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Two pendulums have time periods T and 5T/4. They are in phase at their mean positions at some instant of time. What will be their phase difference when the bigger pendulum completes one oscillation?
- a)300
- b)450
- c)600
- d)900
Correct answer is option 'D'. Can you explain this answer?
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Given: Two pendulums with time periods T and 5T/4 are in phase at their mean positions.
To find: Phase difference between the two pendulums when the bigger pendulum completes one oscillation.
Solution:
Let the two pendulums be P1 and P2 with time periods T and 5T/4 respectively.
Let the amplitudes of the two pendulums be A1 and A2 respectively.
Let the initial phase of P1 be zero.
Since the two pendulums are in phase at their mean positions, the initial phase of P2 will also be zero.
Let the phase difference between the two pendulums be φ at any instant of time.
When P1 completes one oscillation, its phase will change by 2π radians.
The time taken by P1 to complete one oscillation is T.
Hence, the angular frequency of P1 is ω1 = 2π/T.
Similarly, the angular frequency of P2 is ω2 = 2π/(5T/4) = 8π/5T.
The time taken by P2 to complete one oscillation is 5T/4.
Hence, the phase difference between the two pendulums after time t is given by
φ = ω1t - ω2t
When P2 completes one oscillation, its phase will change by 2π radians.
Hence, we need to find the value of t such that P2 completes one oscillation.
The time taken by P2 to complete one oscillation is 5T/4.
Hence, we need to find the value of t such that 5T/4 = t.
Substituting this value of t in the expression for φ, we get
φ = ω1(5T/4) - ω2(5T/4)
φ = (5ω1 - 4ω2)T/4
φ = (5(2π/T) - 4(8π/5T))T/4
φ = 2π/5
Converting this to degrees, we get
φ = 72 degrees
Therefore, the phase difference between the two pendulums when the bigger pendulum completes one oscillation is 72 degrees or 900.
To find: Phase difference between the two pendulums when the bigger pendulum completes one oscillation.
Solution:
Let the two pendulums be P1 and P2 with time periods T and 5T/4 respectively.
Let the amplitudes of the two pendulums be A1 and A2 respectively.
Let the initial phase of P1 be zero.
Since the two pendulums are in phase at their mean positions, the initial phase of P2 will also be zero.
Let the phase difference between the two pendulums be φ at any instant of time.
When P1 completes one oscillation, its phase will change by 2π radians.
The time taken by P1 to complete one oscillation is T.
Hence, the angular frequency of P1 is ω1 = 2π/T.
Similarly, the angular frequency of P2 is ω2 = 2π/(5T/4) = 8π/5T.
The time taken by P2 to complete one oscillation is 5T/4.
Hence, the phase difference between the two pendulums after time t is given by
φ = ω1t - ω2t
When P2 completes one oscillation, its phase will change by 2π radians.
Hence, we need to find the value of t such that P2 completes one oscillation.
The time taken by P2 to complete one oscillation is 5T/4.
Hence, we need to find the value of t such that 5T/4 = t.
Substituting this value of t in the expression for φ, we get
φ = ω1(5T/4) - ω2(5T/4)
φ = (5ω1 - 4ω2)T/4
φ = (5(2π/T) - 4(8π/5T))T/4
φ = 2π/5
Converting this to degrees, we get
φ = 72 degrees
Therefore, the phase difference between the two pendulums when the bigger pendulum completes one oscillation is 72 degrees or 900.
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Two pendulums have time periods T and 5T/4. They are in phase at their mean positions at some instant of time. What will be their phase difference when the bigger pendulum completes one oscillation?a)300b)450c)600d)900Correct answer is option 'D'. Can you explain this answer?
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