Home
Class 11
PHYSICS
A particle is executing simple harmonic ...

A particle is executing simple harmonic motion with time period of 4 s and amplitude 10 cm. Calculate displacement, velocity after `(1)/(2)` second starting from mean position.

Text Solution

Verified by Experts

T = 3s, A = 10 cm, `t = (1)/(2)s`
(i) As x = A `sin omegat`
` = 10 sin (2pi)/(T)t`
` = 10 sin (2pi)/(4) xx (1)/(2)`
` = 10 xx (1)/(sqrt(2)) = 7.07` cm
`V = (dx)/(dt) = Aomegacos omegat`
` = A(2pi)/(T)cos (2pi)/(T)t`
` = 10*(2pi)/(4) cos (2pi)/(4) xx (1)/(2)`
` = 5pi cos (pi)/(4)`
` = (5pi)/(sqrt(2)) = 5 xx 3.14 xx 0.7`
` = 10.99` cm/s
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • OSCILLATIONS

    MODERN PUBLICATION|Exercise Competition File (Multiple choice questions)|20 Videos

  • OSCILLATIONS

    MODERN PUBLICATION|Exercise Competition File (Multiple choice questions) AIPMT/NEET other State Boards for Medical Entrance|14 Videos

  • OSCILLATIONS

    MODERN PUBLICATION|Exercise Revision Exercises (Long Answer Questions)|5 Videos

  • MOTION IN A STRAIGHT LINE

    MODERN PUBLICATION|Exercise CHAPTER PRACTICE TEST|16 Videos

  • PHYSICAL WORLD

    MODERN PUBLICATION|Exercise Revision exercises (Long answer questions)|6 Videos

Similar Questions

Explore conceptually related problems

A particle is executing simple harmonic motion with a time period of 2s and amplitude 10 cm. Calculate its velocity and acceleration 2 seconds after starting from mean position.

A particle is executing simple harmonic motion with an amplitude of 10 cm and time period of 4s. Calculate its displacement, velocity and acceleration after 1/2 seconds, starting from the mean position.

Knowledge Check

  • A particle executes simple harmonic motion with a of period of 6 s and amplitude of 3 cm. Its maximum speed (in cm/s) i

    A
    `pi`
    B
    `2pi`
    C
    `3pi`
    D
    `4pi`

  • A particle is executing simple harmonic motion with a period of T seconds and amplitude a metre. The shortest time it takes to reach alpha/sqrt2 point m from its mean position in seconds is

    A
    T
    B
    `T/4`
    C
    `T/8`
    D
    `T/16`

  • A particle is executing simple harmonic motion with a period of T seconds and amplitude a metre . The shortest time it takes to reach a point a/sqrt2 from its mean position in seconds is

    A
    T
    B
    T /4
    C
    T/8
    D
    T/16

    • Similar Questions

    Explore conceptually related problems

    A particle executes SHM with a time period of 2s and amplitude 5 cm. Find (i) displacement (ii) velocity (iii) acceleration, after 1//3 second, starting from mean position.

    A particle executed SHM iwht a time period of 3s and amplitude 6cm. Find (i) displacement (ii) velocity and (iii) acceleration, after 1//2 second, starting from the mean position.

    A particle executes SHM with a time period of 2s and amplitude 10 cm . Find its (i) Displacement (ii) Velocity (iii) Acceleration after 1/6 s, Starting from mean position.

    A particle executes SHM with a time period of 2 s and amplitude 5 cm. What will be the displacement and velocity of that particle at t = 1/3 ?

    A particle is executing simple harmonic motion with an angular at a 4cm . At the mean position the velocity of the particle is 10cm//sec .The particle from the mean position when its speed becomes 2cm//s is

    Home
    Profile