Vibration and energy

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  1. Vibration
  2. Hooke's Law
  3. Energy
  1. When a family of three with a total mass of 150kg steps into their 1100kg car, the car's spring compresses 2.0cm.
    1. What is the spring constant of the car's spring?
    2. How far will the car lower if loaded with 400kg?
  2. A spring is 45cm long when a weight of 65N hangs from it but is 96cm when a weight of 190 N hangs from it. What is the spring constant?
    1. A spring stretches 0.120m vertically when a mass of 0.400kg is attached to its end. Then spring is set up horizontally with the 0.400kg resting on a frictionless surface. The mass is pulled so that the spring is stretched 0.200m from the equilibrium position.
      1. Find the spring constant
      2. Find the amplitude of the horizontal oscillation
      3. What is the magnitude of the maximum velocity?
      4. Find the magnitude of the velocity when the mass is 0.04m from the equilibrium.
      5. Determine the maximum acceleration of the mass.
    2. An object with mass 4.0kg is attached to a spring with spring stiffness constant kk = 200N/m and is performing the simple harmonic motion. When the object is 0.04m from its equilibrium position, it is moving with a speed of 0.62m/s.
      1. Calculate the amplitude of the motion.
      2. Calculate the maximum velocity attained by the object.
    3. A mass sitting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. 2.0J of work is required to compress the spring by 0.14m.

      If the mass experiences a maximum acceleration of 18 m/s2;
      1. Find the spring constant
      2. Find the mass
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    Topic Notes

    In this lesson, we will learn:

    • Vibrating and oscillating systems
    • Hooke’s law
    • How to find the amount of energy stored in a vibrating system?


    • There are many examples of vibrating objects around us, an object attached to the end of spring, guitar strings, ruler held firmly at the end of the table.
    • During vibration (oscillation) the object moves back and forth about a fixed position called β€œequilibrium position”.
    • Let’s consider a mass vibrating at the end of a uniform spring.;
    \qquad \quad (a) The mass is at its equilibrium position, there is no force exerted.

    Vibration and energy

    \qquad \quad (b) The mass is oscillating; the restoring force tries to pull the mass back to its equilibrium position.
    Vibration and energy

    \qquad \quad (c) The mass is oscillating; the restoring force tries to push the mass back to its equilibrium position.
    Vibration and energy

    Hooke’s Law
    • The magnitude of the restoring force is directly proportional to the displacement xx;
      F∝xF \propto x
    • The direction of the restoring force is always opposite to the displacement which is indicated by a minus sign in the equation.

      F=βˆ’kxF = - kx (Hooke’s Law)

    FF: Force exerted by the spring on the mass
    KK: Spring constant
    xx: Displacement

    • As we know the mechanical energy of a system is the sum of kinetic and potential energies.
    • In the case of the spring-mass system, the potential energy would be in the form of elastic potential energy in the spring which is calculated using the following equation;

    PE=12kx2 PE = \frac{1}{2}kx^{2}

    Therefore; the mechanical energy of the system is;

    E=12mv2 + 12kx2E = \frac{1}{2}mv^{2} \, + \, \frac{1}{2}kx^{2}

    \qquad \quad (a) At the extreme points where the mass stops momentarily to change the direction; vv = 0, xx = AA (amplitude, maximum displacement)

    E=12kA2 E = \frac{1}{2}kA^{2} \, (1)

    \qquad \quad (b) At the equilibrium position the mass moves with maximum velocity;

    V=Vmax,x=0V = V_{max},x = 0

    E=12mvmax2E = \frac{1}{2} mv^{2} _{max} (2)

    From (1) and (2);

    12kA2=12mvmax2β‡’Vmax2=(km) A2 \frac{1}{2}kA^{2} = \frac{1} {2}mv^{2} _{max} \quad \Rightarrow \quad V^{2} _{max} = (\frac{k}{m}) \, A^{2}

    \qquad \quad (c) At intermediate points, the energy is a part kinetic and part potential;

    12mv2 + 12kx2=12kA2 \frac{1}{2}mv^{2} \, + \, \frac{1}{2}kx^{2} = \frac{1}{2}kA^{2}
    • From the above equation we can find velocity as the function of position;

    Vibration and energy

    This gives the velocity of the object at any position.