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A person measures the time period of a simple pendulum inside a stationary lift and finds it to beT. If the lift starts accelerating upward with an acceleration of g/3, the time period of pendulum will be
- a)T/3
- b)T /√3
- c)√3T/2
- d)2T √3
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A person measures the time period of a simple pendulum inside a statio...
In a stationary lift,
T¹= 2π √l/√g
Accelerating upward with accln. g/3
So,
g effective = g + g/3
T² = 2π √3l/√4g
then T¹/T² = 2/√3
Hence T² = √3 T¹/2.
T¹= 2π √l/√g
Accelerating upward with accln. g/3
So,
g effective = g + g/3
T² = 2π √3l/√4g
then T¹/T² = 2/√3
Hence T² = √3 T¹/2.
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Community Answer
A person measures the time period of a simple pendulum inside a statio...
The time period of a simple pendulum is given by 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.
In the stationary lift, the acceleration due to gravity is g. Therefore, the time period is T.
When the lift starts accelerating upward with an acceleration of g/3, the effective acceleration acting on the pendulum will be the sum of the acceleration due to gravity and the acceleration of the lift. Therefore, the effective acceleration will be g + g/3 = 4g/3.
Using the formula for the time period of a simple pendulum, we can calculate the new time period:
T' = 2π√(L/(4g/3))
= 2π√(3L/4g)
= (2π/√3)√(L/g)
Therefore, the time period of the pendulum when the lift is accelerating upward with an acceleration of g/3 is (2π/√3)√(L/g), which is option c) in the given choices.
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.
In the stationary lift, the acceleration due to gravity is g. Therefore, the time period is T.
When the lift starts accelerating upward with an acceleration of g/3, the effective acceleration acting on the pendulum will be the sum of the acceleration due to gravity and the acceleration of the lift. Therefore, the effective acceleration will be g + g/3 = 4g/3.
Using the formula for the time period of a simple pendulum, we can calculate the new time period:
T' = 2π√(L/(4g/3))
= 2π√(3L/4g)
= (2π/√3)√(L/g)
Therefore, the time period of the pendulum when the lift is accelerating upward with an acceleration of g/3 is (2π/√3)√(L/g), which is option c) in the given choices.
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A person measures the time period of a simple pendulum inside a stationary lift and finds it to beT. If the lift starts accelerating upward with an acceleration of g/3, the time period of pendulum will bea)T/3b)T /√3c)√3T/2d)2T √3Correct answer is option 'C'. Can you explain this answer?
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