PYQ Vault
EbooksTutor

Amplitude, displacement or period from v = omega times root of (A squared - x squared)

This type of question, asked 7 times: 2021 Paper 12, 2021 Paper 17, 2021 Paper 26, 30 Jan 2023, 27 Jan 2024, 9 Apr 2024, 4 Apr 2026 S1. All 7 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 12

    11
    A particle is performing SHM along X-axis, such that its velocity is v1, when its displacement from mean position is X1X_{1} and V2V_{2} when its displacement from mean position is X2X_{2}. Time period of oscillation is

    Tap an option to check your answer.

  2. Question 2

    2021 Paper 17

    25
    A particle executes simple harmonic motion represented by displacement function as
    x(t)=Asin(ωt+ϕ)x(t) =Asin(\omega t+\phi)
    If the position and velocity of the particle at t=0 st = 0\text{ }s are 2 cm2\text{ }cm and 2ω cm s−12\omega\text{ }cm\text{ }s^{- 1} respectively, then its amplitude is x2 cmx\sqrt{2}\text{ }cm where the value of xx is

  3. Question 3

    2021 Paper 26

    23
    A particle excutes S.H.M with amplitude ' aa ' and time period TT. The displacement of the particle when its speed is half of maximum speed is xa2\frac{\sqrt{x}a}{2}. The value of xx is

  4. Question 4

    30 Jan 2023

    120
    The velocity of a particle executing SHM varies with displacement (x)\left( x \right) as 4v2=50−x24v^{2}= 50 -x^{2}. The time period of oscillations is x7s\frac{x}{7}s. The value of xx is____ (Take  π=227)\left. \ \pi =\frac{22}{7} \right)

  5. Question 5

    27 Jan 2024

    27
    A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s5\text{ }cm/s is αcm\sqrt{\alpha}cm, where α=\alpha = ________ .

  6. Question 6

    9 Apr 2024

    29
    The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of 4 m,2 ms−14\text{ }m,2{\text{ }ms}^{- 1} and 16 ms−216{\text{ }ms}^{- 2} at a certain instant. The amplitude of the motion is xm\sqrt{x}m where xx is_______ .

  7. Question 7

    4 Apr 2026 S1

    23
    The velocity of a particle executing simple harmonic motion along x -axis is described as v2=50−x2v^{2} = 50 - x^{2}, where x represents displacement. If the time period of motion is x7s\frac{x}{7}s, the value of xx is ____\_\_\_\_ .