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A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by :-
  • a)
    10–3 rad/s
  • b)
    10–1 rad/s
  • c)
    1 rad/s
  • d)
    10–5 rad/s
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A simple pendulum of length 1 m is oscillating with an angular frequen...
Angular frequency of pendulum



s = angular frequency of support]

Δω = 10-3 rad/sec.
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A simple pendulum of length 1 m is oscillating with an angular frequen...
To analyze the motion of the pendulum, we need to consider the motion of both the support and the pendulum itself separately.

The motion of the support can be described as simple harmonic motion with an angular frequency of 1 rad/s and an amplitude of 10. This means that the support oscillates up and down with a maximum displacement of 10 units from its equilibrium position.

The motion of the pendulum can also be described as simple harmonic motion with an angular frequency of 10 rad/s. The angular frequency of the pendulum is independent of the motion of the support. It is determined by the length of the pendulum and the acceleration due to gravity.

The pendulum will oscillate back and forth along a circular arc with a radius equal to the length of the pendulum (1 m in this case). The period of the pendulum is given by T = 2π/ω, where ω is the angular frequency. Therefore, the period of the pendulum is T = 2π/10 = π/5 seconds.

The amplitude of the pendulum's motion is determined by its initial conditions. If the pendulum is released from rest at an angle of 0 degrees (vertical position), then the amplitude of its motion will be 1 meter.

In summary, the support of the pendulum oscillates up and down with an angular frequency of 1 rad/s and an amplitude of 10 units. The pendulum itself oscillates back and forth with an angular frequency of 10 rad/s, a period of π/5 seconds, and an amplitude of 1 meter.
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A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by :-a)10–3 rad/sb)10–1 rad/sc)1 rad/sd)10–5 rad/sCorrect answer is option 'A'. Can you explain this answer?
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A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by :-a)10–3 rad/sb)10–1 rad/sc)1 rad/sd)10–5 rad/sCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by :-a)10–3 rad/sb)10–1 rad/sc)1 rad/sd)10–5 rad/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by :-a)10–3 rad/sb)10–1 rad/sc)1 rad/sd)10–5 rad/sCorrect answer is option 'A'. Can you explain this answer?.
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