Many objects vibrate or oscillate – an object on the end of a spring, a tuning fork,  a pendulum, a plastic ruler held firmly over the edge of a table and gently struck, the strings of a guitar or piano. Spiders detect prey by the vibrations of their webs; cars oscillate up and down when they hit a bump; buildings and bridges vibrate when heavy trucks pass or the wind is fierce. Indeed, because most solids are elastic, they vibrate (at least briefly) when given an impulse. Electrical oscillations are necessary in radio and television sets. At the atomic level, atoms vibrate within a molecule, and the atoms of a solid vibrate about their relatively fixed positions. Because it is so common in everyday life and occurs in so many areas of physics, oscillatory motion is of great importance. Mechanical oscillations are fully described on the basis of Newtonian mechanics.

Vibrations and wave motion are intimately related subjects. Waves – whether ocean waves, waves on a string, earthquake waves, or sound waves – have as their source a vibration.  In the case of sound not only is the source a vibrating object, but so is the detector. Indeed, when a wave travels through a medium, the medium vibrates (such as air for sound waves). After we discuss vibrations we will discuss  simple waves such as those on water or on a string, and we will encounter other forms of wave motion, including electromagnetic waves and light.

Oscillations of a Spring

When an object vibrates or oscillates back and forth, over the same path, each oscillation taking the same amount of time, the motion is periodic. The simplest form of periodic motion is represented by an object oscillating on the end of a uniform coil spring. Because many other types of oscillatory motion closely resemble this system, we will look at it in detail. We assume that the mass of the spring can be ignored, and that the spring is mounted horizontally, as shown in Fig. 1, so that the object of mass m slides without friction on the horizontal surface. Any spring has a natural length at which it exerts no force on the mass m. The position of the mass at this point is called the equilibrium position. If the mass is moved either to the left, which compresses the spring, or to the right, which stretches it, the spring exerts a force on the mass that acts in the direction of returning the mass to the equilibrium position; hence it is called a restoring force. We consider the common situation where we can assume the restoring force F is directly proportional to the displacement x the spring has been stretched or compressed from the equilibrium position:

F = – kx     [force exerted by spring]                                                                                               (1)

Note that the equilibrium position has been chosen at x = 0 and the minus sign in Eq. 1 indicates that the restoring force is always in the direction opposite to the displacement x. For example, if we choose the positive direction to the right in Fig. 1, x is positive when the spring is stretched (Fig. 1-1b), but the direction of the restoring force is to the left (negative direction). If the spring is compressed, x is negative (to the left) but the force F acts toward the right (Fig. 1-1c).

A mass oscillating at the end of a uniform spring.

Equation (1) is often referred to as Hooke’s law, and is accurate only if the spring is not compressed to where the coils are close to touching, or stretched beyond the elastic region. Hooke’s law works not only for springs but for other oscillating solids as well; it thus has wide applicability, even though it is valid only over a certain range of F and x values.
The proportionality constant k  is called the spring constant for that particular spring, or its spring stiffness constant. To stretch the spring a distance x, one has to exert an (external) force on the free end of the spring with a magnitude at least equal to

Fext = + kx           [external force on spring]

The greater the value of k, the greater the force needed to stretch a spring a given
distance. That is, the stiffer the spring, the greater the spring constant k.

Note that the force F in Eq. (1) is not a constant, but varies with position. Therefore the acceleration of the mass m is not constant, so we cannot use the equations for constant acceleration.

Force and acceleration are not constant !

Let us examine what happens when our uniform spring is initially compressed a distance x = – A, as shown in Fig. 2a, and then released on the frictionless surface. The spring exerts a force on the mass that pushes it toward the equilibrium position. But because the mass has inertia, it passes the equilibrium position with considerable speed. Indeed, as the mass reaches the equilibrium position, the force on it decreases to zero, but its speed at this point is a maximum, Vmax . As the mass moves farther to the right, the force on it acts to slow it down, and it stops for an instant at x = A (Fig. 2c). It then begins moving back in the opposite direction, accelerating until it passes the equilibrium point (Fig. 14-2d), and then slows down until it reaches zero speed at the original starting point, x = – A (Fig. 2e). It then repeats the motion, moving back and forth symmetrically between x = A and x = – A.

A mass at different positions of its oscillation cycle

To discuss oscillatory motion, we need to define a few terms. The distance x of the mass from the equilibrium point at any moment is called the displacement. The maximum displacement-the greatest distance from the equilibrium point-is called the amplitude, A. One cycle refers to the complete to-and-fro motion from some initial point back to that same point-say, from x = – A to x = A and back to x = – A. The period, T, is defined as the time required to complete one cycle. Finally, the frequency, f, is the number of complete cycles per second. Frequency is generally specified in hertz (Hz), where 1 Hz = 1 cycle per second. It is easy to see, from their definitions, that frequency and period are inversely related

f = 1/T       and      T = 1/f                                                                                                                    (2)

for example, if the frequency is 5 cycles per second, then each cycle takes 0,2 s. Often the frequency f is denoted with the Greek letter  ν.

The oscillation of a spring hung vertically is essentially the same as that of a horizontal spring. Because of gravity, the length of a vertical spring with a mass m on the end will be longer at equilibrium than when that same spring is horizontal, as shown in Fig. 3. The spring is in equilibrium when ΣF = 0 = mg – kxo, so the spring stretches an extra amount Xo = mg / k to be in equilibrium. If x is measured from this new equilibrium position, Eq. (1) can be used directly with the same value of k.

Vertical spring-min

Example: Car springs. 

When a family of four with a total mass of  200 kg step into their 1200-kg car, the car’s springs compress 3.0 cm. (a) What is the spring constant of the car’s springs,   assuming they act as a single spring? (b) How far will the car lower if loaded with 300 kg rather than 200 kg?

Solution: We use Hooke’s law: the weight of the people, mg, causes a 3.0-cm displacement.

(a) The added force of (200 kg)(9.8 m/s^2) = 1960 N causes the springs to compress 3.0 X 10^(-2) m. Therefore (Eq. 1), the spring constant is

k = F/x = 6.5 X 10^{4} N/m.

(b) If the car is loaded with 300 kg, Hooke’s law gives

x = F/k = 4,5 . 10^{-2} m   or 4,5 cm. We could have obtained x without solving for k: since x is proportional to F, if 200 kg compresses the spring 3.0 cm, then 1.5 times the force will compress the spring 1.5 times as much, or 4.5 cm.

Simple Harmonic Motion

Any oscillating system for which the net restoring force is directly proportional to the negative of the displacement (as in Eq. 1, F = -kx) is said to exhibit simple harmonic motion (SHM). Such a system is often called a simple harmonic oscillator (SHO).  Most solid materials stretch or compress according to Eq. 1 as long as the displacement is not too great. Because of this, many natural oscillations are simple harmonic or close to it.

EXERCISE Which of the following represents a simple harmonic oscillator: (a) F = -0, 4 . x^{2};   (b) F = -1,3z;   (c) F = 5.6x;    (d) F = -7θ?

RESPONSE Both (b) and (d) represent simple harmonic oscillators because they give the force as minus a constant times a displacement.  The displacement need not be x,  but the minus sign is required to restore the system to equilibrium, which is why (c) is not a SHO.

Energy in the Simple Harmonic Oscillator*

When forces are not constant, as is the case here with simple harmonic motion, it is often convenient and useful to use the energy approach.

Potential energy  When a force is applied to stretch a spring, such as by hanging an object on its end, there is a direct linear relationship between the exerted force and the displacement, as shown by the graph in Figure 4.

Force vs. Displacement for linear spring

The work W (in joules) done by a constant force F is equal to the product of that force and the distance d over which it acts:

W = F. d

The area under the curve represents the work done to stretch the spring, and therefore equals the elastic potential energy that is stored in the spring as a result of that work. We know that the area under a force-displacement curve for a linear spring will always be a right triangle, and the area of a right triangle is:

Area = \frac{1}{2}. Base. Height

Physics Springs Tutorial Potential Energy as area under force-displacement curve for linear spring.

Figure 5. The area under the force-displacement curve for a linear spring                              forms a right triangle, and the area of that triangle is used to calculate the                            potential energy stored in the spring.

The base of the triangle is x, and the height is the force, which, according to the equation for Hooke’s law, is equal to kx, so the potential energy in the spring is given by the following equation:

PE = \frac{1}{2} F.x = \frac{1}{2} (kx).x = \frac{1}{2} k.x^2                                         (2)

The total mechanical energy is the sum of the kinetic and potential energies:
E =\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}                                                                        (3)

where v is the velocity of the mass m when it is a distance x from the equilibrium position. SHM can occur only if there is no friction, so the total mechanical energy E
remains constant. As the mass oscillates back and forth, the energy continuously changes
from potential energy to kinetic energy, and back again. At the extreme points, x = A and x = – A, all the energy is stored in the spring as potential energy (and is the same whether the spring is compressed or stretched to the full amplitude). At these extreme points, the mass stops for an instant as it changes direction, so v = 0 and:

E =\frac{1}{2}m0^{2}+\frac{1}{2}kA^{2}=\frac{1}{2}kA^{2}                                      (4)

Thus, the total mechanical energy of a simple harmonic oscillator is proportional to the
square of the amplitude. At the equilibrium point, x = 0, all the energy is kinetic:

E =\frac{1}{2}mv^{2}+\frac{1}{2}k0^{2}=\frac{1}{2}mv_m^{2}                               (5)

where v_ m is the maximum velocity during the motion. At intermediate points the energy is part kinetic and part potential, and because energy is conserved

E = \frac{1}{2}kA^{2} = \frac{1}{2}mv_m^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}


 v_m=A(k/m)^{1/2}                                                                                                              (6)


v=\pm v_m(1-x^{2}/A^{2})^{1/2}                                                                                     (7)

The Period and Sinusoidal Nature of SHM

The period of a simple harmonic oscillator is found to depend on the stiffness of the spring and also on the mass m that is oscillating. But-strange as it may seem-the period does not depend on the amplitude. You can find this out for yourself by using a watch and timing 10 or 20 cycles of an oscillating spring for a small amplitude and then for a large amplitude.

We can derive a formula for the period of simple harmonic motion (SHM) by comparing SHM to an object rotating in a circle.  It turns out that simple harmonic motion has a simple relationship to a particle rotating in a circle with uniform speed. From this same “referecnce circle” we can obtain a second useful result-a formula for the position of an
oscillating mass as a function of time. There is nothing actually rotating in a circle when a spring oscillates linearly,  but it is the mathematical similarity that we find useful.

Period and Frequency

Consider a small object of mass m  revolving counterclockwise in a circle of radius A, with constant  speed v_m on top of a table as shown in Fig. 6. As viewed from above the motion is a circle in the xy plane. But a person who looks at the motion from the edge of the table sees an oscillatory motion back and forth, and this one-dimensional motion corresponds precisely to simple harmonic motion, as we shall now see.

shm-circular motion comparison
Fig. 6 a. Circular motion of a small object b. Side view of circular motion (x component) is simple harmonic motion

What the person sees, and what we are interested in, is the projection of the circular motion onto the x axis (Fig. 6b).  To see that this x-motion is analogous to SHM, let us calculate the magnitude of the x component of the velocity   v_m which is labeled v in Fig. 6.  The two triangles  involving θ  in Fig. 6 are similar, so

 \frac{v}{v_m} =  \frac{(A^{2}-x^{2})^{1/2}}{A}  


v=\pm v_m(1-x^{2}/A^{2})^{1/2}

This is exactly the equation for the speed of a mass oscillating with SHM, as we saw. Thus the projection on the x axis of an object revolving in a circle has the same motion as a mass at the end of a spring.

We can now determine the period of SHM because it is equal to that of the revolving object making one complete revolution. First we note that the the velocity v_m is equal to tile circumference of the circle (distance) divided by the period T:

 v_m=  \frac{2\pi A}{T} =2πAf                                                                                          (8)

We solve for the period T:

T = \frac{2\pi A}{v_m}

using eq. 6 we get

T=2\pi (m/k)^{1/2}                                                                                                               (9)

And this is the formula we were looking for. The period depends on the mass m and the spring stiffness constant k, but not on the amplitude A. We see from eq. 9  that the larger the mass, the longer the period; and the stiffer the spring (larger k), the shorter the period. This makes sense since a larger mass means more inertia and therefore slower response (smaller acceleration). And larger k means greater force and therefore quicker response (larger acceleration). Notice that Eq. 9 is not a direct proportion: the period varies as the square root of m/k. For example, the mass must be quadrupled to double the period.

Equation 9 is fully in accord with the experiment and is valid not only for a spring, but for all kinds of simple harmonic motions – that is, for motion subject to restoring force proportional to displacement, Eq. 1.

We can write the frequency using f=1/T:

f=\frac{1}{T}=\frac{1}{2\pi}(k/m)^{1/2}                                                                        (10)

EXERCISE Does a car bounce faster on its springs when empty or fully loaded?

Example: Spider web.  A spider of mass 0.30 g waits in its web of negligible mass. A slight movement causes the web to vibrate with a frequency of about 15 Hz. (a) Estimate the value of the spring stiffness constant k for the web. (b) At what frequency would you expect the web to vibrate if an insect of mass 0.1 g were trapped in addition to the spider?

SOLUTION We use SHM as an approximate model.

(a) The frequency is given by Eq. 10, solving for k:

 k =  (2\pi f)^{2} m    or k = 2.7 N/m

(b) The total mass is now 0,1g +0,3g=0,4g and we could substitute it into Eq. 10  Instead we  notice that the frequency decreases with the square root of the mass. Since the new mass is 4/3 time the first mass, the frequency changes by a factor of 1/(4/3)^{1/2}=(3/4)^{1/2}. Thus f = (15Hz)(3/4)^{1/2}=13 Hz

The Simple Pendulum

A simple pendulum consists of a small object (the pendulum bob) suspended from the end of a lightweight cord. We assume that the cord does not stretch and that its mass can be ignored relative to that of the bob. The motion of a simple pendulum moving back and forth with negligible friction resembles simple harmonic motion: the pendulum oscillates along the arc of a circle with equal amplitude on either side of its equilibrium point and as it passes through the equilibrium point (where it would hang vertically) it has its maximum speed. But is it really undergoing SHM? That is, is the restoring force proportional to its displacement? Let us find out.

Simple pendilum

The displacement of the pendulum along the arc is given by x = lθ, where θ is the angle (in radians) that the cord makes with the vertical and l is the length of the cord (Fig. 7). If the restoring force is proportional to x or to θ, the motion will be simple harmonic. The restoring force is the net force on the bob, equal to the component of the weight, mg, tangent to the arc:

F = – mg sinθ

where g is the acceleration of gravity. The minus sign here, as in Eq. 1, means that the force is in the direction opposite to the angular displacement θ. Since F is proportional to the sine of a and not to a itself, the motion is not SHM. However, if θ is small, then sinθ is very nearly equal to θ when the latter is specified in radians.  For angles less than 15°, the difference between θ (in radians) and sinθ  is less than 1%. Thus, to a very good approximation for small angles,

F = – mgsinθ ≈ – mgθ

Substituting x = lθ or  θ = x/l, we have

F = - \frac{mg}{l} x

Thus, for small displacements, the motion is essentially simple harmonic, since this equation fits Hooke’s law, F = -kx. The effective force constant is k = mg/l. If we substitute k=mg/l into Eq. 9 we obtain the period of simple pendulum:

T = 2\pi (\frac{l}{g})^{1/2}                                                                            [small  θ]  (11)

The frequency is f=1/T, so

f =\frac{1}{2\pi}(g/l)^{1/2}                                                                          [small  θ]  (12)

The mass m of the pendulum bob does not appear in these formulas for T and f. Thus we have the surprising result that the period and frequency of a simple pendulum do not depend on the mass of the pendulum bob. You may have noticed this if you pushed a small child and then a large one on the same swing.

We also see from Eq. 11 that the period of a pendulum does not depend on the amplitude (like any SHM), as long as the amplitude a is small. Galileo is said to have first noted this fact while watching a swinging lamp in the cathedral at Pisa. This discovery led to the invention of the pendulum clock, the first really precise timepiece, which became the standard for centuries.

Damped Harmonic Motion

The amplitude of any real oscillating spring or swinging pendulum slowly decreases in time until the oscillations stop altogether. Figure 8 shows a typical graph of the displacement as a function of time. This is called damped harmonic motion.  To “damp” means to diminish, restrain, or extinguish, as to “dampen one’s spirits.” The damping  is generally due to the resistance of air and to internal friction within the oscillating system. The energy that is dissipated to thermal energy is reflected in a decreased amplitude of oscillation.

Damped harmonic motion

Since natural oscillating systems are damped in general, why do we even talk about (undamped) simple harmonic motion? The answer is that SHM is much easier to deal with mathematically. And if the damping is not large, the oscillations can be thought of as simple harmonic motion on which the damping is superposed. The decrease in amplitude shown by the dashed curves in Fig. 8 represents the damping.  Although damping does alter the frequency of vibration, the effect is usually small if the damping is small.

In many systems, the oscillatory motion is what counts, as in clocks and watches, and damping needs to be minimized. In other systems, oscillations are the problem, such as a car’s springs, so a proper amount of damping is desired. Well-designed damping is needed for all kinds of applications. Large buildings, especially in California, are now built (or retrofitted) with huge dampers to reduce earthquake damage.

Forced Oscillations; Resonance

When an oscillating system is set into motion, it oscillates at its natural frequency (Eqs. 10 and 12). However, a system may have an external force applied to it that has its own particular frequency and then we have a forced oscillation.

For example, we might pull the mass on the spring of Fig. 1 back and forth at a frequency f. The mass then oscillates at the frequency f of the external force, even if this frequency is different from the natural frequency of the spring, which we will now denote by f_0 where


In a forced oscillation the amplitude of oscillation, and hence the energy transferred to the oscillating system, is found to depend on the difference between f and f_0  as
well as on the amount of damping, reaching a maximum when the frequency of the external force equals the natural frequency of the system-that is, when f = f_0 . The amplitude is plotted in Fig. 9 as a function of the external frequency f. Curve A represents light damping and curve B heavy damping. The amplitude can become large when the driving frequency f is near the natural frequency, f≈f_0 , as long as the damping is not too large. When the damping is small, the increase in amplitude near f = f_0  is very large (and often dramatic). This effect is known as resonance. The natural frequency f_0  of a system is called its resonant frequency.

Resonance for lightly damped (A) and heavily damped (B) systems.

A simple illustration of resonance is pushing a child on a swing. A swing, like any pendulum, has a natural frequency of oscillation that depends on its length L. If you push on the swing at a random frequency, the swing bounces around and reaches no great amplitude. But if you push with a frequency equal to the natural frequency of the swing, the amplitude increases greatly. At resonance, relatively little effort is required to obtain a large amplitude.

The great tenor Enrico Caruso was said to be able to shatter a crystal goblet by singing a note of just the right frequency at full voice. This is an example of resonance, for the sound waves emitted by the voice act as a forced oscillation on the glass. At resonance, the resulting oscillation of the goblet may be large enough in amplitude that the glass exceeds its elastic limit and breaks.

Since material objects are, in general, elastic, resonance is an important phenomenon in a variety of situations. It is particularly important in structural engineering, although the effects are not always foreseen. For example, it has been reported that a railway bridge collapsed because a nick in one of the wheels of a crossing train set up a resonant oscillation in the bridge. Indeed, marching soldiers break step when crossing a bridge to avoid the possibility that their normal rhythmic march might match a resonant frequency of the bridge. The famous collapse of the Tacoma Narrows Bridge  in 1940 occurred as a result of strong gusting winds driving the span into large-amplitude oscillatory motion. The Oakland freeway collapse in the 1989 California earthquake
involved resonant oscillation of a section built on mudfill that readily transmitted
that frequency.

Large-amplitude oscillations of the Tacoma Narrows Bridge, due to gusty winds, led to its collapse (1940).

Tacoma Narrows Bridge in Washington

In 1940, four months after being completed, the Tacoma Narrows Bridge in Washington State was destroyed by wind-generated resonance. A mild gale produced an irregular force in resonance with the natural frequency of the bridge, steadily increasing the amplitude of vibration until the bridge collapsed.