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    Question

    The kinetic energy of a particle executing  SHM is maximum, when its displacement is equal to

    A

    zero

    B

    amplitude /4

    C

    amplitude /2

    D

    amplitude

    Medium Open in App

    Updated on : 2022-09-05

    Solution Verified by Toppr

    Correct option is A)

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    The kinetic energy of a particle executing SHM is $\\;16J$ , when it is at its mean position. If the mass of the particle is $\\;0.32kg$ , then what is the maximum velocity of the particle? (A) $5m\/s$(B) $15m\/s$(C) $10m\/s$(D) $20m\/s$

    The kinetic energy of a particle executing SHM is $\\;16J$ , when it is at its mean position. If the mass of the particle is $\\;0.32kg$ , then what is the maximum velocity of the particle? (A) $5m\/s$(B) $15m\/s$(C) $10m\/s$(D) $20m\/s$. Ans: Hint:...

    The kinetic energy of a particle executing SHM is

    16J 16J

    , when it is at its mean position. If the mass of the particle is

    0.32kg 0.32kg

    , then what is the maximum velocity of the particle?

    (A) 5m/s 5m/s (B) 15m/s 15m/s (C) 10m/s 10m/s (D) 20m/s 20m/s Answer Verified 212.7k+ views

    Hint: Here we have to use the concepts of simple harmonic motion. Simple harmonic motion can be defined as an oscillatory motion in which the acceleration of the particle at any point is directly proportional to the displacement of the mean position.Complete step-by-step solution:

    The acceleration is the variation in velocity with respect to time. If the velocity of the simple harmonic motion is maximum, the acceleration must be equal to zero. The acceleration is equal to zero only when the particle or object is at the initial position or if the displacement of the particle is zero. We know that the displacement of a particle is zero when the particle doesn’t change its initial position or it comes to the initial position after a certain time period.

    Therefore, the particle will have a maximum velocity at the central position and minimum at the extreme positions.

    At mean positions, the Kinetic energy ( K.E.) is given by,

    KE= 1 2 m. v max 2 KE=12m.vmax2 Here, m m

    = mass of the particle,

    v max vmax

    = velocity of the particle at its mean position

    The given values in question are,

    Mass of object (m)=0.32kg (m)=0.32kg K.E. of object = 16J 16J

    So here we put these values in the given formula to calculate the velocity of the object:

    16= 1 2 ×0.32× v max 2 16=12×0.32×vmax2 v max 2 = 32 0.32 vmax2=320.32

    On further solving the above equation we get,

    v max 2 =100 vmax2=100

    Take the square root on both sides,

    v max = 100 − − − √ vmax=100 v max =10m/s vmax=10m/s

    So the maximum velocity of the particle at its mean position is

    Hence, the correct answer is option (C)

    10m/s 10m/s . 10m/s 10m/s .

    Additional Information:

    Not all oscillatory motions are simple harmonic whereas all harmonic motions are periodic and oscillatory. Oscillatory motion is often referred to as the harmonic motion of all oscillatory movements, the most important of which is basic harmonic motion.

    In a simple harmonic motion, the coming back force or acceleration acting on the particle should always be equal to the displacement of the particle and be oriented towards the equilibrium state.

    In other words, the same equation states the position of an object facing basic harmonic motion and to one component of the position of an object facing uniform circular motion.

    In basic harmonic motion, the speed and displacement of the target are zero at an extreme location.

    Note: Kinetic energy is zero at the extreme position as velocity is zero and potential energy is maximum. Potential energy is minimum at the mean position which is also the lowest position. If the motion of the pendulum makes a very small angle, then this motion can be approximated to Simple harmonic motion. One will need this information to solve SHM and pendulum involving questions.

    स्रोत : www.vedantu.com

    What should be the displacement of a particle, executing SHM, from its position of equilibrium so that the kinetic energy of the particle is half of its maximum kinetic energy?

    Maximum kinetic energy of a particle executing SHM = 1/2momega^(2)A^(2) When the displacement of the particle is x from the position of equilibrium, kinetic energy of the particle = 1/2momega^(2)(A^(2)-x^(2)) According to the question, 1/2momega^(2)(A^(2)-x^(2))=1/2xx1/2momega^(2)A^(2) or, A^(2)-x^(2)=A^(2)/2 or, x^(2)=A^(2)-A^(2)/2=A^(2)/2 or, x=pmA/sqrt2, this is the required displacement.

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    What should be the displacemen...

    What should be the displacement of a particle, executing SHM, from its position of equilibrium so that the kinetic energy of the particle is half of its maximum kinetic energy?

    Updated On: 27-06-2022

    00 : 30

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    Text Solution Open Answer in App Solution

    Maximum kinetic energy of a particle executing SHM =

    1 2 m ω 2 A 2 12mω2A2

    When the displacement of the particle is x from the position of equilibrium, kinetic energy of the particle =

    1 2 m ω 2 ( A 2 − x 2 ) 12mω2(A2-x2)

    According to the question,

    1 2 m ω 2 ( A 2 − x 2 )= 1 2 × 1 2 m ω 2 A 2

    12mω2(A2-x2)=12×12mω2A2

    or, A 2 − x 2 = A 2 2 A2-x2=A22 or, x 2 = A 2 − A 2 2 = A 2 2 x2=A2-A22=A22 or, x=± A 2 – √ x=±A2

    , this is the required displacement.

    Answer

    Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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