Simple Harmonic Motion- with Examples, Problems, Visuals, MCQ Quiz Questions- Force Law, Pendulums, Phase, Amplitude, Damped Oscillations


 
 



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Simple Harmonic Motion - With Examples, Solved Problems and Visualizations

Target Audience: High School Students, College Freshmen and Sophomores, students preparing for the International Baccalaureate (IB), AP Physics B, AP Physics C, A Level, Singapore/GCE A-Level; 

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A tutorial with solved problems


Introduction

A motion which repeats itself after a definite interval of time is called periodic motion. The time interval after which the motion repeats is called the time period. Circular motion is one example of periodic motion which we come across everyday like the hands of a clock undergo circular motion with different time periods. Similarly, if a body moves to and fro on the same path, it is said to perform oscillations. Simple harmonic motion is a type of oscillatory motion in which the displacement x of the particle from the origin is given by

x = Asin(ωt +ф)

where A, ω and ф are constants. This kind of motion where displacement is a sinusoidal function of time is called simple harmonic motion. The quantities A, ω and ф which determine the shape of the graph (displacement curve) are:


Related Visualizations




Visualizing a Block Spring System

Simple Harmonic Motion - A Block-Spring System



Amplitude

The quantity A is called the amplitude of the motion. As sin(ωt +ф) can take values between -1 and +1, the displacement x can take values between –A and +A. This gives the physical significance of the constant A. It is the maximum displacement of the particle from the center of oscillation i.e. the amplitude of oscillation.

Phase Angle (ф)

The time varying quantity (ωt +ф) is called the phase of the motion and the quantity ф is called the phase angle or phase constant. It depends on the choice of the instant t = 0. To describe the motion quantitatively, a particular instant should be called zero and measurement of time should be made from this instant. If we choose to call the instant when the particle is passing through its mean position as the time t = 0, then ф = 0 as at that time x has to be zero. But on the other hand, if for the convenience of the problem, t = 0 has to be considered at a point when the particle at an extreme position, then ф would be ±п/2 depending on the direction of velocity of the particle at that position.

Angular frequency (ω)

To understand angular frequency, we first try to understand frequency of simple harmonic motion. We know that the motion repeats after a certain time called the time period (T) of the motion i.e x should have the same value at time t and t+T.
Thus, sin(ωt + ф) = sin(ω(t+T) + ф)
Also, the velocity should have the same value at t+T, i.e.
ωcos(ωt + ф) = ωcos(ω(t+T) + ф)
As T is the smallest time for repetition both the above equations are true for ωT = 2п,     n = 1,2,3,……
or ω = 2п/T
As (1/T) is the frequency of oscillation i.e. the number of oscillations per unit time, (2п/T) is the angular frequency i.e. the number of radians covered per unit time (where 1 complete oscillation corresponds to 2п radians).


Velocity

The expression for velocity can be derived by differentiating the displacement w.r.t. time i.e.

v(t) = dx(t)/dt

                                                                                = d[Asin(ωt + ф)]/dt
                                                                                = Aωcos(ωt + ф)

From the expression it can be seen that the velocity varies between the extreme values –Aω and +Aω. Also as we know that sin(90 + θ) = cosθ, we can conclude that the velocity curve is shifted to the left by 90o w.r.t. the displacement curve. When the magnitude of displacement is least, that of the velocity is maximum and vice versa.

Acceleration

Knowing the velocity we can arrive at acceleration for simple harmonic motion by differentiating the velocity, i.e.

a(t) =  dv(t)/dt

                                                                                = d[Aωcos(ωt + ф)]/dt
                                                                                = -Aω2sin(ωt + ф)
Thus the acceleration of the particle varies between the limits -Aω2 and +Aω2. Also the acceleration curve is shifted to the left by 90o w.r.t. the velocity curve.
Combining the equations for acceleration and displacement, we get

a(t) = -ω2x(t)

which implies that the acceleration is proportional to the displacement in simple harmonic motion and the two related by the square of the angular frequency.

Force Law for SHM

From Newton’s second law we know that F = ma and that for SHM, a = -Aω2sin(ωt + ф). Thus, for simple harmonic motion, F = -mAω2sin(ωt + ф) = -mω2x(t)
This force law is familiar. It is Hooke’s law

F = -kx   where k = mω2

For a spring, spring constant being k = mω2
Thus the spring-block system forms a simple harmonic oscillator with angular frequency, ω = √(k/m) and time period, T = 2п/ω = 2п√(m/k).

Energy of SHM

Simple Harmonic motion is defined by the equation F = -kx. The work done by the force F during a displacement from x to x + dx is

dW = Fdx

                                                                                    = -kxdx
The work done in a displacement from x = 0 to x is

W = 0x-kxdx = -kx2/2

As the change in potential energy corresponding to a force is negative of the work done by the force, U(x) – U(0) = -W = ½ kx2
Let us choose the origin to be zero of potential energy, then U(0) = 0 and U(x) = ½ kx2
The kinetic energy at a time t is K = ½ mv2
The total mechanical energy at time t is E = U + K
= ½ kx2 + ½ mv2 = ½ kA2sin2(ωt + ф) + ½ mA2ω2cos2(ωt + ф)
Putting k = mω2, we get
E = ½ 2[A2sin2(ωt + ф) + cos2(ωt + ф)] = ½ mω2A2
Thus the total mechanical energy remains constant and is independent of time.

Simple Pendulum

A simple pendulum consists of a heavy particle suspended from a fixed support through a light, inextensible string. This system can stay in equilibrium if the string is vertical. This is called the mean position or the equilibrium position. If the particle is pulled aside and released, it oscillates in a circular arc with the center at the point of suspension ‘O’.
At angle θ, the forces acting on the particle are:
(1) the weight mg downwards
(2) The tension T in the string
As the particle makes pure rotation about a horizontal axis through O (say OA), let us find the torques of the forces acting on it.
The torque of T is zero as it intersects OA at point O.
The torque of mg about OA is
τ = mg(l sinθ)
As this torque tries to bring the particle back towards θ = 0, we can write
τ = -mgl sinθ
According to this equation, the resultant torque is not proportional to the angular displacement, hence the motion is not angular simple harmonic.
However, if the angular displacement is small, sinθ is approximately equal to θ (expressed in radians) and the equation may be written as
τ = -mgl θ
Thus, if the amplitude of oscillation is small, the motion of the particle is approximatelt angular simple harmonic.
The moment of inertia of the particle about the rotation axis OA is ml 2
Hence, angular acceleration, α = τ/I = -mgl θ/ml 2 = -(g/l )θ
Also, in SHM, α = -ω2θ
Therefore, ω = √(g/l )
and the time period is T = 2п/ω = 2п√(l /g).

Physical Pendulum

Any rigid body suspended from a fixed support constitutes a physical pendulum. The motion of such a system can be followed very nearly in the same way as that of a simple pendulum. The distance of the center of mass of the body from the fixed suspension point acts as the effective length of the pendulum and the total mass being the mass of the particle situated at the center of the body. This system can now be treated in exactly the same way as a simple pendulum.
The restoring torque acting on the body when rotated through an angle θ is
τ = -mgl sinθ
Let the moment of inertia of the body about the horizontal rotation axis through point of suspension be ‘I’. Thus, angular acceleration of the body is
α = τ/I = -(mgl /I)sinθ
For small displacements, sinθ ≈ θ, therefore,
α = -(mgl /I)θ
Hence the body undergoes simple harmonic oscillations, with angular frequency, ω = √(mgl /I) and time period, T = 2п/ω = 2п√(I/mgl )

Torsional Pendulum



In a torsional pendulum, an extended body is suspended by a light wire. The body is rotated about the wire as the axis of rotation. The upper end of the wire remains fixed with the support and the lower end of the wire is rotated through an angle with the body, thus a twist θ is produced in the wire.
The twisted wire exerts a restoring torque on the body to bring it back to its original position i.e. θ = 0. This torque has a magnitude proportional to the angle of twist, θ, i.e. τ = -kθ,                              where k is called the torsional constant of the wire.
If the moment of inertia of the body about the vertical axis is ‘I’,
the angular acceleration, α = τ/I = -(k/I)θ
Thus the motion of the body is simple harmonic with angular frequency, ω = √(k/I) and time period, T = 2п/ω = 2п√(I/k)

Damped Oscillations

When the motion of an oscillator is reduced by an external force, the oscillator and its motion are said to be damped. The damping force is a function of speed and is directed opposite to the velocity. Energy is lost due to the negative work done by the damping force and the system comes to a halt in due course.
The damping may be a complicated function of speed but in several cases of practical interest the damping force is proportional to the speed and can be written as

F = -bv

The equation of motion becomes

F = -kx –bv

or                                                             mdv/dt = -kx –bv
or                                                          md2x/dt2 = -kx – bdx/dt
or                                                          md2x/dt2 + bdx/dt + kx = 0
The solution of this differential equation is x(t) = Ae-bt/2msin(ω′t + δ)
where ω′ = √((k/m) – (b/2m)2) = √(ω02 – (b/2m)2)
For small b, the angular frequency ω′ ≈ √(k/m) = ω0. Thus, the system oscillates with almost the natural angular frequency and with amplitude decreasing with time according to

A = A0e-bt/2m

The amplitude decreases and finally becomes zero.
The energy of a damped oscillator (if the damping is small) can be approximated as
E(t) = ½ mω2A02e-bt/m = ½ kA02e-bt/m




Here are some of the problems which have been solved in this tutorial :

Q: A loudspeaker produces a musical sound by means of the oscillation of a diaphragm whose amplitude is limited to 1.00 μm. (a) At what frequency is the magnitude a of the diaphragm’s acceleration equal to g? (b) For greater frequencies, is ‘a’ greater than or less than g? Take g = 10 ms-2.

Q: What is the phase constant for the harmonic oscillator with the velocity function v(t) given in figure if the position function x(t) has the form x = Acos(ωt +ф)? The vertical axis scale is set by vs = 4 cm/s.

Q: The equation of motion of a particle started at t=0 is given by x = 10sin(10t + п/3), where x is in centimeter and t in second. When does the particle (a) first come to rest (b) first have zero acceleration (c) first have maximum speed?

Q: Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement to change value from half the amplitude to the amplitude (b) velocity to change value from half its maximum to maximum.

Q: A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 6 s. If the oscillations are stopped and the body hangs in equilibrium, find the potential energy stored in the spring. Use g = 10 ms-2.


Q: A particle is subjected to two simple harmonic motions, one along the x-axis and the other on a line making an angle of 45o with the x-axis. The two motions are given by

x= x0sinωt and s = s0sinωt

Find the amplitude of the resultant motion.

Q: In a damped oscillator with m = 500 g, k = 100 N/m, and b = 75 g/s, what is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 20 cycles?


Q: A block of mass ‘m’ is suspended from the ceiling of a stationary standing elevator through a spring of spring constant ‘k’. Suddenly the cable breaks and the elevator starts falling freely. Show that the block now executes a simple harmonic motion in the elevator. Find the amplitude.

Q: A uniform rod of mass ‘m’ and length ‘l’ is suspended through a light wire of length ‘l’ and torsional constant ‘k’ as shown in figure. Find the time period if the system makes (a)small oscillations in the vertical plane about the suspension point and (b) angular oscillations in the horizontal plane about the center of the rod. 

Q: A simple pendulum is suspended from the ceiling of a car accelerating uniformly on a horizontal road. If the acceleration is ‘a0’ and the length of the pendulum is ‘l ’, find the time period of small oscillations about the mean position.

Q: A damped harmonic oscillator consists of a block (m = 2 kg), a spring (k = 30 N/m), and a damping force (F = -bv). Initially, it oscillates with an amplitude of 25 cm; because of the
damping, the amplitude falls to three-fourths of this initial value at the completion of four oscillations. (a) What is the value of b? (b) How much energy has been “lost” during these four oscillations?

Q: What is the frequency of a simple pendulum 2 m long (a) in a room, (b) in an elevator accelerating upward at a rate of 2 m/s2, and (c) in free fall? Use g = 10 ms-2

Q: A uniform disc of radius ‘r’ is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the center of it to have minimum time period? 

Q: A small block oscillates back and forth on a smooth concave surface of radius R. Find the time period of small oscillations.

Q: A rectangular block, with face lengths a = 35 cm and b = 45 cm, is to be suspended on a thin horizontal rod running through a narrow hole in the block. The block is then to be set swinging about the rod like a pendulum, through small angles so that it is in SHM. The given figure shows one possible position of the hole, at distance r from the block’s center, along a line connecting the center with a corner. (a) For what value of r does the pendulum have minimum time period? There is actually a line of points around the block’s center for which the period of swinging has the same minimum value. (b) What shape does that line make?

Q: Assume that a narrow tunnel is dug between two diametrically opposite points of the earth. Treat the earth as a solid sphere of uniform density. Show that if a particle is released in this tunnel, it will execute a simple harmonic motion. Calculate the time period of the motion.


Q: For a simple pendulum, find the angular amplitude at which the restoring torque required for simple harmonic motion deviates from the actual restoring torque by 2%. Use g = 10 ms-2.

Q: Hanging from a horizontal beam are nine simple pendulums of the following lengths: (a) 0.10, (b) 0.30, (c) 0.40, (d) 0.80, (e) 1.2, (f) 2.8, (g) 3.5, (h) 5.0, and (i) 6.2 m. Suppose the beam undergoes horizontal oscillations with angular frequencies in the range from 2.00 rad/s to 4.00 rad/s. Which of the pendulums will be (strongly) set in motion? Use g = 10 ms-2.

Q: A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motions, what is the phase difference between the individual motions?


Complete tutorial document with examples, figures and solved problems :




MCQ Quiz: Simple Harmonic Motion- Challenging Problems

Companion MCQ Quiz- test how much you know about the topic. Your score will be e-mailed to you at the address you provide.

Simple Harmonic Problems- Challenging Problems