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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...
Solution:
Period of a simple pendulum of length 'L' is given by:
T = 2π√(L/g)
where g is acceleration due to gravity.
Let the shorter pendulum complete 'n' oscillations in time 't'.
Then, the longer pendulum will complete (n/20) oscillations in the same time 't', as the period of the longer pendulum is 20 times that of the shorter pendulum.
Now, the pendulums will be in phase again when both complete an integral number of oscillations in the same time 't'.
Therefore, we need to find the smallest value of 'n' for which both pendulums complete an integral number of oscillations in the same time 't'.
As given, both pendulums are given small linear displacement in one direction at the same time. This means that they start oscillating in phase.
Hence, the pendulum of shorter length will complete one oscillation in the time taken by the longer pendulum to complete 1/20th of an oscillation.
We need to find the value of 'n' for which this time is equal to the time period of the shorter pendulum.
Let T1 and T2 be the periods of the shorter and longer pendulums respectively.
Then, we have:
T1 = 2π√(0.5/g)
T2 = 2π√(20/g) = 4π√(0.5/g)
Time taken by the longer pendulum to complete 1/20th of an oscillation is given by:
t = (1/20) × T2
= (1/5) × π√(0.5/g)
Equating this time with the time period of the shorter pendulum, we get:
t = T1
⇒ (1/5) × π√(0.5/g) = 2π√(0.5/g)
⇒ n = 1
Hence, both pendulums will be in phase again when the pendulum of shorter length has completed one oscillation. Therefore, the correct answer is option B.
Period of a simple pendulum of length 'L' is given by:
T = 2π√(L/g)
where g is acceleration due to gravity.
Let the shorter pendulum complete 'n' oscillations in time 't'.
Then, the longer pendulum will complete (n/20) oscillations in the same time 't', as the period of the longer pendulum is 20 times that of the shorter pendulum.
Now, the pendulums will be in phase again when both complete an integral number of oscillations in the same time 't'.
Therefore, we need to find the smallest value of 'n' for which both pendulums complete an integral number of oscillations in the same time 't'.
As given, both pendulums are given small linear displacement in one direction at the same time. This means that they start oscillating in phase.
Hence, the pendulum of shorter length will complete one oscillation in the time taken by the longer pendulum to complete 1/20th of an oscillation.
We need to find the value of 'n' for which this time is equal to the time period of the shorter pendulum.
Let T1 and T2 be the periods of the shorter and longer pendulums respectively.
Then, we have:
T1 = 2π√(0.5/g)
T2 = 2π√(20/g) = 4π√(0.5/g)
Time taken by the longer pendulum to complete 1/20th of an oscillation is given by:
t = (1/20) × T2
= (1/5) × π√(0.5/g)
Equating this time with the time period of the shorter pendulum, we get:
t = T1
⇒ (1/5) × π√(0.5/g) = 2π√(0.5/g)
⇒ n = 1
Hence, both pendulums will be in phase again when the pendulum of shorter length has completed one oscillation. Therefore, the correct answer is 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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