Contents
Physics Topics are often described using mathematical equations, making them precise and quantifiable.
What are the Characteristics of SHM?
Relation Between SHM and Uniform Circular Motion
SHM is the simplest form of linear periodic motion. Again, uniform circular motion is the simplest form of rotational periodic motion.
The relation between SHM and uniform circular motion can be shown by a mechanical example.
A wheel is rotating with uniform speed about its centre [Fig.]. A rod is connected between the circumference of the wheel and the handle of a frictionless piston fitted within a cylinder. With the uniform rotation of the wheel the piston moves to and fro along the shown path uniformly. As the motion of the wheel is uniform, the motion of the piston will be simple harmonic.
Geometrical proof: Suppose a particle is moving with a uniform angular velocity in an anticlockwise direction. This is shown by an arrow, along the circumference A’C’B’D’ of a circle having its centre at O [Fig.]. The foot of the perpendicular, drawn from different positions of the particle on the diameter B’OA’ (or C’OD’), will execute simple harmonic motion. This is proved below
When the particle is at P the foot of the perpendicular drawn from P on the diameter B’OA’ is at N. Now, as the particle moves around the circle, the foot of the perpendicular moves along the diameter A’OB’. Let the particle starting from A’, move around the circle in an anticlockwise direction and come back to A’. Its projection on A’OB’ (N) also moves along A’OB’, reverses its direction and then comes back along the same path to its starting point simultaneously. This to and fro motion of N along A’B’ is a linear periodic motion.
Similarly, if we draw a perpendicular from P on the diameter C’OD’, the foot of the perpendicular will execute a linear periodic motion along the diameter C’OD’. This linear periodic motion will be simple harmonic if it can be shown that the acceleration of N is proportional to its displacement from O and is directed towards the position of equilibrium O.
Let ω be the uniform angular velocity of the particle and be the radius r of the circular path.
ar = \(\frac{v^2}{r}\)
∴ The uniform linear speed of the particle is, v = ωr
The centripetal acceleration of the particle at P along PO is,
ar = \(\frac{v^2}{r}\) = ω2r
Let at any instant, OP make an angle θ with the diameter B’OA’ (taken as x-axis) and the displacement of the foot of the perpendicular (N) be x from the position of equilibrium.
∴ x = ON = OPcosθ = rcosθ
Again, the component of ar along the diameter A’OB’, i.e., the acceleration of N at that instant is
a = arcosθ = ω2rcosθ = ω2x
Hence, the acceleration of N is proportional to its displacement from its position of equilibrium.
The component of ar along A’OB’ is related to the motion of N. Obviously, if the particle is at P, this component of acceleration is directed towards O, the position of equilibrium. So the motion of N is simple harmonic.
Therefore, when a particle Is In a uniform circular motion, the motion of its projection on any diameter of the circular path is simple harmonic.
The circle in the example above is called the circle of reference and the particle is called the reference particle.
Some Quantities Related to SHM
Displacement: From section 1.2.1. we know that the general equation for displacement of a particle executing SHM is.
x = Asin(ωt + α) ……….. (1)
Special cases:
i) If the particle starts its motion from one of the extremities B or C of its path [Fig.], then the equation becomes
x = Acosωt ……….. (2)
ii) If the particle starts its motion from O, the position of equilibrium, then the equation becomes
x = Asinωt ………. (3)
Velocity: From equation (1) we get,
sin(ωt + α) = \(\frac{x}{A}\)
∴ cos(ωt + α) = ±\(\sqrt{1-\frac{x^2}{A^2}}\)
So, the velocity of the particle executing SRM is
v = \(\frac{d x}{d t}\) = Aωcos(ωt + α) = ± Aω\(\sqrt{1-\frac{x^2}{A^2}}\)
or, v = ±ω\(\sqrt{A^2-x^2}\) ……. (4)
It shows that the velocity of the particle depends on its displacement.
Special cases:
i) When x = 0, i.e., when the particle is at O, the position of equilibrium,
v = ±ωA, which is the maximum velocity.
i.e., Vmax = ±ωA
ii) When x = ±A, i.e., when the particle is at B or C, the two extremities of its path,
v = ±ω\(\sqrt{A^2-A^2}\) = 0, which is the minimum velocity.
i.e., vmin = 0.
So, a particle executing SHM has different velocities at different points on its path. It passes the position of equilibrium with maximum velocity. The magnitude of its velocity gradually decreases as the particle moves towards its extremities from the equilibrium position and it momentarily comes to rest at the extreme points of its path.
Acceleration: From equation (1), we get
velocity, v = \(\frac{d x}{d t}\) = ωAcos(ωt + α)
∴ Acceleration, a = \(\frac{d v}{d t}\) = -ω2Asin(ωt + a)
or, a = -ω2x ……….. (5)
From this equation. we see that the acceleration of the particle depends on its displacement from its mean position.
The negative sign indicates that acceleration and displacement are mutually opposite in direction.
Special cases:
i) When x = 0, i.e., when the particle is at O, the position of equilibriums
a = 0, which is the minimum acceleration.
ii) When x = ±A, i.e., when the particle is at B or C, the extremities of its path,
a = ∓ω2A, which is the maximum acceleration.
A particle undergoing SHM possesses different accelerations at different points on its path. It has zero acceleration at equilibrium and attains the maximum acceleration at the extreme positions of its path.
Instead of equation (1) if we use equation (2) or (3), we get the same equations (4) and (5) for velocity and acceleration.
Time period and frequency:
Let the equation of a simple harmonic motion: x = Acosωt. In this case, at the beginning of the motion. i.e., at time t = 0, ωt = 0 and x = Acos0 = A. This implies that, the particle begins its motion from one extreme end of its path.
Now, as time advances, when t = \(\frac{2 \pi}{\omega}\), again we get, x = Acos2π = A, i.e., the particle returns to the initial point from which it began its motion. Thus, an oscillation is completed.
So, for one complete oscillation, change in ωt = 2π – 0 = 2π, i.e., for one complete oscillation, time taken is \(\frac{2 \pi}{\omega}\) (∴ ωt = 2π).
As the total time elapsed for one complete oscillation is called the time period (T) of a SHM, we have,
T = \(\frac{2 \pi}{\omega}\) ……… (6)
Definition: Frequency is defined as the number of complete oscillations per second of a particle executing SHM.
In time T the number of oscillations is 1. Hence in unit time, the number of oscillations is \(\frac{1}{T}\)
∴Frequency, n = \(\frac{1}{T}\) = \(\frac{\omega}{2 \pi}\) ………… (7)
or, ω = 2πn
ω is called the angular frequency.
Considering only the magnitude of acceleration of a particle executing SHM, we get from equation (5)
Amplitude: In Fig., the amplitude of simple harmonic motion = OB = OC. Since, -1 ≤ sinθ ≤ +1 and -1 ≤ cosθ ≤ +1, we get from equation (1), the maximum value of displacement x = amplitude = |± A| = A
∴ \(\overrightarrow{O B}\) = \(\vec{A}\) and \(\overrightarrow{O C}\) = –\(\vec{A}\)
From equations (8) and (9) it is evident that time period and frequency of a SHM do not depend on the amplitude A. So, if the amplitude of oscillation of a simple pendulum diminishes gradually due to air resistance, its time period remains unchanged (law of isochronism). For this, simple harmonic motion is called an isochronous motion.
Phase: The phase of a particle executing SHM at any instant, is defined as its state of motion at that instant. The term ‘state of motion’ indicates displacement, velocity, acceleration, etc., of the particle at any instant.
Suppose in equation (1), θ = ωt + α ……….. (10)
So, displacement, x = Asinθ;
velocity, a = ±ω\(\sqrt{A^2-x^2}\) = ±ωAcosθ;
acceleration, a = -ω2x = -ω2Asinθ
Now, ω and A are both constants. So displacement, velocity and acceleration of the particle at any instant depend entirely on the angle θ. The angle θ, expressed by equation (10), is called the phase angle of the simple harmonic motion. It is seen from equation (10), that, θ depends on time t. Thus, the phase of a particle executing SHM changes continuously with respect to time.
Special cases:
i) If θ = 0, then x = Asinθ = 0, i.e., the particle is at O [Fig.].
If the value of θ becomes 90°, then x = Asin90° = A, i.e., the particle is at the end B of its path, i.e., the change of phase = 90° = \(\frac{\pi}{2}\)
ii) if θ = 270°, then x = Asin270° = -A, i.e.. the particle is at C. So, when the particle moves from B to C, change of phase angle = 270° – 90° = 180° = π. The positions B and C are then said to be in opposite phase.
iii) If θ = 450° = 360° + 90°, then
x = Asin450° = Asin90° = A;
i.e., the particle is at B. So, when the particle starts its motion from B, goes to C and then returns to B, change of phase angle = 450° – 90° = 360° = 2π. In this case, the initial and the final positions are inihe same phase.
Epoch: It is defined as the initial phase of motion (i.e., at t = 0) of the particle executing SHM.
If the equation of SHM is x = Asin(ωt + α), then phase angle θ = ωt + α.
Putting t = 0 in this equation, we get epoch, θ = α
As a special case, if the particle starts its motion from one extremity of its path we have,
x = Acosωt = Asin(ωt + \(\frac{\pi}{2}\))
Here phase angle, θ = ωt + \(\frac{\pi}{2}\). If t = 0, epoch = \(\frac{\pi}{2}\).
Again, if the particle starts its motion from the position of equilibrium, we have x = Asinωt
Here, phase angle θ = ωt. Putting t = 0, epoch = 0.
Phase difference: In case of two particles executing SHM, if the phase angle of the first particle is θ1 and that of the second particle is θ2 at an instant, then the phase difference of the SHMs is θ = θ2 – θ1 (or θ1 – θ2). Two simple harmonic motions having the sanie time period frequency and amplitude may have different phases. For example, in Fig, if at an instant when the first particle reaches C, the second particle is at B, then their phase difference = 180° = π.
If the phase difference between two SHMs remains constant, i.e., it does not change with time, they are said to be coherent; the particles executing these SHMs are said to be in coherent motion.
Relations among displacement velocity and acceleration: Suppose a particle executing SHM starts its motion from the extremity B of its path [Fig.]. If T is the time period, then the particle crosses the position of equilibrium O in time \(\frac{T}{4}\); reaches C, the other extremity of the path in time \(\frac{T}{2}\); again on its way back crosses O, the position of equilibrium in time \(\frac{3 T}{4}\); and returns to B in time T.
Using v = ± ω\(\sqrt{A^2-x^2}\) and a = -ω2x, we can calculate displacement, velocity and acceleration of the particle at different times as shown in the table below.
The graph of displacement and velocity against time is shown in Fig., and the graph of displacement and acceleration against time in Fig.
Phase difference of velocity and acceleration with displacement: Displacement of a particle executing a SHM is x = Acosω t,
velocity, v = -Aωsinωt = Aωcos(ωt + \(\frac{\pi}{2}\))
and acceleration, a = -Aω2cosωt = Aω2cos(ωt + π)
So, the phase difference between velocity and displacement is \(\frac{\pi}{2}\) or 90°. [Fig.] and the difference between acceleration and displacement is π or 180° [Fig.].
Characteristics of SHM
- Simple harmonic motion is a kind of linear periodic motion, i.e., in this motion, the particle moves to and fro following the same path repeatedly at regular time intervals.
- The acceleration of the particle executing SHM is always directed towards the position of equilibrium.
- The acceleration of the particle is proportional to its displacement from the position of equilibrium at any instant.
- When the particle passes the position of equilibrium, its velocity becomes maximum. Velocity of the particle gradually reduces and momentarily comes to zero at the extremities of its path.
- The time period of SHM does not depend on the amplitude. Though the amplitude decreases gradually due to various external resistances, the time period remains unchanged.
It is to be noted that SHM is a special form of periodic motion. If the periodic motion is
1. linear and
2. the acceleration of the particle is proportional to its displacement from the position of equilibrium and is directed towards it, only then the motion of the particle is called simple harmonic. The motion of the hands of a clock or the motions of the planets and satellites are periodic, but as these motions do not satisfy the above two conditions. they are not considered as simple harmonic. So, it can be said that all simple harmonic motions are periodic, but all periodic motions are not simple harmonic.
Numerical Examples
Example 1.
A particle of mass 0.5 g is executing SHM with a time period of 2 s and an amplitude of 5 cm. Calculate its
(i) maximum velocity,
(ii) maximum acceleration and
(iii) velocity, acceleration and force acting on the particle when it is at a distance of 4 cm from its position of equilibrium.
Solution:
Amplitude, A = 5 cm; time period, T = 2 s
∴ ω = \(\frac{2 \pi}{T}\) = \(\frac{2 \pi}{2}\) = π rad ᐧ s-1
i) Maximum velocity, vmax = ωA = π ᐧ 5 = 5π
= 5 × 3.14 = 15.7 cm ᐧ s-1 = 0.157 m ᐧ s-1
ii) Maximum acceleration, amax = ω2A = π2 ᐧ 5
= 5 × (3.14)2 = 49.298 cm ᐧ s-2 ≈ 0.493 m ᐧ s-2
iii) When x = 4 cm,
velocity, v = ω\(\sqrt{A^2-x^2}\) = π\(\sqrt{5^2-4^2}\)
= 3.14 × 3 = 9.42 cm ᐧ s-1
= 0.094 m ᐧ s-1
acceleration, a = ω2x = (3.14)2 × 4
= 39.438 cm ᐧ s-2
= 0.394 m ᐧ s-2
and force, F = ma = 0.5 × 39.438
= 19.72 dyn ≈ 0.197 × 10-3N.
Example 2.
A particle executing SHM possesses velocities 20 cm ᐧ s-1 and 15 cm ᐧ s-1 at distances 6 cm and 8 cm respectively from its mean position. Calculate the amplitude and the time period of the particle.
Solution:
Velocity of the particle executing SHM,
v = ω\(\sqrt{A^2-x^2}\)
In the first case, 20 = ω\(\sqrt{A^2-6^2}\) ………. (1)
In the second case, 15 = ω\(\sqrt{A^2-8^2}\) ………. (2)
Dividing (1) by (2) we get,
\(\frac{20}{15}\) = \(\frac{\omega \cdot \sqrt{A^2-6^2}}{\omega \sqrt{A^2-8^2}}\) or, \(\frac{4}{3}\) = \(\frac{\sqrt{A^2-36}}{\sqrt{A^2-64}}\) or, \(\frac{A^2-36}{A^2-64}\) = \(\frac{16}{9}\)
or, 16A2 – 1024 = 9A2 – 324
or 7A2 = 700 or, A2 = 100, A = 10 cm = 0.1 m
From equation (1) we get,
20 = ω\(\sqrt{10^2-6^2}\) = 8ω or, ω = \(\frac{20}{8}\) = \(\frac{5}{2}\) rad ᐧ s-1
∴ T = \(\frac{2 \pi}{\omega}\) = \(\frac{2 \pi}{5}\) × 2 = \(\frac{4}{5}\) × 3.14 = 2.51 s.
Example 3.
The time period and amplitude of a particle executing SHM are 10 s and 0.12 m respectively. Find its velocity at a distance 0.04 m from its position of equilibrium.
Solution:
Here T = 10 s; A = 0.12 m
Example 4.
The frequency of a vibrating wire is 200 Hz. The velocity of a particle on the wire is 4.35 m ᐧ s-1 when it is at a distance of half its amplitude. Calculate the acceleration of the particle at that instant.
Solution:
Here, ω = 2πn = 2π ᐧ 200 = 400π rad ᐧ s-1
[∵ n = 200 Hz]
Let A be the amplitude of the particle.
The acceleration of a rapidly vibrating object may reach an exceptionally high value.
Example 5.
Two particles executing SHM possess the same frequency. When the first particle just passes the mean position of its path, the second particle moving in the same direction is at a distance of 3 cm from its mean position. If the amplitude of vibration of the second particle is 6 cm, what is the phase difference of the two particles?
Solution:
If A is the amplitude and θ is the phase angle, then displacement, x = Acosθ
For the first particle,
x1 = A1cosθ1 or, 0 = A1cosθ1
or, cosθ1 = 0 or, θ1 = ±90°
For the second particle,
x2 = A2θ2, or cosθ2 = \(\frac{x_2}{A_2}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
or, θ2 = ±60°
At the given instant, the two particles are in motion in the same direction
So,
(i) θ1 = 90°, then θ2 = 60°;
phase difference = θ2 – θ1 = -30°
or, (ii) if θ1 = -90°, then θ2 = -60°;
phase difference θ2 – θ1 = + 30°
∴ The required phase difference is ± 30°.
Example 6.
A particle executing SHM possesses velocities v1 and v2 when it is at distances x1 and x2 respectively from its mean position. Show that, the time period of oscillation is given by T = 2π\(\left(\frac{x_2^2-x_1^2}{v_1^2-v_2^2}\right)^{1 / 2}\).
Solution:
We know, v = ω\(\sqrt{A^2-x^2}\).
According to the question,
Example 7.
The equation of a simple harmonic motion is x = 10sin(\(\frac{\pi}{3}\)t – \(\frac{\pi}{12}\))cm. Calculate its
(i) amplitude,
(ii) time period,
(iii) maximum speed,
(iv) maximum acceleration
(v) epoch and
(vi) speed after 1 s of initiation of motion.
Solution:
x = 10sin(\(\frac{\pi}{3}\)t – \(\frac{\pi}{12}\)) cm.
Comparing this equation with the equation of SHM, x = A sin(ωt + α) we get,
i) amplitude, A = 10 cm.
ii) ω = \(\frac{\pi}{3}\); so time period, T = \(\frac{2 \pi}{\omega}\) = \(\frac{2 \pi \times 3}{\pi}\) = 6 s.
iii) maximum speed, ωA = \(\frac{\pi}{3}\) × 10 = \(\frac{10 \pi}{3}\) cm ᐧ s-1.
iv) maximum acceleration,
ω2A = \(\left(\frac{\pi}{3}\right)^2\) × 10 = \(\frac{10 \pi^2}{9}\) cm ᐧ s-1
v) epoch = –\(\frac{\pi}{12}\) = -15°.
vi) displacement after 1 s of initiation of motion,
Example 8.
Write down the equation of a simple harmonic motion whose amplitude is 5 cm, epoch is 0° and the number of vibrations per minute is 150.
Solution:
According to the problem,
amplitude, A = 0.05 m; epoch, α = 0°; frequency, n = \(\frac{150}{60}\) = \(\frac{5}{2}\)Hz
∴ ω = 2πn = 2π ᐧ \(\frac{5}{2}\) = 5π rad ᐧ s-1
So, the equation of the simple harmonic motion is
x = A sin(ωt + α) or, x = 0.05sin5πt m.
Example 9.
The displacement of a vibrating particle at time t is given by x = A’sin(\(\frac{\pi}{6}\)t) + B’cos (\(\frac{\pi}{6}\)t) where A’ = 0.03 m, B’ = 0.04 m. Calculate the
(i) amplitude
(ii) epoch,
(iii) displacement, velocity and acceleration of the particle after 2 seconds.
Solution:
From this equation we get,
i) amplitude, A = \(\sqrt{A^{\prime 2}+B^{\prime 2}}\)
= \(\sqrt{(0.03)^2+(0.04)^2}\) = 0.05 m
ii) epoch, α = tan-1\(\frac{B^{\prime}}{A^{\prime}}\) = tan-1\(\frac{4}{3}\)
iii) when t = 2 s, we get from equation (1)
displacement, x = A’ sin(\(\frac{\pi}{6}\)t) + B’ cos(\(\frac{\pi}{6}\)r)
= 0.03 sin\(\frac{\pi}{3}\) + 0.04 cos\(\frac{\pi}{3}\)
= 0.03 × \(\frac{\sqrt{3}}{2}\) + 0.04 × \(\frac{1}{2}\)
= 0.04598 m ≈ 0.046 m
From equation (2) we get, ω = \(\frac{\pi}{6}\) rad ᐧ s-1
So, velocity after 2 s,
v = ω\(\sqrt{A^2-x^2}\)
= \(\frac{\pi}{6} \sqrt{(0.05)^2-(0.046)^2}\) = 1.03 m ᐧ s-1
Again, acceleration after 2s,
α = ω2x = \(\left(\frac{\pi}{6}\right)^2\) × 0.046 = 1.26 m ᐧ s2.
Example 10.
The equation of motion of a particle executing SHM is expressed by x = 10 sin(10t – \(\frac{\pi}{6}\)). Establish an equation to express its velocity and also calculate the magnitude of its maximum acceleration.
Solution:
Given, x = 10sin(10t – \(\frac{\pi}{6}\))
∴ Velocity, v = \(\frac{d x}{d t}\) = 10 × 10cos(10t – \(\frac{\pi}{6}\))
= 100cos(10t – \(\frac{\pi}{6}\)) unit.
Again, acceleration, a = \(\frac{d v}{d t}\) = -10 × 100sin(10t – \(\frac{\pi}{6}\))
= -1000sin(10t – \(\frac{\pi}{6}\)) unit.
Acceleration will be maximum when,
sin(10t – \(\frac{\pi}{6}\)) = 1
∴ Magnitude of maximum acceleration, amax = 1000 units.