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    Two Body system: Motion of two masses attached by spring

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    Here it is not told that m2 is placed on a horizontal surface and thus we will consider that at t = 0 m1 is moving with velocity v, m2 is at rest and just released.
    Let at the initial position of center of mass of the system is the origin.
    As initially velocity v is imparted to m1 only the initial velocity of center of mass of this two body system is given by

    As the only external force on the system is its weight (m1 +m2)*g downwards, the position of center of mass at any subsequent time is given by

    Or
    Now let the extension in the spring at time t is
    Then the tension in the spring is given by F = k
    Now the motion of the two bodies relative to their center of mass is due to internal forces i.e. the tension in the string. Let at time t the position of the two bodies are y1 and y2 relative to their center of mass then

    As we know that the motion of one body relative to the other in a two body system can be considered as a single body system replacing the mass by reduced mass given by

    The motion of the two

    me t the position of m1 is y1 and position of m2 is y2.
    The extension in the spring in this case will be
    l = y1 - y2 - l
    Thus the equation of motion for the two masses can be written as

    And
    Or
    And
    Subtracting we get

    Now as y1 - y2 = l + l and l is natural length of the spring,
    Hence above equation can be written as

    Or
    Where

    Above equation is similar to the equation of a simple harmonic motion for a particle of mass  (called reduced mass) and force constant k. Thus we can say that the system will oscillate about its center of mass and the angular frequency of oscillation is given by

    Now the initial kinetic energy given to the system is and the energy acquired by the system corresponding to the motion of center of mass is . The rest of energy will remain in the system as the energy of oscillations and hence the energy corresponding to oscillation is given by

    Or
    Or
    Now the tension in the spring will increase till the velocity of m1 remains greater than that of m2 and will be maximum when both will move with the velocity of center of mass of the system thus the maximum extension in the spring can be calculated by conservation of energy of oscillation (relative to center of mass) as
    Loss in energy = gain in elastic potential energy
    Or
    Or
    Or
    Hence the amplitude of oscillation of the particle of mass m1 relative to its initial position will be

    And its displacement from its equilibrium position (relative to system) as a function of time is given by

    And hence the position of the particle of mass m1 as a function of time with initial position of center of mass as origin is given by
    = position of center of mass + initial distance of m1 from center of mass with no extension in spring + extension in spring at time t
    Or
    Similarly
    the amplitude of oscillation of the particle of mass m1 relative to its initial position will be

    And its displacement from its equilibrium position (relative to system) as a function of time is given by

    And hence the position of the particle of mass m1 as a function of time with initial position of center of mass as origin is given by
    = position of center of mass - initial distance of m1 from center of mass with no extension in spring - extension in spring at time t
    Or

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 7:46 pm ad1c9bdddf>
    https://brainmass.com/physics/internal-energy/two-body-system-motion-two-masses-attached-spring-217781

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