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A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string is
- a)5 grams
- b)10 grams
- c)20 grams
- d)40 grams
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A hollow pipe of length 0.8 m is closed at one end. At its open end a ...
Frequency of 2nd harmonic of string = Fundamental frequency produced in the pipe
The mass of the string = μ l1
= 0.02 × 0.5 kg = 10g
= 0.02 × 0.5 kg = 10g
Most Upvoted Answer
A hollow pipe of length 0.8 m is closed at one end. At its open end a ...
To find the fundamental frequency of the pipe, we can use the formula:
f = v/λ
where f is the frequency, v is the speed of sound, and λ is the wavelength.
Since the pipe is closed at one end, the fundamental frequency is given by:
f₁ = v/2L
where L is the length of the pipe.
Given that the length of the pipe is 0.8 m and the speed of sound is 320 m/s, we can substitute these values into the formula to find the fundamental frequency:
f₁ = 320/(2*0.8) = 200 Hz
Since the string resonates with the fundamental frequency of the pipe, the second harmonic of the string must also have a frequency of 200 Hz.
The frequency of the second harmonic of a string is given by:
f₂ = 2f₁
Substituting the value of f₁, we can find the frequency of the second harmonic:
f₂ = 2*200 = 400 Hz
To find the wavelength of the second harmonic of the string, we can use the formula:
λ = 2L/n
where λ is the wavelength, L is the length of the string, and n is the harmonic number.
Given that the length of the string is 0.5 m and the harmonic number is 2, we can substitute these values into the formula to find the wavelength:
λ = 2*0.5/2 = 0.5 m
Now, we can find the speed of the second harmonic of the string using the formula:
v = f₂ * λ
Substituting the values of f₂ and λ, we can find the speed of the second harmonic:
v = 400 * 0.5 = 200 m/s
Finally, we can find the tension in the string using the formula:
v = √(T/μ)
where v is the speed of the wave, T is the tension in the string, and μ is the linear mass density of the string.
Given that the speed of the wave is 200 m/s and the tension is 50 N, we can substitute these values into the formula to find μ:
200 = √(50/μ)
Squaring both sides of the equation, we get:
40000 = 50/μ
Rearranging the equation, we can solve for μ:
μ = 50/40000 = 0.00125 kg/m
Therefore, the linear mass density of the string is 0.00125 kg/m.
f = v/λ
where f is the frequency, v is the speed of sound, and λ is the wavelength.
Since the pipe is closed at one end, the fundamental frequency is given by:
f₁ = v/2L
where L is the length of the pipe.
Given that the length of the pipe is 0.8 m and the speed of sound is 320 m/s, we can substitute these values into the formula to find the fundamental frequency:
f₁ = 320/(2*0.8) = 200 Hz
Since the string resonates with the fundamental frequency of the pipe, the second harmonic of the string must also have a frequency of 200 Hz.
The frequency of the second harmonic of a string is given by:
f₂ = 2f₁
Substituting the value of f₁, we can find the frequency of the second harmonic:
f₂ = 2*200 = 400 Hz
To find the wavelength of the second harmonic of the string, we can use the formula:
λ = 2L/n
where λ is the wavelength, L is the length of the string, and n is the harmonic number.
Given that the length of the string is 0.5 m and the harmonic number is 2, we can substitute these values into the formula to find the wavelength:
λ = 2*0.5/2 = 0.5 m
Now, we can find the speed of the second harmonic of the string using the formula:
v = f₂ * λ
Substituting the values of f₂ and λ, we can find the speed of the second harmonic:
v = 400 * 0.5 = 200 m/s
Finally, we can find the tension in the string using the formula:
v = √(T/μ)
where v is the speed of the wave, T is the tension in the string, and μ is the linear mass density of the string.
Given that the speed of the wave is 200 m/s and the tension is 50 N, we can substitute these values into the formula to find μ:
200 = √(50/μ)
Squaring both sides of the equation, we get:
40000 = 50/μ
Rearranging the equation, we can solve for μ:
μ = 50/40000 = 0.00125 kg/m
Therefore, the linear mass density of the string is 0.00125 kg/m.
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A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string isa)5 gramsb)10 gramsc)20 gramsd)40 gramsCorrect answer is option 'B'. Can you explain this answer?
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A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string isa)5 gramsb)10 gramsc)20 gramsd)40 gramsCorrect 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 hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string isa)5 gramsb)10 gramsc)20 gramsd)40 gramsCorrect 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 hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string isa)5 gramsb)10 gramsc)20 gramsd)40 gramsCorrect answer is option 'B'. Can you explain this answer?.
A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string isa)5 gramsb)10 gramsc)20 gramsd)40 gramsCorrect 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 hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string isa)5 gramsb)10 gramsc)20 gramsd)40 gramsCorrect 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 hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string isa)5 gramsb)10 gramsc)20 gramsd)40 gramsCorrect answer is option 'B'. Can you explain this answer?.
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