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Two pendulums of length 1 m and 16 m 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 pendulum will be one behind the other is
- a)2T/5
- b)T/3
- c)T/4
- d)4T/3
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Two pendulums of length 1 m and 16 m start vibrating one behind the ot...
T1=2π√1/g=T, T2=2π√16/g=4T
(ω1−ω2)t=2π
(2π/T−2π/4T)t=2π
t=4T/3
(ω1−ω2)t=2π
(2π/T−2π/4T)t=2π
t=4T/3
Most Upvoted Answer
Two pendulums of length 1 m and 16 m start vibrating one behind the ot...
Solution:
Given, length of shorter pendulum, l₁ = 1 m
and length of longer pendulum, l₂ = 16 m
Let the time period of shorter pendulum be T
The time period of a simple pendulum is given by the formula,
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) ...(1)
and time period of longer pendulum, T' = 2π √(16/g) = 8π √(1/g) ...(2)
Let the shorter pendulum complete n oscillations before the longer pendulum completes one oscillation.
In n time periods of the shorter pendulum, the angle covered by it is θ₁ = 2π n
In the same time period, the angle covered by the longer pendulum is θ₂ = 2π (n/16)
When the two pendulums are in the mean position and in the same phase, the angle between them is zero.
To bring both the pendulums back in the same phase, the shorter pendulum must complete one more oscillation than the longer pendulum.
So, we can equate the angles covered by both pendulums as follows:
θ₁ - θ₂ = 2π
=> 2π n - 2π (n/16) = 2π
=> 2π (15n/16) = 2π
=> n = 16/15
The time taken for the shorter pendulum to complete 16/15 oscillations is given by
t = (16/15) × T = 16T/15
The time taken for the longer pendulum to complete one oscillation is given by
t' = T'
The minimum time after which two pendulums will be one behind the other is the time taken by the longer pendulum to complete (16/15) oscillations which is given by
t' × (16/15) = (8π/15)√(1/g) × (16/15) = (32/15)π√(1/g)
From equation (1), we know that T = 2π √(1/g)
So, (32/15)π√(1/g) = (64/15)T = 4T/3
Therefore, the correct answer is option (d) 4T/3.
Given, length of shorter pendulum, l₁ = 1 m
and length of longer pendulum, l₂ = 16 m
Let the time period of shorter pendulum be T
The time period of a simple pendulum is given by the formula,
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) ...(1)
and time period of longer pendulum, T' = 2π √(16/g) = 8π √(1/g) ...(2)
Let the shorter pendulum complete n oscillations before the longer pendulum completes one oscillation.
In n time periods of the shorter pendulum, the angle covered by it is θ₁ = 2π n
In the same time period, the angle covered by the longer pendulum is θ₂ = 2π (n/16)
When the two pendulums are in the mean position and in the same phase, the angle between them is zero.
To bring both the pendulums back in the same phase, the shorter pendulum must complete one more oscillation than the longer pendulum.
So, we can equate the angles covered by both pendulums as follows:
θ₁ - θ₂ = 2π
=> 2π n - 2π (n/16) = 2π
=> 2π (15n/16) = 2π
=> n = 16/15
The time taken for the shorter pendulum to complete 16/15 oscillations is given by
t = (16/15) × T = 16T/15
The time taken for the longer pendulum to complete one oscillation is given by
t' = T'
The minimum time after which two pendulums will be one behind the other is the time taken by the longer pendulum to complete (16/15) oscillations which is given by
t' × (16/15) = (8π/15)√(1/g) × (16/15) = (32/15)π√(1/g)
From equation (1), we know that T = 2π √(1/g)
So, (32/15)π√(1/g) = (64/15)T = 4T/3
Therefore, the correct answer is option (d) 4T/3.
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Two pendulums of length 1 m and 16 m 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 pendulum will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 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 1 m and 16 m 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 pendulum will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two pendulums of length 1 m and 16 m 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 pendulum will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'D'. Can you explain this answer?.
Two pendulums of length 1 m and 16 m 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 pendulum will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 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 1 m and 16 m 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 pendulum will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two pendulums of length 1 m and 16 m 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 pendulum will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'D'. Can you explain this answer?.
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Two pendulums of length 1 m and 16 m 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 pendulum will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Two pendulums of length 1 m and 16 m 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 pendulum will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Two pendulums of length 1 m and 16 m 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 pendulum will be one behind the other isa)2T/5b)T/3c)T/4d)4T/3Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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