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NEET Physics: Simple pendulum time period derivation Practice Questions

12 free multiple-choice questions on Simple pendulum time period derivation (Oscillations and Waves) for NEET Physics, each with the correct answer and a step-by-step explanation. Attempt a question, then reveal the solution to check your reasoning.

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  1. Q1. Which physical quantity is NOT required to determine the time period of a simple pendulum swinging under gravity?

    • AMass of the bob
    • BLength of the pendulum
    • CThe amplitude of swing
    • DAcceleration due to gravity
    Show answer & explanation

    Correct answer: C — The amplitude of swing

    The time period T of a simple pendulum depends only on the length of the string (l) and the acceleration due to gravity (g). The mass of the bob and the amplitude of oscillation do not affect the time period of the simple harmonic motion.

  2. Q2. For a pendulum, if the acceleration due to gravity (g) is decreased, what happens to the time period of oscillation?

    • AIt remains unchanged
    • BIt becomes zero
    • CIt decreases
    • DIt increases
    Show answer & explanation

    Correct answer: D — It increases

    Since T is inversely proportional to √g (T ∝ 1/√g), if g decreases, the value of √g decreases, and therefore the time period T increases.

  3. Q3. If a pendulum executes oscillations for a very small angle, the time period formula can be derived using which approximation?

    • ASmall angle approximation
    • BBinomial theorem
    • CFourier analysis
    • DTaylor series expansion
    Show answer & explanation

    Correct answer: A — Small angle approximation

    For small angular displacements (θ is small), the mathematical expression for the period involves terms that are approximated using the small angle approximation, leading to the simple harmonic motion result derived from the equation of motion.

  4. Q4. When analyzing the mathematical derivation of the time period for a simple pendulum, the resulting formula involves terms related to the length (L) and the acceleration due to gravity (g). Which of the following statements correctly describes the role of these parameters in determining the oscillation period?

    • AThe period is linearly proportional to L and inversely proportional to g.
    • BThe period is proportional to the square root of the ratio of L to g.
    • CThe period is directly proportional to the square of L and inversely proportional to g.
    • DThe period is inversely proportional to the ratio of L to g.
    Show answer & explanation

    Correct answer: B — The period is proportional to the square root of the ratio of L to g.

    The derived formula is T = 2π√(L/g). This shows that the time period is proportional to the square root of the ratio (L/g).

  5. Q5. What is the relationship between the time period (T) and the length of the pendulum (L)?

    • AT is inversely proportional to the square of L (T ∝ 1/L² )
    • BT is inversely proportional to L (T ∝ 1/L)
    • CT is directly proportional to L (T ∝ L)
    • DT is proportional to the square root of L (T ∝ √L )
    Show answer & explanation

    Correct answer: D — T is proportional to the square root of L (T ∝ √L )

    From the formula T = 2π√(L/g), the time period T is directly proportional to the square root of the length L.

  6. Q6. A simple pendulum is undergoing simple harmonic motion. If the length of the pendulum is doubled while keeping the acceleration due to gravity constant, how does the time period of oscillation change?

    • AThe time period remains the same
    • BThe time period is halved
    • CThe time period increases by a factor of √2
    • DThe time period is doubled
    Show answer & explanation

    Correct answer: D — The time period is doubled

    The time period T of a simple pendulum is given by the formula T = 2π√(L/g). Since T is proportional to the square root of the length (T ∝ √L), if the length L is doubled (L' = 2L), the new time period T' will be T' = 2π√(2L/g) = √2 (2π√(L/g)) = √2 T. Thus, the time period increases by a factor of √2.

  7. Q7. The time period of oscillation for a pendulum is inversely proportional to which physical quantity?

    • AThe amplitude of swing
    • BThe mass of the bob
    • CThe length of the string
    • DThe acceleration due to gravity
    Show answer & explanation

    Correct answer: D — The acceleration due to gravity

    The formula for the time period is T = 2π√(l/g). Since T is inversely proportional to √g, the time period is inversely proportional to the square root of the acceleration due to gravity (g).

  8. Q8. The expression for the time period of a simple pendulum is T = 2π√(L/g). If the length L is kept constant, how does the time period change if the acceleration due to gravity g is tripled?

    • AIt is reduced by a factor of 3
    • BIt remains unchanged
    • CIt is tripled
    • DIt is reduced by a factor of √3
    Show answer & explanation

    Correct answer: D — It is reduced by a factor of √3

    Since T ∝ 1/√g, if g is tripled, the new time period T' = 2π√(L/(3g)) = T/√3.

  9. Q9. If the acceleration due to gravity (g) is increased, the time period of a simple pendulum will:

    • ABecome zero
    • BRemain unchanged
    • CIncrease
    • DDecrease
    Show answer & explanation

    Correct answer: D — Decrease

    Since T ∝ 1/√g, if g increases, the time period T decreases.

  10. Q10. For a simple pendulum undergoing simple harmonic motion, the time period T depends on the length of the pendulum l and the acceleration due to gravity g. Which of the following statements correctly describes the relationship?

    • AT is directly proportional to l and inversely proportional to g
    • BT is directly proportional to the square root of l and inversely proportional to the square root of g
    • CT is directly proportional to l and inversely proportional to √g
    • DT is inversely proportional to √l and directly proportional to √g
    Show answer & explanation

    Correct answer: B — T is directly proportional to the square root of l and inversely proportional to the square root of g

    The time period T of a simple pendulum is derived as T = 2π√(l/g). This formula shows that the time period is directly proportional to the square root of the length (l) and inversely proportional to the square root of the acceleration due to gravity (g).

  11. Q11. The derived relationship for the time period of a simple pendulum is T = 2π√(l/g). If the acceleration due to gravity (g) is tripled, how does the time period change?

    • AIt is reduced by a factor of 3
    • BIt is reduced by a factor of √3
    • CIt is tripled
    • DIt is reduced by a factor of 9
    Show answer & explanation

    Correct answer: B — It is reduced by a factor of √3

    Since T ∝ 1/√g, if g is tripled (3g), the new period T' will be T/√3.

  12. Q12. The derivation of the time period for a simple pendulum requires which approximation?

    • ASmall acceleration approximation
    • BSmall force approximation
    • CSmall time approximation
    • DSmall angle approximation (sin θ≈ θ )
    Show answer & explanation

    Correct answer: D — Small angle approximation (sin θ≈ θ )

    The standard derivation relies on the assumption that the angle of swing (θ) is very small, which allows the use of the small angle approximation, sin θ≈ θ.

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