHM of the particles is zero.
Thus, the amplitude of the SHM of the particles at 9cm from the open end is zero.
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A closed organ pipe is vibrating in its second overtone. The length of the pipe is 10cm and maximum amplitude of vibration of particles of the air in the pipe is 2mm. Then the amplitude of S.H.M. of the particles at 9cm from the open end is:a)√3 mmb)√2 mmc)√3/2 mmd)none of theseCorrect answer is option 'B'. Can you explain this answer?
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A closed organ pipe is vibrating in its second overtone. The length of the pipe is 10cm and maximum amplitude of vibration of particles of the air in the pipe is 2mm. Then the amplitude of S.H.M. of the particles at 9cm from the open end is:a)√3 mmb)√2 mmc)√3/2 mmd)none of theseCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A closed organ pipe is vibrating in its second overtone. The length of the pipe is 10cm and maximum amplitude of vibration of particles of the air in the pipe is 2mm. Then the amplitude of S.H.M. of the particles at 9cm from the open end is:a)√3 mmb)√2 mmc)√3/2 mmd)none of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A closed organ pipe is vibrating in its second overtone. The length of the pipe is 10cm and maximum amplitude of vibration of particles of the air in the pipe is 2mm. Then the amplitude of S.H.M. of the particles at 9cm from the open end is:a)√3 mmb)√2 mmc)√3/2 mmd)none of theseCorrect answer is option 'B'. Can you explain this answer?.
A closed organ pipe is vibrating in its second overtone. The length of the pipe is 10cm and maximum amplitude of vibration of particles of the air in the pipe is 2mm. Then the amplitude of S.H.M. of the particles at 9cm from the open end is:a)√3 mmb)√2 mmc)√3/2 mmd)none of theseCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A closed organ pipe is vibrating in its second overtone. The length of the pipe is 10cm and maximum amplitude of vibration of particles of the air in the pipe is 2mm. Then the amplitude of S.H.M. of the particles at 9cm from the open end is:a)√3 mmb)√2 mmc)√3/2 mmd)none of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A closed organ pipe is vibrating in its second overtone. The length of the pipe is 10cm and maximum amplitude of vibration of particles of the air in the pipe is 2mm. Then the amplitude of S.H.M. of the particles at 9cm from the open end is:a)√3 mmb)√2 mmc)√3/2 mmd)none of theseCorrect answer is option 'B'. Can you explain this answer?.
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