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Two pendulums of length I meter and 16 meters start vibrating one behind the other from the same stand. At some instant the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulums will be one behind the other is
- a)2T/5
- b)T/3
- c)T/4
- d)4T/3
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Two pendulums of length I meter and 16 meters start vibrating one behi...
Solution:
Given, length of shorter pendulum = 1 m
Length of longer pendulum = 16 m
Time period of shorter pendulum = T
Let's assume that after time t, the two pendulums are again in the same phase and the thread of the longer pendulum is one behind the shorter one.
Now, we know that the time period of a pendulum is given by:
T = 2π√(l/g)
where l is the length of the pendulum and g is the acceleration due to gravity.
Therefore, time period of shorter pendulum = T = 2π√(1/g)
Time period of longer pendulum = T₁ = 2π√(16/g) = 8T
Let's consider the displacement of the shorter pendulum as x(t) and that of the longer pendulum as y(t). At t=0, both pendulums are in the same phase and the amplitude of both the pendulums is zero.
Now, we know that the displacement of a simple pendulum is given by:
x(t) = Asin(ωt + φ)
where A is the amplitude, ω is the angular frequency, and φ is the phase constant.
At t=0, both the pendulums are in the same phase and the amplitude of both the pendulums is zero. Therefore, the equation for the displacement of both the pendulums at t=0 is:
x(0) = y(0) = 0
Now, let's assume that after time t, the two pendulums are again in the same phase and the thread of the longer pendulum is one behind the shorter one. Therefore, the displacement of the shorter pendulum at time t is given by:
x(t) = Asin(ωt)
where ω = 2π/T
The displacement of the longer pendulum at time t is given by:
y(t) = Asin(2πt/T₁ + φ)
Now, we need to find the minimum time t after which the thread of the longer pendulum is one behind the shorter one. This means that the displacement of the longer pendulum is π/2 radians behind the displacement of the shorter pendulum.
Therefore, we have:
y(t) = x(t - δ)
where δ = π/2ω
Substituting the values of x(t) and y(t), we get:
Asin(2πt/T₁ + φ) = Asin(ω(t - δ))
Solving this equation, we get:
2πt/T₁ + φ = ω(t - δ) + nπ
where n is an integer
Substituting the values of T₁, ω, and δ, we get:
2πt/8T + φ = πt/T - π/2 + nπ
Simplifying this equation, we get:
t = (2n + 1)T/5
Therefore, the minimum time after which the two threads of the pendulums will be one behind the other is (2n + 1)T/5, where n is an integer.
Since we need to find the minimum time, we take n=0. Therefore, the answer is:
t = T/5 + T
t = 2T/5
Hence, the correct answer is option (a
Given, length of shorter pendulum = 1 m
Length of longer pendulum = 16 m
Time period of shorter pendulum = T
Let's assume that after time t, the two pendulums are again in the same phase and the thread of the longer pendulum is one behind the shorter one.
Now, we know that the time period of a pendulum is given by:
T = 2π√(l/g)
where l is the length of the pendulum and g is the acceleration due to gravity.
Therefore, time period of shorter pendulum = T = 2π√(1/g)
Time period of longer pendulum = T₁ = 2π√(16/g) = 8T
Let's consider the displacement of the shorter pendulum as x(t) and that of the longer pendulum as y(t). At t=0, both pendulums are in the same phase and the amplitude of both the pendulums is zero.
Now, we know that the displacement of a simple pendulum is given by:
x(t) = Asin(ωt + φ)
where A is the amplitude, ω is the angular frequency, and φ is the phase constant.
At t=0, both the pendulums are in the same phase and the amplitude of both the pendulums is zero. Therefore, the equation for the displacement of both the pendulums at t=0 is:
x(0) = y(0) = 0
Now, let's assume that after time t, the two pendulums are again in the same phase and the thread of the longer pendulum is one behind the shorter one. Therefore, the displacement of the shorter pendulum at time t is given by:
x(t) = Asin(ωt)
where ω = 2π/T
The displacement of the longer pendulum at time t is given by:
y(t) = Asin(2πt/T₁ + φ)
Now, we need to find the minimum time t after which the thread of the longer pendulum is one behind the shorter one. This means that the displacement of the longer pendulum is π/2 radians behind the displacement of the shorter pendulum.
Therefore, we have:
y(t) = x(t - δ)
where δ = π/2ω
Substituting the values of x(t) and y(t), we get:
Asin(2πt/T₁ + φ) = Asin(ω(t - δ))
Solving this equation, we get:
2πt/T₁ + φ = ω(t - δ) + nπ
where n is an integer
Substituting the values of T₁, ω, and δ, we get:
2πt/8T + φ = πt/T - π/2 + nπ
Simplifying this equation, we get:
t = (2n + 1)T/5
Therefore, the minimum time after which the two threads of the pendulums will be one behind the other is (2n + 1)T/5, where n is an integer.
Since we need to find the minimum time, we take n=0. Therefore, the answer is:
t = T/5 + T
t = 2T/5
Hence, the correct answer is option (a
Community Answer
Two pendulums of length I meter and 16 meters start vibrating one behi...
Shouldnt the answer be d
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Two pendulums of length I meter and 16 meters start vibrating one behind the other from the same stand. At some instant the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulums will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'A'. Can you explain this answer?
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Two pendulums of length I meter and 16 meters start vibrating one behind the other from the same stand. At some instant the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulums will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Two pendulums of length I meter and 16 meters start vibrating one behind the other from the same stand. At some instant the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulums will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two pendulums of length I meter and 16 meters start vibrating one behind the other from the same stand. At some instant the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulums will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'A'. Can you explain this answer?.
Two pendulums of length I meter and 16 meters start vibrating one behind the other from the same stand. At some instant the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulums will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Two pendulums of length I meter and 16 meters start vibrating one behind the other from the same stand. At some instant the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulums will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two pendulums of length I meter and 16 meters start vibrating one behind the other from the same stand. At some instant the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulums will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'A'. Can you explain this answer?.
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Two pendulums of length I meter and 16 meters start vibrating one behind the other from the same stand. At some instant the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulums will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Two pendulums of length I meter and 16 meters start vibrating one behind the other from the same stand. At some instant the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulums will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Two pendulums of length I meter and 16 meters start vibrating one behind the other from the same stand. At some instant the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulums will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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