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Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :
- a)9
- b)10
- c)8
- d)11
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Two pendulums of length 121 cm and 100 cm start vibrating in phase. At...
(n)Tl = (n + 1)Ts
(n)(1.1) = (n + 1)
0.1(n) = 1
n = 10
No. of oscillation of smaller one
= n + 1
= 10 + 1
= 11
(n)(1.1) = (n + 1)
0.1(n) = 1
n = 10
No. of oscillation of smaller one
= n + 1
= 10 + 1
= 11
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Two pendulums of length 121 cm and 100 cm start vibrating in phase. At...
Given:
- Length of the first pendulum (l₁) = 121 cm
- Length of the second pendulum (l₂) = 100 cm
To find:
The minimum number of vibrations of the shorter pendulum after which the two pendulums are again in phase at the mean position.
Solution:
Let's calculate the time periods of both pendulums using the formula:
T = 2π√(l/g)
where T is the time period, l is the length of the pendulum, and g is the acceleration due to gravity.
Calculating time period of first pendulum (T₁):
l₁ = 121 cm = 1.21 m
g = 9.8 m/s²
T₁ = 2π√(1.21/9.8)
≈ 2π√(0.1235)
≈ 2π * 0.3517
≈ 2.21 s
Calculating time period of second pendulum (T₂):
l₂ = 100 cm = 1 m
T₂ = 2π√(1/9.8)
≈ 2π√(0.102)
≈ 2π * 0.3194
≈ 2.00 s
Since the pendulums start in phase, both pendulums will complete an integral number of vibrations when they meet again at the mean position.
Let's find the least common multiple (LCM) of the time periods T₁ and T₂ to determine the minimum number of vibrations of the shorter pendulum required for both pendulums to be in phase again.
Calculating LCM of T₁ and T₂:
T₁ = 2.21 s
T₂ = 2.00 s
Multiplying both time periods by 100 to remove decimal places:
T₁ = 221 s
T₂ = 200 s
The LCM of 221 and 200 is 22100.
Therefore, the shorter pendulum will complete 22100 / 200 = 110.5 vibrations.
Since we need an integral number of vibrations, the minimum number of vibrations of the shorter pendulum required is 111.
However, we need to subtract the initial phase. As given, the pendulums start in phase at the mean position, so we subtract 1.
Therefore, the minimum number of vibrations of the shorter pendulum required after which the two pendulums are again in phase at the mean position is 111 - 1 = 110 vibrations.
Hence, the correct option is D) 111.
- Length of the first pendulum (l₁) = 121 cm
- Length of the second pendulum (l₂) = 100 cm
To find:
The minimum number of vibrations of the shorter pendulum after which the two pendulums are again in phase at the mean position.
Solution:
Let's calculate the time periods of both pendulums using the formula:
T = 2π√(l/g)
where T is the time period, l is the length of the pendulum, and g is the acceleration due to gravity.
Calculating time period of first pendulum (T₁):
l₁ = 121 cm = 1.21 m
g = 9.8 m/s²
T₁ = 2π√(1.21/9.8)
≈ 2π√(0.1235)
≈ 2π * 0.3517
≈ 2.21 s
Calculating time period of second pendulum (T₂):
l₂ = 100 cm = 1 m
T₂ = 2π√(1/9.8)
≈ 2π√(0.102)
≈ 2π * 0.3194
≈ 2.00 s
Since the pendulums start in phase, both pendulums will complete an integral number of vibrations when they meet again at the mean position.
Let's find the least common multiple (LCM) of the time periods T₁ and T₂ to determine the minimum number of vibrations of the shorter pendulum required for both pendulums to be in phase again.
Calculating LCM of T₁ and T₂:
T₁ = 2.21 s
T₂ = 2.00 s
Multiplying both time periods by 100 to remove decimal places:
T₁ = 221 s
T₂ = 200 s
The LCM of 221 and 200 is 22100.
Therefore, the shorter pendulum will complete 22100 / 200 = 110.5 vibrations.
Since we need an integral number of vibrations, the minimum number of vibrations of the shorter pendulum required is 111.
However, we need to subtract the initial phase. As given, the pendulums start in phase at the mean position, so we subtract 1.
Therefore, the minimum number of vibrations of the shorter pendulum required after which the two pendulums are again in phase at the mean position is 111 - 1 = 110 vibrations.
Hence, the correct option is D) 111.
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Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is : a)9b)10c)8d)11Correct answer is option 'D'. Can you explain this answer?
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Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is : a)9b)10c)8d)11Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is : a)9b)10c)8d)11Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is : a)9b)10c)8d)11Correct answer is option 'D'. Can you explain this answer?.
Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is : a)9b)10c)8d)11Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is : a)9b)10c)8d)11Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is : a)9b)10c)8d)11Correct answer is option 'D'. Can you explain this answer?.
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Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is : a)9b)10c)8d)11Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is : a)9b)10c)8d)11Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is : a)9b)10c)8d)11Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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