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Two simple pendulums of length 0.5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed... oscillations
- a)5
- b)1
- c)2
- d)3
Correct answer is option 'B'. Can you explain this answer?
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Two simple pendulums of length 0.5 m and 20 m respectively are given s...
Given:
Two simple pendulums with lengths 0.5 m and 20 m respectively are given small linear displacements in one direction at the same time.
To find:
The number of oscillations completed by the shorter pendulum when both pendulums are in the same phase again.
Solution:
When two pendulums are in the same phase, it means that they have completed the same number of oscillations.
Formula:
The time period of a simple pendulum is given by:
T = 2π√(L/g)
Where:
T = time period
L = length of the pendulum
g = acceleration due to gravity
Let's calculate the time period of both pendulums:
For the pendulum with length 0.5 m:
T1 = 2π√(0.5/9.8)
T1 ≈ 2π√(0.051)
T1 ≈ 2π * 0.225
T1 ≈ 1.413 seconds (approximately)
For the pendulum with length 20 m:
T2 = 2π√(20/9.8)
T2 ≈ 2π√(2.041)
T2 ≈ 2π * 1.43
T2 ≈ 9.003 seconds (approximately)
Let's calculate the number of oscillations completed by the shorter pendulum:
The number of oscillations completed by a pendulum in time 't' is given by:
N = t / T
Where:
N = number of oscillations
t = time
T = time period
For the shorter pendulum (T1 = 1.413 seconds), let's consider the time 't' when both pendulums are in the same phase again.
N1 = t / T1
For the longer pendulum (T2 = 9.003 seconds), the same number of oscillations would be completed in the same time 't'.
N2 = t / T2
Since both pendulums are in the same phase again, the number of oscillations completed by both pendulums would be equal.
N1 = N2
t / T1 = t / T2
Cross-multiplying:
T2 * t = T1 * t
T2 = T1
Therefore, the number of oscillations completed by the shorter pendulum when both pendulums are in the same phase again is 1 (option B).
Two simple pendulums with lengths 0.5 m and 20 m respectively are given small linear displacements in one direction at the same time.
To find:
The number of oscillations completed by the shorter pendulum when both pendulums are in the same phase again.
Solution:
When two pendulums are in the same phase, it means that they have completed the same number of oscillations.
Formula:
The time period of a simple pendulum is given by:
T = 2π√(L/g)
Where:
T = time period
L = length of the pendulum
g = acceleration due to gravity
Let's calculate the time period of both pendulums:
For the pendulum with length 0.5 m:
T1 = 2π√(0.5/9.8)
T1 ≈ 2π√(0.051)
T1 ≈ 2π * 0.225
T1 ≈ 1.413 seconds (approximately)
For the pendulum with length 20 m:
T2 = 2π√(20/9.8)
T2 ≈ 2π√(2.041)
T2 ≈ 2π * 1.43
T2 ≈ 9.003 seconds (approximately)
Let's calculate the number of oscillations completed by the shorter pendulum:
The number of oscillations completed by a pendulum in time 't' is given by:
N = t / T
Where:
N = number of oscillations
t = time
T = time period
For the shorter pendulum (T1 = 1.413 seconds), let's consider the time 't' when both pendulums are in the same phase again.
N1 = t / T1
For the longer pendulum (T2 = 9.003 seconds), the same number of oscillations would be completed in the same time 't'.
N2 = t / T2
Since both pendulums are in the same phase again, the number of oscillations completed by both pendulums would be equal.
N1 = N2
t / T1 = t / T2
Cross-multiplying:
T2 * t = T1 * t
T2 = T1
Therefore, the number of oscillations completed by the shorter pendulum when both pendulums are in the same phase again is 1 (option B).
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Two simple pendulums of length 0.5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed... oscillationsa)5b)1c)2d)3Correct answer is option 'B'. Can you explain this answer?
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