Transcript SHM Part 1 - Ask Physics

© 2007-08 P K Bharti, IIT Kharagpur
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Please revise these topics before studying this chapter:
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TOPICS
 Kinematics (Displacement, velocity & acceleration)
 Newton’s Laws
 Equilibrium
 Spring force
 Kinetic Energy
 Potential Energy
 Relation of Potential energy with work done in conservative field
 Conservation of Energy
 Differentiation
 Differential Equation
NOTE: You will study differentiation and differential equation in Std. XIIth. For the time
being, you can just mug up the things blindly where you find difficulty.
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We are going to study about the kind of motions where same motion is repeated.
Periodic motion, oscillation, vibration and simple harmonic motion (SHM) are some of the
motions where same motion is repeated over time.
We shall study SHM in detail.
We shall study both the mathematical and graphical aspects of SHM.
SHM is a very important topic from the IIT JEE and other competitive examinations point of
view; so be careful of all concepts.
Before studying this chapter you do require to study almost all chapters of mechanics
specially Newton’s laws and concept of Equilibrium.
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PERIODIC MOTION: A motion which repeats itself after a regular interval of time is called
periodic motion.
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TIME PERIOD (T): The smallest interval of time after which the motion is repeated is called its
time period.
S.I. unit: second (s)
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FREQUENCY (n or f): The number of repetitions that occur per unit time is called frequency of
the periodic motion.
It is denoted by n (Greek nu) or f.
Frequency is the reciprocal of time period T. Therefore,
1
n=f =
T
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[frequency (n or f)]
S. I. Unit: hertz (Hz) which is same as s–1
1 Hz = 1 s–1
Physically, if a body repeats its motion faster, it will said to have higher frequency.
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Q. Is the revolution of the earth around the sun a periodic motion? Find the time period and
frequency of the motion if it is periodic.
SOLUTION:
The earth repeats its revolution around the sun every year. A motion which repeats itself
after a regular intervals of time is called periodic motion. Therefore, revolution of the Earth
around the Sun is a periodic motion.
The time period of this periodic motion is roughly 365 days.
Therefore,
T ≈ 365 days = 365 X 24 X 60 X 60 s = 31536000 s.
Hence, frequency:
n=
1
T
=
1
31536000 s
= 3.17 X 10 -8 Hz
(Ans)
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OSCILLATION/ VIBRATION: Those periodic motion which repeats itself about equilibrium point are
known as oscillation or vibration.
NOTE: Equilibrium point is the point where net force and net torque is zero. (Please see chapter on
equilibrium for more idea about equilibrium).
For example, a ball placed in a U-shaped bowl will be in equilibrium at the point E at bottom. If
displaced a little from this point E, it will perform oscillations in the bowl.
E
(Equilibrium point)
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NOTE: This animation is for illustration only. It does not show practical situation.
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Uniform circular motion is a periodic motion, but it is not oscillatory.
Every oscillatory motion is periodic, but every periodic motion need not be oscillatory.
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DIFFERENCE BETWEEN OSCILLATION & VIBRATION:
When the frequency is small, we call it oscillation.
E.g., the oscillation of a pendulum.
When the frequency is high, we call it vibration.
E.g., the vibration of a string of a guitar.
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Q 1. A simple pendulum is oscillating 20 times a second. Find its time period and frequency.
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Q2. Which has higher frequency:
(a) Revolution earth around the sun
(b) Revolution of moon around the earth?
ANS:
1. Frequency, n = number of oscillations per second = 20 Hz.
Time period, T = 1/ n = 0.05 s.
2. (b) Because moon takes less time (28 days) to revolve around the earth.
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Slideshow to see animation.
Let us consider a mass attached to a spring which in turn,
attached to a rigid wall. The spring-mass system lies on a
frictionless surface.
We know that if we stretch or compress a spring, the
mass will oscillate back and forth about its
equilibrium (mean) position.
Equilibrium position is the point where net force and net torque is zero.
Here, equilibrium position corresponds to the point at which spring is at its natural length, i.e.,
the position at which spring is neither compressed nor stretched.
The point at which the spring is fully compressed or fully stretched is known as extreme position.
The maximum displacement of the body executing oscillation on either side of the equilibrium
position is called the amplitude.
In other language, we can say that amplitude is the distance between mean position and
extreme position.
Amplitude is denoted by letter A.
As amplitude is maximum displacement from the equilibrium position, its SI unit is m.
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Click F5 to see animation.
Let us consider a mass attached to a spring which in turn,
attached to a rigid wall. The spring-mass system lies on a
frictionless surface.
We know that if we stretch or compress a spring with a
mass on the end and let it go, the mass will oscillate back
and forth about its equilibrium (mean) position.
If you observe motion of the block carefully, you will find that speed i.e., velocity magnitude is
maximum at mean position.
Similarly speed is minimum i.e., zero at extreme positions as block stops momentarily at
extreme positions.
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Since, equilibrium position is the point where net force and net torque is zero.
Therefore, acceleration of the mass is zero at equilibrium point.
Also, acceleration magnitude is maximum at extreme positions.
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All these points are shown in the next slide.
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Click F5 to see animation.
Extreme
position
Equilibrium
position Extreme
position
A
A
m
x=–A
x=A
O
v=0
vmax
v =0
amax
a=0
amax
x
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Q . A mass (m = 1 kg) is attached to a vertical spring (natural length 1 m and spring constant
k = 100 N/m). Find the length of the spring in equilibrium position. Assume g =10 m/s 2.
• SOLUTION:
• It is clear that spring elongates when we attach a mass to it in
a vertical position as shown.
• Let the elongation of the spring from the natural length be x.
1m
(Natural length,
• Now, we know that equilibrium position is the point where net
without mass)
k
force and net torque is zero.
• Clearly forces on the mass are:
kx
x
Weight mg downward
m
Spring force kx upward (don’t ask why upward)
• Using Newton’s 2nd law at equilibrium in the vertical direction,
we get,
mg
mg – kx = 0
 x = mg/k = 1 X 10/100 = 0.1 m
• It can easily be checked out that net torque is zero about this position.
• Therefore, equilibrium position is at a distance 0.1 m below the natural length of the spring.
• Therefore, length of the spring in equilibrium position = 1 + 0.1 = 1.1 m
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Q . A mass (m = 1 kg) is placed at top of a vertical spring (natural length 1 m and spring
constant k = 100 N/m). Find the length of the spring in equilibrium position. Assume g =10
m/s 2.
m
ANS: 0.9 m
k
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Before giving the proper definition of Simple Harmonic Motion (SHM), let us have a look at one
example.
Let us consider a mass attached to a spring which in turn, attached to a rigid wall. The springmass system lies on a frictionless surface.
We know that if we stretch a spring with a mass on the end and let it go, the mass will oscillate
Equilibrium
back and forth about its equilibrium (mean) position.
position
Let us displace a spring by a distance x towards right.
As we displace it towards right, spring force will try to
bring that mass m towards left.
m
• Thus at a displacement x, a spring force develops in
the spring in the left direction.
• We also say this force F as restoring force as it tries to
F
bring back mass m towards the mean position.
• As this restoring force F is opposite to that of
displacement, therefore, we can write from Hooke’s Law
m
F = – kx (negative sign because F is opposite to x)
 Fa–x
(because k is a constant)
O x
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Thus, the resultant restoring force F acting on the body is proportional to the displacement x
and is directed opposite to the displacement, i.e., towards the equilibrium point.
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From last slide, we have seen that restoring force F is given by F a – x for a spring-mass system.
Thus, the resultant restoring force F acting on the body is proportional to the displacement x
and is directed opposite to the displacement, i.e., towards the equilibrium point.
A motion that arises when the force on the oscillating body is directly proportional to its
displacement from the equilibrium position and is directed opposite to the displacement, is
known as Linear Simple Harmonic Motion (SHM).
Fa–x
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(Condition for linear SHM)
We are using word Linear to emphasize that particle moves on straight line.
The body performing SHM is known as a simple harmonic oscillator ( SHO ).
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Q . Which of the following restoring force F vs. displacement x graph represents SHM. ‘A’ is
amplitude.
F
F
a)
b)
A
–A
A
F
c)
x
–A
x
F
d)
A
–A
A
x
–A
x
ANS: Next slide
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Solution:
You require a little knowledge of coordinate geometry.
We know that a motion is SHM when F = –k x. Thus restoring force varies linearly with x.
Therefore, F vs. graph should be a straight line.
Since, there is no constant term in eqn. F = –k x, therefore this line should pass through
origin.
Also, slope of this line is –k, which is negative.
Therefore correct option is (b).
F
A
–A
x
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Q . Find the spring constant from F vs. x graph for SHM.
F
4N
5 cm
x
–5 cm
–4N
ANS: k = 80 N/m
Explanation:
For SHM, F = –k x.
Since, at x = 5cm = 0.05 m, F = – 4 N
Therefore, from equation, F = –k x,
– 4 = – k (0.05)
k = 80 N/m
You can check it for F = 4N at x = – 5 cm.
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From last two slides, we have seen that an oscillatory motion will be SHM if F = – kx.
We also know from Newton’s 2nd Law that
F = ma
Equilibrium
position
Therefore, – kx = ma
 a = – (k/m) x
…(1)
m
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Since k and m are constants, acceleration a of the
oscillating body is directly proportional to its
displacement from the equilibrium position and is
directed opposite to the displacement, i.e.,
a a –x
a
F
m
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Therefore we can also define SHM as a motion in
which acceleration of the oscillating body is directly
proportional to its displacement from the equilibrium
position and is directed opposite to the displacement.
aa–x
(Condition for linear SHM)
O
x
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From last slide, we have seen that an oscillatory motion will be SHM if
a a –x
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If we put k/m = w2 in this eqn. (1) of last slide, i.e., eqn. a = – (k/m) x, becomes
a = – w2x
where, w is known as angular frequency of SHM.
(why we are introducing this concept of angular momentum, you will be able to
understand in the topic SHM as a projection of uniform circular motion)
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Clearly, angular frequency is related to k and m by:
w =
k
m
(angular frequency)
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We shall study about angular frequency when we learn SHM as the projection of uniform
circular motion.
From last slide we have seen that, angular frequency is given by:
k
w =
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Relation between angular frequency (w) with time period (T) and frequency (f)
Loosely speaking, we can consider angular frequency to be the angular velocity when a body
moves in uniform circular motion.
Clearly, the particle covers an angular displacement 2π rad in a time equal to its time period T.
Therefore,
w =
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m
2p
(angular frequency in terms of time period T)
T
Clearly, SI unit of ω is rad/s same as that of angular velocity.
Since, frequency f is given by f = 1/T, therefore we can write
w =
2p
T
=2pf
(angular frequency in terms of time period T and frequency f)
• A motion is Linear SHM if given conditions are satisfied:
1.
2.
3.
Motion must be oscillatory and hence periodic.
Force or acceleration of the particle is directly proportional to its displacement
from the equilibrium position.
Force or acceleration is always directed opposite to the displacement.
• Thus, any oscillatory motion will be SHM if
or,
F=–kx

F a–x
a = – w2 x

a a–x
where,
w =
k
m
=
2p
T
=2pf
where,
w = angular frequency (rad/s)
T = time period (s)
f = frequency (Hz or s –1)
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Q . Which of the following acceleration a vs. displacement x graph represents SHM. ‘A’ is
amplitude.
a
a
a)
b)
A
–A
A
a
c)
x
–A
x
a
d)
A
–A
A
x
–A
x
ANS: (b)
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Q . Find the time period from a vs. x graph for SHM.
a
5 m/s2
5 cm
ANS: T = 0.63 s
–5 cm
– 5 m/s2
Explanation:
For SHM, a = – w2x.
Since, at x = 5cm = 0.05 m, a = – 5 m/s2
Therefore, from equation, a = – w2x,
– 5 = – w2 (0.05)
w2 = 100
w = 10 rad/s (considering positive value only)
Also, w = 2p/T
Therefore, T = 2p / w = 2 x 3.14/10 = 0.628 s ≈ 0.63 s
x
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We have seen the condition for SHM in the previous slides. Now we state step by step method
to use SHM concepts in solving problems. Generally you are asked to find out the time period
of SHM.
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STEP BY STEP METHOD TO FIND OUT TIME PERIOD
STEP I: Find out the equilibrium position.
At equilibrium position net force and net torque is zero. For linear SHM net force should be
zero at equilibrium position.
STEP II: Assume x = 0 at the equilibrium position. Displace particle at a distance x from the
equilibrium position.
STEP III: Draw FBD of the particle when the particle is at a distance x from the equilibrium.
STEP IV: Write Newton’s 2nd law using FBD of STEP III. Write this equation in the form of
a = – w2 x and find out w.
STEP V: Use T = 2p/w to find out time period.
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Let us see some examples to illustrate the idea.
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Q . A mass m is attached to a vertical spring of spring constant k. Suppose the mass is displaced
from the equilibrium position vertically. Find the time period of the resulting oscillation.
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SOLUTION:
Let us use step by step method given in previous slide
to find out the time period of the oscillation.
(Natural length,
without mass)
k
STEP I: Find out the equilibrium position.
Let the elongation of the spring be y at the equilibrium position.
ky
y
Now, we know that equilibrium position is the point where net
Equilibrium
m
force and net torque is zero.
position
• We draw FBD to find out equilibrium position.
• Clearly forces on the mass are:
mg
Weight mg downward
Spring force ky upward (don’t ask why upward)
• Using Newton’s 2nd law at equilibrium in the vertical direction, we get,
mg – ky = 0
 y = mg/k
…(1)
• Therefore, equilibrium position is at a distance y = mg/k below the natural length of the spring.
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STEP II: Assume x = 0 at the equilibrium position. Displace particle at a distance x from the
equilibrium position.
(Natural length,
without mass)
(Natural length,
without mass)
y
m
y
k(x + y)
x=0
m
x
Equilibrium
position
mg
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STEP III: Draw FBD of the particle when the particle is at a distance x from the equilibrium.
Forces are:
Weight mg (downward)
Spring force k(x + y) upward.
(because net elongation from natural length is (x + y) in this case)
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STEP IV: Write Newton’s 2nd law using FBD of STEP III. Write this equation in the form of
a = – w2 x and find out w.
Using Newton’s 2nd Law in the downward direction, we have,
mg – k (x + y) = ma
…(2)
Now, we have t write this equation in the form of a = – w2 x. What to do?
We divide both sides of eqn. (2) by m which gives,
g – k (x + y)/m = a
 a = g – k (x + y)/m
…(3)
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But right hand side of this equation is not in the – w2 x. What to do?
Actually there is a term y in right hand side, which we have to eliminate.
How can we eliminate y?
Yes! We can use equation (1) for the equilibrium position to eliminate y.
Therefore, putting y = mg/k from eqn. (1) in eqn. (3) we get,
a = g – k (x + y)/m = g – k (x + mg/k)/m = g – (k/m)x – g
 a = – (k/m)x
…(4)
• Clearly, equation (4) is in the form of a = – w2 x. Therefore, motion is SHM.
• Therefore, comparing a = – w2 x with equation (4), we get
w2 = k/m  w = √(k/m) …(5)
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STEP V: Use T = 2p/w to find out time period.
• From equation (5), we have, w = √(k/m)
• Therefore, time period, T = 2p/w
 T = 2p √(m/k)
(Ans).
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DISCUSSION:
In this example we have explained the way how SHM problems are solved.
First we have to find out equilibrium position by equating forces to zero. We take x = 0 at
equilibrium position.
Then, we displace particle at a distance x from the origin and draw FBD at this displaced
position. We apply Newton’s 2nd law at this position and simplify this equation in the form of
a = – w2 x. This step may require little calculations and approximations too. Then, we find out
angular frequency w and hence, time period T.
Let us see another example to further clarify the idea. This problem is from IIT-JEE 2008.
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Q. A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs
of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the
axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and
both the springs and the axle are in a horizontal plane. The unstretched length of each spring is L. The
disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall.
The disk rolls without slipping with velocity v0 = v0i . The coefficient of friction is µ.
Answer the following questions.
(1). The net external force acting on the disk when its
centre of mass is at displacement x with respect to its
equilibrium position is :
(A) – kx
(B) – 2kx
(C) – 2kx/3
(D) –4kx/3
(2). The centre of mass of the disk undergoes SHM with
angular frequency w equal to :
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SOLUTION:
This question is simple problem. It requires little knowledge of Rotational Motion.
We will follow step by step method to find out the time period of the oscillation of centre of mass of
the disc. Slideshow to view animation.
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STEP I: Find out the equilibrium position.
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As the springs mass system lies on a horizontal plane,
springs will neither be compressed nor be stretched
at equilibrium.
Therefore equilibrium position corresponds to the
natural length of the springs.
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y
STEP II: Assume x = 0 at the equilibrium position. Displace
particle at a distance x from the equilibrium position.
x
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STEP III: Draw FBD of the particle when the particle is at a distance x from the equilibrium.
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Forces are: Spring force kx from each spring, friction f from ground, normal force N and weight mg.
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We can redraw FBD as shown.
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STEP IV: Write Newton’s 2nd law using FBD of STEP III.
Write this equation in the form of a = – w2 x and find out w.
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Along vertical net force = 0. (why?)
Along horizontal,
net force = f – 2kx = ma …(1)
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a
We have to find out f using rotational Mechanics.
As it is in pure rolling motion, we have
a = αr
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a
mg
f
…(2)
We arbitrarily choose clockwise direction to be positive.
Taking torque about centre of mass of the disc we get,
y
– fr = Iα
 – fr = (mr2/2) α
 – fr = (mr2/2)a/r (using (2))
 f = – ma/2 …(3)
Putting f = – ma/2 in (1), we get,
ma = – 4kx/3 which is equal to net force.
x
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Therefore, net force = ma = – 4kx/3.
…(4) (Ans)
(Please note, net force = net force along horizontal; because net force along vertical = 0)
• Using equation (4), we get,
 ma = – 4kx/3
 a = – (4k/3m) x
…(5)
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Clearly, equation (4) is in the form of a = – w2 x. Therefore, motion is SHM.
Therefore, comparing a = – w2 x with equation (4), we get
w2 = k/m  w = √(4k/3m)
…(6)
(Ans)
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Therefore, correct options:
(1). d
(2). d
• Appropriate form of Linear SHM
• WE know that a motion is linear SHM when
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a = – w2 x
• We can write a =
d2x
dt2
• Therefore, appropriate form of Linear SHM s given by
d2x
dt2
= – w2 x
(appropriate form of Linear SHM )
• We shall use this form when we study Simple Pendulum.
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A simple pendulum is an idealized model consisting of a point mass (which is known as bob)
suspended by a massless, unstretchable string. In nature there is no massless string, therefore
point mass is very heavy relative to string so that the mass of the string can be assumed
negligible compared to the mass of the bob.
When the point mass is pulled to one side of its straight-down
equilibrium position and released, it oscillates in a circular arc
q
about the equilibrium position.
We shall show that, provided the angle is small (less than about
L
10°), the motion is that of a simple harmonic oscillator. Before
that please revise examples 5 and 6 of Circular Motion.
If the amplitude oscillation is small, the path of the particle is
xm
approximately a straight line and the motion can be described
m
as linear SHM.
Equilibrium
position
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Let us consider the bob of mass m is suspended by a light string of length L that is fixed at the
upper end.
Clearly the equilibrium position is the lowest position of the bob. The path of the point mass is
not a straight line but the arc of a circle with radius L equal to the length of the string. We use as
our coordinate the distance x measured along the arc from the equilibrium position. If the
motion is simple harmonic, the acceleration must be directly proportional to x.
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The tangential acceleration at of the particle at time t is given by
at =
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d2x
dt2
…(1)
q
Forces acting on the particle are:
L
weight mg downward and tension T along the string.
T
L
We take positive direction along the displacement x from
the mean position. Therefore component of mg along
at
x m
tangential direction is –mg sin q. (because opposite to x)
Using Newton’s Law along the radial direction, we get,
mg sinq
– mg sinq = m at  at = – m sinq …(2)
q
As the amplitude is small (less than about 10°), sin q ≈ q.
(How? You will study about it in Mathematics.)
mg cosq
Hence, eqn. (2) becomes
mg
2
dx
at = – g q 
= – g q ...(3)
dt2
d2x
Now, we have to write this in the form of
= – w2 x . What to do to express q in terms of x
dt2
in eqn. (3) ?
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Yes! We can write using a little Mathematics,
q =
•
 q =
radius
x
L
q
Putting this into eqn. (3), we get,
d2x
dt2
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arc
= –
g
L
x
Clearly, eqn. (4) is in the form of
w=
L
…(4).
d2x
dt2
mg sinq
L
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Hence, motion is SHM.
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Time period is given by: T = 2p/w

T = 2p
L
g
T
x m
= – w2 x, where
g
L
at
q
mg mg cosq
(time period of a Simple Pendulum)
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Discussion:
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Time period of Simple pendulum, when amplitude is very small is given by:

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T = 2p
L
g
(time period of a Simple Pendulum; small amplitude)
Note that these expressions do not involve the mass of the particle. For small oscillations, the
period of a pendulum for a given value of g is determined entirely by its length.
The dependence of time period T on L and g in is just what we should expect. A long pendulum
has a longer period than a shorter one. Increasing g increases the restoring force, causing the
frequency to increase and the period to decrease.
We emphasize again that the motion of a pendulum is only approximately simple harmonic.
When the amplitude is not small, the departures from simple harmonic motion can be
substantial.
© 2007-08 P K Bharti, IIT Kharagpur