2.5: Spring-Mass Oscillator (2024)

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    Spring-Mass Force

    Take a look at the three scenarios depicted below for a horizontal spring-mass system. In picture A) the spring-mass is at its equilibrium position (x=0), which means that it is not experiencing a net force. The equilibrium position is where the spring-mass system "wants" to naturally be. When the mass is displaced from the equilibrium, the force, known as restoring force, will always point back toward equilibrium. In picture B) the spring is compressed and the force pushes the mass back toward equilibrium, where in picture C) the spring is stretched, and the restoring force pulls the mass back to the left.

    Figure 2.5.1: forces experienced by a spring-mass system.

    2.5: Spring-Mass Oscillator (2)

    2.5: Spring-Mass Oscillator (3)The force with which a spring pulls back when stretched (or pushes back when compressed) is proportional to the amount of stretch from equilibrium, provided the spring is not stretched too far. Historically, this linear proportionality between the force and amount of stretch is referred to as Hooke’s Law behavior, which is true if the mass of the spring is negligible. We write the restoring force as:

    \[F = -k x\]

    where k is the “spring constant” or “force constant” (and depends on the stiffness of the particular spring), and x is the displacement from equilibrium. The minus sign indicates that the force is the opposite direction to the direction the spring was stretched or compressed. In other words, if you are applying a force on the spring-mass that displaces it from equilibrium, the restoring force will act in the opposite direction of the applied force. An important point to notice in the expression for the force of the spring is that x is measured from the un-stretched position of the free end of the spring.

    In order for the force to have units of newtons, the units of k must be newtons per meter (N/m). The force that you (an external agent) have to exert on the spring to stretch it a distance \(x\) is in the opposite direction to the restoring force and is equal to \(F_{applied}=+k x\).

    Hooke's Law breaks down at the extremes of a spring's motion. For example, when stretched to the point of breaking or permanent deformation, a spring's behavior will begin to deviate substantially from the linear expectation. Also, when compressed so far that it begins to touch itself, and the forces at play change. For these reasons, it is usually assumed that springs stretch only within a small portion of their maximum deformation.

    Spring-Mass Potential Energy

    Let us refer back to Figure 2.3.4. Work is done when the mass is pulled away from equilibrium. Assuming the kinetic energy stays constant (spring-mass is motionless at equilibrium and held in place when stretched), the work done contributes only to increasing the potential energy of the spring-mass system. We call this type of energy, the spring-mass potential energy, \(P_{sm}\). The change in potential energy depends on the amount of work done. Work is proportional to force which depends on the distance away from equilibrium. This suggests that the indicator for \(P_{sm}\) is displacement from equilibrium, denoted by d or |x| or sometimes |y| for vertically hanging springs. Note, the absolute value sign around "|x|" implies that the potential energy does not depend on whether the spring is compressed or stretched, since Hooke's Law is symmetric about the equilibrium position. Assuming the spring-mass is stretched a distance d from equilibrium, the Energy-Interaction diagram for the interval depicted in Figure 2.3.4 is:

    Figure 2.5.2: Energy-Interaction diagram for stretching a spring-mass system.

    2.5: Spring-Mass Oscillator (4)

    The restoring force, \(F = -kx\), is not constant since it depends on displacement x. That is, the force is proportional to the distance the spring has been pushed or pulled. So we can no longer used the simplified version of work, but will use the general Equation 2.3.1 for work. The applied force does work when the spring is stretched from distance of zero to a max of d, in this example. Thus, the work done on this system:

    \[W = \int \limits_0^d kxdx= \dfrac{1}{2}kd^{2}\]

    Notice that while the force scales linearly with deformation, the work required to deform the spring scales as the square of deformation. This also confirms that the potential energy will depend on the magnitude of the displacement only, and not the direction. Using the energy-conservation equation from Figure 2.5.2 we get \(\Delta PE_{sm}= \dfrac{1}{2}kd^{2}\). Generally, the spring-mass potential energy is given by:

    \[PE_{sm} = \dfrac{1}{2}kx^{2}\]

    where x is displacement from equilibrium. Upon stretching the spring, energy is stored in the springs' bonds as potential energy. This potential energy is released when the spring is allowed to oscillate. The maximum speed is accomplished when the spring returns to its equilibrium position (PEsm=0), and all energy is kinetic energy.

    Example \(\PageIndex{1}\)

    A spring of negligible mass and a spring constant of 120 N/m is fixed to a wall and is free to oscillate. On the other end, a ball with a mass of 1.5 kg is attached. The spring-mass is then stretched 0.4 m and released.

    a) What is the maximum speed of the attached ball? Neglect any effects of friction.

    b) The spring-mass looses 2 Joules of energy per oscillation. Find the maximum displacement of the spring-mass after 4 oscillations.


    a) Let us focus on the interval once the ball is released until it is at equilibrium when the speed is maximum.The spring-mass system is closed since it is not interacting with other systems in this interval. As the spring-mass oscillates potential and the kinetic energies change, \(\Delta E_{tot}= \Delta PE_{sm}+\Delta KE=0\). The Energy-Interaction Diagram for this process is shown below:

    2.5: Spring-Mass Oscillator (5)

    Using definitions of potential and kinetic energies:


    Plugging in values into the equation:

    \[\frac{1}{2}(120 N/m)(0-0.4^2m^2)+\frac{1}{2}(1.5 kg)(v_{max}^2-0)=0\nonumber\]

    and solving for \(v_f\), we get \(v_f=3.58 m/s\).

    b) In 4 oscillations the spring-mass will loose 8 Joules. Let's choose an interval from initial maximum displacement of 0.4m to the final maximum displacement after 4 oscillations. Since the speed is zero when the displacement is maximum, there will be no change in kinetic energy over this interval. The Energy-Interaction Diagram for this process is shown below:

    2.5: Spring-Mass Oscillator (6)

    The equation can be rewritten as:


    Plugging in values into the equation:

    \[\frac{1}{2}(120 N/m)(d_{max}^2-0.4^2m^2)=-8 J\nonumber\]

    Solving for \(d_{max}\), we find the maximum displacement after 4 oscillations to be 0.16m.

    Vertical Spring-Mass Systems

    A spring with a mass hanging down acts exactly like a horizontal spring, except that the end of the spring has a different equilibrium position. For a vertically hanging spring-mass there are two forces acting on the mass as it oscillates: the force from the spring pulling up or pushing down, and the force from the Earth always pulling down. Let us take a closer look why the horizontal and vertical spring-mass systems can still be treated identically.

    For the vertical spring (as for the horizontal one), the force with which a spring-mass pulls back when stretched (or pushes back when compressed) is proportional to the amount of stretch from the equilibrium determined with the mass attached (provided the spring is not stretched too far). For a vertical spring, we will usually write the restoring force using the symbol “y” instead of with an “x”, but this is just a convention:

    \[F = -k y\]

    Similarly to the horizontally attached spring in Example 2.5.1, let us analyze a situation of finding the maximum speed of the mass at equilibrium. In the example, we found the following general equation for an interval when the mass is released from maximum displacement to its equilibrium position, assuming thermal dissipation is negligible:


    Let us now see what we get for a vertical spring given the same interval. We will start by measuring the potential energy of this system from (y=0) in the figure below, the equilibrium position of the spring without the mass attached.

    Figure 2.5.3: Vertical spring-mass system.

    2.5: Spring-Mass Oscillator (7)

    When the mass is initially attached it stretches by a distance \(d_1\) to a new spring-mass equilibrium. At this position the spring-mass is not moving, so the forces acting on it must be balanced. That is, the gravitation force must be equal and opposite to the spring-mass restoring force:


    The spring-mass is then stretched another distance \(d_2\), released, and allowed to oscillate. Recall, all the vertical heights used in calculating PEg are measured from the y=0 defined in Figure 2.5.3. Thus, the values of y for the equilibrium and stretched positions (center and right configurations in the figure) will be negative. The Energy-Interaction Diagram for calculating the maximum speed when the spring-mass returns back to equilibrium after release is shown in the figure below.

    Figure 2.5.4: Energy-Interaction diagram for a vertical spring-mass system.

    2.5: Spring-Mass Oscillator (8)

    Expanding the equation from the Energy-Interaction diagram and plugging in the variables for the energy indicators:





    Using Equation \ref{forces.vertical.SM} for d1, the equation simplifies further to:


    The above equation is identical to the horizontal spring-mass result in Equation \ref{energy.SM}, as long as the displacement for the vertical spring-mass is measured from the new equilibrium position once the mass is attached to the spring.


    The result above shows that when a spring-mass is vertically attached the effect of the gravitation force is automatically taken into account, as long as the new equilibrium position is defined at the position of the spring with the mass attached. Using the new equilibrium position the potential energy of the spring-mass is described by the behavior of the spring-mass oscillator only. Imagine if the spring-mass system was far in outer space away from gravity, mounted on a wall of a spaceship. The spring-mass would still oscillate, but the orientation of "horizontal" vs. "vertical" would lose its meaning.


    2.5: Spring-Mass Oscillator (2024)


    What is the oscillation of a spring mass? ›

    A mass suspended on a spring will oscillate after being displaced. The period of oscillation is affected by the amount of mass and the stiffness of the spring. This experiment allows the period, displacement, velocity and acceleration to be investigated by datalogging the output from a motion sensor.

    What is the formula for the mass spring oscillator? ›

    my + by + ky = Fext. This is the differential equation that governs the motion of a mass-spring oscillator. To start, we consider on external force and no friction, my + ky = 0.

    What is the aim of the spring mass oscillator? ›

    Aim: To determine the force constant (k) of the given spring. To determine the mass of the spring. Apparatus: Light spiral spring with clamping arrangement and attached pointer, meter scale, hanger, weight box, stop watch. Where, F-is restoring force.

    What is the motion of a mass oscillating on a spring? ›

    In simple harmonic motion, the acceleration of the system, and therefore the net force, is proportional to the displacement and acts in the opposite direction of the displacement. A good example of SHM is an object with mass m attached to a spring on a frictionless surface, as shown in (Figure).

    What is the effective mass of an oscillating spring? ›

    The oscillations of a spring-mass oscillator is one classic example of simple harmonic motion in which the period of oscillations T is expressed as 6 (1) where m is the suspended mass, k the spring constant, and m eff(s) is the effective mass of the spring and is equal to onethird of the mass of the spring for the case ...

    What is the formula for the period of oscillation of a spring? ›

    Mass on a spring - Where a mass m attached to a spring with spring constant k, will oscillate with a period (T). Described by: T = 2π√(m/k).

    How do you calculate spring mass? ›

    The spring constant of a spring can only be calculated experimentally. We hold it upright, attach a mass m and note its elongation x . We then use the formula F=−kx F = − k x , where k is the spring constant to be obtained. Force F here is the weight of the mass m and is given by mg .

    What is the equation for the spring mass vibration? ›

    The equation of motion for a spring-mass system excited by a harmonic force is. M x ¨ + k x = F cos ⁡ Where M is the mass, K is the spring stiffness, F is the force amplitude and ω is the angular frequency of excitation.

    What is the frequency of an oscillating mass on a spring? ›

    Consider a mass M hanging from a spring of unstretched length l, spring constant k, and mass m. If the mass of the spring is neglected, the oscillation frequency would be ω = √k/M. The quoted rule suggests that the effect of the spring mass would be to replace M by M + m/3 in the equation for ω.

    What is the principle of spring-mass oscillator? ›

    If the object, or mass, is pulled downwards a distance A from the equilibrium position and then released, the spring restoring force will initially cause the object to accelerate upwards. This would continue until the object moves above the equilibrium position, and the spring compresses past that point.

    What reduces spring oscillation? ›

    A: Damping reduces or dissipates the energy of oscillations in a spring system to control and minimize vibrations.

    Why would an ideal mass spring system oscillate indefinitely? ›

    Short Answer

    The energy is constantly converted from potential energy (when the spring is compressed or extended) to kinetic energy (when the mass is moving), and back again, with no energy lost to friction or any other forces. Hence, the oscillation continues indefinitely.

    How do you find the oscillation of a spring? ›

    Step 1: Identify the angular frequency and amplitude A of the spring and plug into the equation for the position of an oscillating spring: x ( t ) = A cos ⁡ . If the angular frequency is not given, calculate it using the equation for frequency to angular frequency: ω = 2 π f .

    What is the effect of spring mass on oscillation? ›

    IB Physics Tutor Summary: The time period of a mass-spring system depends mainly on the mass (m) and spring constant (k), shown in the formula T = 2π√(m/k). A heavier mass or a stiffer spring (higher k) changes the time it takes to complete one oscillation.

    What are the oscillations of springs? ›

    A spring moves downwards and upwards, which creates oscillatory movement. Oscillations can be divided into two types – damped oscillation and undamped oscillation. Alternating current or AC waves is an example of undamped oscillation. Imagine a spring with no force applied to it.

    What is meant by oscillation? ›

    Oscillation is going back and forth repeatedly between two positions or states. An oscillation can be a periodic motion that repeats itself in a regular cycle, such as the side-to-side swing of a pendulum, or the up-and-down motion of a spring with a weight.

    What is the law of spring oscillation? ›

    Springs to a large extent obey Hooke's law, which says the force that an elastic object exerts when it is stretched or compressed is directly proportional to the distance over which it is compressed/stretched, namely F = -kx, where k is called the spring constant or stiffness, measured in N/m, and x is measured from ...

    What does an oscillating mass spring system have? ›

    An oscillating mass spring system has a mechanical energy 1 joule, when it has an amplitude 0.1m and maximum speed of 1ms−1.


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