Vibration and energy - Simple Harmonic Motion (SHM)

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Vibration and energy



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 “equilibrium 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;
    FxF \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=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=12kA2E = \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=12mvmax2Vmax2=(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.
  • Intro Lesson
  • 1.
    When a family of three with a total mass of 150kg steps into their 1100kg car, the car's spring compresses 2.0cm.
  • 3.
    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.
  • 4.
    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.
  • 5.
    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;
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Vibration and energy

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