SIMPLE HARMONIC MOTION (SHM)
Simple harmonic motion is a special type of oscillatory motion in which the acceleration or force on the particle is directly proportional to its displacement from a fixed point and is always directed towards that fixed point. In one dimensional case, let x be the displacement of the particle and ax be the acceleration of the particle, then
where b is a constant which measures acceleration per unit displacement and dimensionally it is equal to T−2. By multiplying by mass of the particle on both sides of equation (10.2) and from Newton’s second law, the force is
where k is a force constant which is defined as force per unit length. The negative sign indicates that displacement and force (or acceleration) are in opposite directions. This means that when the displacement of the particle is taken towards right of equilibrium position (x takes positive value), the force (or acceleration) will point towards equilibrium (towards left) and similarly, when the displacement of the particle is taken towards left of equilibrium position (x takes negative value), the force (or acceleration) will point towards equilibrium (towards right). This type of force is known as restoring force because it always directs the particle executing simple harmonic motion to restore to its original (equilibrium or mean) position. This force (restoring force) is central and attractive whose center of attraction is the equilibrium position.
In order to represent in two or three dimensions, we can write using vector notation
where is the displacement of the particle from the chosen origin. Note that the force and displacement have a linear relationship. This means that the exponent of force and the exponent of displacement are unity. The sketch between cause (magnitude of force | |) and effect (magnitude of displacement | |) is a straight line passing through second and fourth quadrant as shown in Figure 10.5. By measuring slope 1/k , one can find the numerical value of force constant k.
Consider a particle of mass m moving with uniform speed v along the circumference of a circle whose radius is r in anti-clockwise direction (as shown in Figure 10.6). Let us assume that the origin of the coordinate system coincides with the center O of the circle. If ω is the angular velocity of the particle and θ the angular displacement of the particle at any instant of time t, then θ = ωt. By projecting the uniform circular motion on its diameter gives a simple harmonic motion.
This means that we can associate a map (or a relationship) between uniform circular (or revolution) motion to vibratory motion. Conversely, any vibratory motion or revolution can be mapped to uniform circular motion. In other words, these two motions are similar in nature.
Let us first project the position of a particle moving on a circle, on to its vertical diameter or on to a line parallel to vertical diameter as shown in Figure 10.7. Similarly, we can do it for horizontal axis or a line parallel to horizontal axis.
As a specific example, consider a spring mass system (or oscillation of pendulum) as shown in Figure 10.8. When the spring moves up and down (or pendulum moves to and fro), the motion of the mass or bob is mapped to points on the circular motion.
Thus, if a particle undergoes uniform circular motion then the projection of the particle on the diameter of the circle (or on a line parallel to the diameter ) traces straightline motion which is simple harmonic in nature. The circle is known as reference circle of the simple harmonic motion. The simple harmonic motion can also be defined as the motion of the projection of a particle on any diameter of a circle of reference.
The distance travelled by the vibrating particle at any instant of time t from its mean position is known as displacement. Let P be the position of the particle on a circle of radius A at some instant of time t as shown in Figure 10.9. Then its displacement y at that instant of time t can be derived as follows In ∆OPN
But θ = ωt, ON = y and OP = A
The displacement y takes maximum value (which is equal to A) when sin ωt = 1. This maximum displacement from the mean position is known as amplitude (A) of the vibrating particle. For simple harmonic motion, the amplitude is constant. But, in general, for any motion other than simple harmonic, the amplitude need not be constant, it may vary with time.
The rate of change of displacement is velocity. Taking derivative of equation (10.6) with respect to time, we get
For circular motion (of constant radius), amplitude A is a constant and further, for uniform circular motion, angular velocity ω is a constant. Therefore,
From equation (10.6),
From equation (10.8), when the displacement y = 0, the velocity v = ωA (maximum) and for the maximum displacement y = A, the velocity v = 0 (minimum).
As displacement increases from zero to maximum, the velocity decreases from maximum to zero. This is repeated.
Since velocity is a vector quantity, equation (10.7) can also be deduced by resolving in to components.
The rate of change of velocity is acceleration.
From the Table 10.1 and figure 10.10, we observe that at the mean position
(y = 0), velocity of the particle is maximum but the acceleration of the particle is zero. At the extreme position (y = ±A), the velocity of the particle is zero but the acceleration is maximum ±Aω2 acting in the opposite direction.
Which of the following represent simple harmonic motion?
(i) x = A sin ωt + B cos ωt
(ii) x = A sin ωt+ B cos 2ωt
(iii) x = A eiωt
(iv) x = A ln ωt
(i) x = A sin ωt + B cos ωt
This differential equation is similar to the differential equation of SHM (equation 10.10).
Therefore, x = A sin ωt + B cos ωt represents SHM.
(ii) x =A sin ωt + B cos2ωt
This differential equation is not like the differential equation of a SHM (equation 10.10). Therefore, x = A sin ωt + B cos 2ωt does not represent SHM.
This differential equation is like the differential equation of SHM (equation 10.10). Therefore, x = A eiωt represents SHM.
(iv) x = A ln ωt
This differential equation is not like the differential equation of a SHM (equation 10.10). Therefore, x = A ln ωt does not represent SHM.
Consider a particle undergoing simple harmonic motion. The velocity of the particle at position x1 is v1 and velocity of the particle at position x2 is v2. Show that the ratio of time period and amplitude is
The time period is defined as the time taken by a particle to complete one oscillation. It is usually denoted by T. For one complete revolution, the time taken is t = T, therefore
Then, the displacement of a particle executing simple harmonic motion can be written either as sine function or cosine function.
T represents the time period. Suppose the time t is replaced by t + T, then the
Thus, the function repeats after one time period.
This y(t) is an example of periodic function.
The number of oscillations produced by the particle per second is called frequency. It is denoted by f. SI unit for frequency is s−1 or hertz (In symbol, Hz).
Mathematically, frequency is related to time period by
The number of cycles (or revolutions) per second is called angular frequency. It is usually denoted by the Greek small letter ‘omega’, ω. Comparing equation (10.11) and equation (10.12), angular frequency and frequency are related by
SI unit for angular frequency is rad s−1. (read it as radian per second)
The phase of a vibrating particle at any instant completely specifies the state of the particle.
It expresses the position and direction of motion of the particle at that instant with respect to its mean position (Figure 10.11).
where ωt + φ0 = φ is called the phase of the vibrating particle. At time t = 0s (initial time), the phase φ = φ0 is called epoch (initial phase) where φ0 is called the angle of epoch. Phase difference: Consider two particles executing simple harmonic motions. Their equations are y1 = A sin(ωt + φ1) and y2 = A sin(ωt + φ 2), then the phase difference ∆φ= (ωt + φ2) − (ωt + φ1) = φ2 −φ1.
A nurse measured the average heart beats of a patient and reported to the doctor in terms of time period as 0.8s. Express the heart beat of the patient in terms of number of beats measured per minute.
Let the number of heart beats measured be f. Since the time period is inversely proportional to the heart beat, then
Calculate the amplitude, angular frequency, frequency, time period and initial phase for the simple harmonic oscillation given below
a. y = 0.3 sin (40πt + 1.1)
b. y = 2 cos (πt)
c. y = 3 sin (2πt − 1.5)
Simple harmonic oscillation equation is y = A sin(ωt + φ0) or y =A cos(ωt + φ0)
Show that for a simple harmonic motion, the phase difference between
a. displacement and velocity is π/2 radian or 90°.
b. velocity and acceleration is π/2 radian or 90°.
c. displacement and acceleration is π radian or 180°.
a. The displacement of the particle executing simple harmonic motion
y = A sinωt
Velocity of the particle is
v = Aωcos ωt = Aωsin(ωt+ π /2)
The phase difference between displacement and velocity is π/2.
b. The velocity of the particle is
v = A ω cos ωt
Acceleration of the particle is
a = Aω2sinωt = Aω2cos(ωt+ π /2)
The phase difference between velocity and acceleration is π/2.
c. The displacement of the particle is y = A sinωt
Acceleration of the particle is
a = − A ω2 sin ωt = A ω2 sin(ωt + π)
The phase difference between displacement and acceleration is π.