Physics Notes of Chap 10, Simple hormonic motion, for all board of kpk
CHAPTER 10 NOtes for kpk studetns
SIMPLE HARMONIC MOTION & WAVES
Q1:
Define oscillatory motion. Explain with examples.
Ans. Oscillatory Motion.
“The to and fro motion
of a body about its mean position is known as oscillatory motion”.
(OR) “The back and forth
motion of a body about its equilibrium position is known as oscillatory
motion”.
Explanation
The oscillatory motion
is also known as vibratory motion. If a body moves back and forth over the same
path, then such motion of the body is known as oscillatory motion.
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Examples
The examples of
oscillatory motion are given below:
1.
Motion
of simple pendulum.
2.
Motion
of wires of sitar and violin.
3.
Motion
of a body attached to an elastic spring.
Q2.
Define periodic motion. Explain with examples. Find the relation between the
frequency and time period of a periodic motion.
Ans. Periodic Motion.
“The motion which
repeats its self in equal intervals of time is known as periodic motion.”
Examples
1. Motion of simple
pendulum.
2. Motion of mass-spring
system.
3. Motion of metallic
strip.
Relation between
frequency and time period.
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If “T” represents the
time period and “f” represents the frequency, then we have.
f=
(1)
Equation (1) represents
the relation between frequency and time period of periodic motion.
Q3:
Define simple harmonic motion. Describe its characteristics features.
Ans. Simple Harmonic
Motion.
“The motion in which the
acceleration of the body is directly proportional to the displacement from its
mean position is known as simple harmonic motion”.
Explanation
If “a” represents the
acceleration of the body and “x” represents the displacement of the body from
its then we have,
a α
(-x) ------------------ 1
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relation (1) represents
the mathematical form of simple harmonic motion. The negative sign shows that,
the body is always directed towards the mean position.
Characteristics of
Simple Harmonic Motion: Consider a body of mass “m” attached to an
elastic spring as shown in the figure. If the body is displaced towards right
position “A” through a displacement “O” then on releasing the body at position
“A”, the body comes back towards the mean position “O” due to restoring force
“F”.
On the way, its velocity
increases due to which the body does not stop at mean position “O” and moves
towards left position “A”. Again due to restoring force “F” the body comes back
towards the mean position “O”. In this way the body performs simple harmonic
motion. During this motion the following characteristics can be observed:
1. In simple harmonic
motion, the body oscillates about its mean position “O”.
2. The distance of the body
from mean position is known as displacement. It is represented by “X”.
3. The maximum displacement
of the body from its mean position is known as amplitude.
It
is represented by “Xm” or “Xo”.
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4. In simple harmonic
motion, the acceleration of the body is directly proportional to its
displacement from mean position “O” i.e. a α (-x)
5. The velocity of the body
is zero at extreme position and maximum at mean position “O”.
6. At mean position, the
energy of the body is in the form of kinetic energy (K.E).
7. At extreme position, the
energy of the body is in the form of potential energy (P.E).
8. The total energy remains
constant during simple harmonic motion.
9. Time period of mass
spring system is given by T = 2Ï€
10. The frequency of mass
spring system is given by f=
Q4. Define the following terms: vibration /
Cycle, time period, displacement, frequency and amplitude.
Ans. Vibration
or Cycle: A complete round trip
of an oscillating body about its mean position is known as vibration or cycle.
Time Period: The time during which
one vibration is complete is known as time period. It is denoted by “T”.
Frequency: The number of vibrations
per seconds is known as frequency. It is represented by “f” its units are
vibration/sec, cycle/sec, hertz etc.
Displacement: The straight distance of
a vibrating body from mean position is known as displacement. It is represented
by “x”.
Amplitude: The maximum displacement
of a vibrating body from mean position is known as amplitude. It is represented
by “Xo”. The displacement and amplitude are measured in meter.
Relation between
frequency and time period: The
relation between frequency and time period is given by f =
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Q5. What is simple harmonic motion (S.H.M)?
What are the conditions for an object to oscillate with simple harmonic motion?
Ans. Simple Harmonic
Motion: “The
motion in which the acceleration of the oscillating body is directly
proportional to the displacement from its mean position is known as simple
harmonic motion.”
Explanation: If “a” represents the
acceleration of the oscillating body and “x” represents the displacement from
its mean position, then we have
a α (-x) ------------------- 1
Relation (1) represents the mathematical form of simple harmonic
motion. The negative sign shows that, the body is directed towards the mean
position due to restoring force.
Condition for an object to oscillate with S.H.M:
For a body executing S.H.M the following conditions must be
satisfied:
i.
The displacement of the oscillating
body should be kept small.
ii.
The motion should be performed along
same straight path about the mean position.
iii.
The motion should be performed under
the action of restoring force.
iv.
The relation a α (-x) should be satisfied.
v.
The motion should be repeated in equal
intervals of time.
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Q6. Show
that the mass spring system executes simple harmonic motion?
Ans. Consider a body of
mass “m” attached to the end of an elastic spring as shown in figure. Now if we
displace the body from its mean position “O” to the extreme position “A”, then
on releasing the body at “A” the body comes back towards the mean position “O”
due to the restoring force “F. on the way, its velocity increases due to which
the body does not stop at mean position “O” and moves towards left side of mean
position “O”.
On the way, again its velocity increases due to which it
does not stop at “O” and moves towards right of mean position. In this way, the
body performs S.H.M about mean position. Now according to Hooke’s law we have,
F α (-x) à F = - Kx ------------------ 1
The negative sign shows that the body is always directed
towards the mean position. Now according to Newton 2nd Law we have
F = ma ---------------- 2
Comparing equation 1 and equation 2 we get
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ma = -kx à a = - (
x ------------- 3
a = constant (-x)
a a (-x) --------------- 4
Relation 4 represents
the mathematical form of S.H.M, so the mass spring system executes simple
harmonic motion.
Time Period: The time period in terms
of angular frequency “w”
is given by
T = 2p / w ------------ 5
In equation 5 “ w” represents the angular
frequency i.e. the measure of angular displacement per unit. The acceleration
in terms of “w”
is given by
a = - w2 x ----------------- 6
comparing equation 3 and
6 we get
w2 = k/m or
w =
------------- 7
Putting equation 7 in
equation 5 we get
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T = 2p/
T = 2p
---------------- 8
Equation 8 represents
the time period of mass spring system.
Frequency:
We know that f =
--------------- 9
Putting equation 8 in equation
9 we get,
f =
----------------- 10
Equation 10 represents
the frequency of mass spring system.
Q7.
What is simple pendulum? Diagrammatically show the forces acting on the simple
pendulum. Also show that simple pendulum execute simple harmonic motion.
Ans. Simple
Pendulum: A
simple pendulum consists of a small and heavy metallic spherical bob which is
suspended by a light and inextensible from a fixed support.
Explanation: Consider a simple
pendulum as shown in the figure.
At mean position “O“ the
weight “W” of the pendulum is acting in downward direction while the tension
“T” in the string is acting in upwards direction, as shown in figure. Now if we
bring the bob from mean position “O” to extreme position “A”, then its weight “
W” is resolved into two components I,e “mg cosÓ©” and “ mg sinÓ© “.
The component “mg cosÓ©”
balance the tension “T” in the string. While the component “mg sinÓ© “ is acting
towards the mean position “O”. So “mg
sinÓ© “acts as restoring force. So we have,
mg cosÓ©=T _____________ (1)
And F= -mgsinÓ© ____________ (2)
Now according to newton’s 2nd
law, we have,
F = ma ____________ (3)
Comparing equation (2) and
equation(3), we get,
ma = -mg sinÓ©
= a = -g sinÓ©
_____________ (4)
Now in triangle “
OPA”, we
have
SinÓ© = perpendicular/
hypotenuse
SinÓ© =
Putting equation (5) in
equation (4), we get,
a = - g
= a =
( -x) _________ (6)
a = constant (-x)
a = α (- x) _________ (7)
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Relation (7) represents
the mathematical form of simple harmonic motion, so the simple harmonic
pendulum executes simple harmonic motion.
Time period: The time period in
terms of angular frequency “w “ is given by,
T =
_________ (8)
The acceleration in terms of “w “ is given by,
a = - w2 x _________ (9)
Comparing equation (6)
and equation (9), we get
w2 =
= w2
________ (10)
Putting equation (10)
in equation (8), we get
T =
T =
=
_________ (11)
Equation (11)
represents the time period of simple pendulum.
Frequency:
We know that,
F =
__________ (12)
Putting equation (11)
in equation (12), we get
F =
________ (13)
Equation (13) represent
the frequency of simple pendulum.
Q8. What do you mean by damped oscillations?
Also write the applications of damped oscillations.
Ans. Damped
Oscillations: “The
Oscillations, in which the amplitude decreases
steadily with time are called damped oscillations”.
OR
The process in which
energy is dissipated from the oscillating system is called oscillations.
Explanation:
This is a common everyday
experience that a balance wheel of a watch will cease to oscillate unless some
energy can be supplied to it from the watch spring. The effect of frictional
forces is to reduce the total mechanical energy of the oscillating system. A
note played on a piano is loud immediately after it has been played and then
gradually fades away.
Application of Damped
Oscillations
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i.
The shock
absorber of a car is the practical application of damped oscillations. It
provides a damping force to prevent excessive oscillation and thus provides
comfortable journey. A shock absorber consists of a piston in a reservoir of
oil. When the piston moves in response to a bump in the road, holes in the
piston head permit the piston to pass through the oil. During this movement,
viscous forces arises which cause the damping. As a result we enjoy comfortable
journey. Another application of damped oscillation is the shock absorber system
of human body. Skier’s body moves over the bumpy snow smoothly while his or her
thighs and calves act like a damping spring.
Q9. What is wave motion? How waves can be
categorized?
Ans. Wave and Wave Motion: A wave is a traveling
disturbance which carries energy from one place to another. The motion or
propagation of waves occurs in the form of transverse waves or longitudinal
waves.
Types of Waves:
There are two main types
of waves which are given below:
1. Mechanical Waves
2. Electromagnetic waves
Mechanical Waves: Those waves which are
produced due to the oscillation of material particles are called mechanical
waves.
Examples: Water waves, sound
waves, seismic waves. Waves produced in a stretched string etc are examples of
mechanical waves.
Electromagnetic Waves: The waves that propagate
by oscillation of electric and magnetic fields are called tromagnetic waves.
The electromagnetic waves needs no material medium for their propagation. The
electromagnetic waves are the combinations of travelling electric and magnetic
fields. Both fields varies in magnitude and both oscillate at right angle to
each other and to the direction of propagation of the wave.
Example: Visible light waves,
ultraviolet light waves, X-rays, micro waves and radio waves etc are the
example of electromagnetic waves.
Q10. How
waves transport energy without carrying the material medium? Also describe the
connection between oscillatory motion and waves.
Ans. A wave is a disturbance which
carries energy from one place to another. Waves transport energy without
carrying the material medium. It can be explain with the help of following
example:
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Example: Take a tub and filled it with
water. Now take a pencil and move it up and down at one edge of tub. As a
result, a disturbance is produce in the water. The particles of water begin to
oscillate about their mean position. The disturbance spread out in the form of
waves on the surface of water.
During the propagation of these
waves, the particle of water [medium] does not change its position permanently.
They perform only oscillatory motion.
For example, if we place a
piece of cork in the middle of the tub, the waves will spread out and will pass
through the cork and will reach the edges. But the cork will still remains in
the middle of the tub and will perform only up and down motion i.e. oscillatory
motion. This shows that during propagation of waves, the particle of the medium
does not change its position permanently and perform oscillatory motion only.
Q11. What are mechanical waves? Discuss the
types of mechanical waves.
Ans. Mechanical waves:
Those
waves which require a material medium for their propagation are called
mechanical waves.
Types of Mechanical waves: There are two types of
mechanical waves i.e.
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1. Transverse waves
2. Longitudinal waves
(1) Transverse waves: “These are those waves in
which the disturbance occurs perpendicular to the direction of motion of the
wave”.
Explanation: The transverse waves consist
of crests and trough which are produce one after the other in a certain order.
The crest represents the part of wave above the equilibrium position while the
trough represents the part of waves below the equilibrium position.
The
distance between two nearest crests or two nearest trough is equal to
wavelength “l”, while
the distance between nearest crest and trough is equal to half of the
wavelength i.e “
”. The velocity, frequency and wavelength of
transverse waves are related to each other by the following equation i.e.
V = fl
Production
of Transverse Waves in Slinky: The transverse waves can be produced in a
stretch string. They can be also generated in slinky. Slinky represents a long
loosely coiled spring. If one end of the slinky is moved up and down continuously,
then transverse waves is generated which move towards right as shown in the
figure.
(2). Longitudinal Waves:
“These
are those in which the disturbance occurs parallel to the line of travel of the
waves”.
Explanation: The longitudinal waves
consist of compressions and rarefactions which are produced one after the other
is certain order. Compression represents the regions of high density or
pressure relative to the mean density or pressure of the medium. The
rarefaction represents the regions of low density or pressure relative to the
mean density or pressure of the medium. The distance between two nearest
compressions or two nearest compressions or two nearest rarefaction is equal to
wavelength “λ”, while the distance between nearest compression and rare
fraction is equal to half of the wavelength i.e.
. The velocity, frequency and
wavelength of these waves are related to each other by the following i.e.
V = f λ
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Production of Longitudinal
Waves in Slinky: If we
push one end of the slinky forward and backward along its length, then two
regions are formed. The region where the parts of the slinky are compressed
together is called compression and it move towards right. While the region
where the parts of slinky are stretched apart is called rarefaction and it also
move towards right. In this way, longitudinal waves are produced in slinky, as
shown as in the figure.
Q12. Clearly distinguish between transverse and
longitudinal waves.
Transverse Waves
|
Longitudinal Waves
|
The disturbance occur
perpendicular to the direction of motion of the wave.
|
The disturbance occurs parallel to the direction of motion
of the wave.
|
Consist of crests and
trough
|
Consists of
compression and rarefaction
|
The distance between two
nearest crests or tow nearest toughs is equal to wavelength “λ”
|
The distance between
two nearest compressions or two rarefactions is equal to wavelength “λ”
|
Water waves produced in a
stretched string arte the examples of transverse waves
|
Sound waves produced
in a compressed spring are the examples of longitudinal waves.
|
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Q13. Define the following terms: wavelength,
frequency, time period, amplitude, velocity of waves.
Ans. Wavelength: The shortest distance between points where the
wave pattern repeats itself is called wavelength. OR
The distance covered by a
wave in one time period “T” is called wavelength. The wavelength is represented
by “λ”. Its unit is meter.
Frequency: The number of wave cycles “N”
passing through a certain point “P” in unit time T is called frequency i.e.
F
------------------ 1
In the given figure, if the
three cycles C1,C2 and C3 cross point “P” in one second, then according to
equation 1 frequency will be 3 hertz i.e.
F
= 3HZ
F = 3Hz
Time Period: “The time required for one
cycle to pass through a certain point is called time period. It is represented
by “T”. its unit is second. The time period is equal to the reciprocal of
frequency i.e.
T=
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Amplitude: The maximum displacement of
particle of the medium from its mean position is known as amplitude. It is
represented by “X0” its unit is meter.
Speed of a Wave: The distance travelled by a
wave in a unit is called speed of a wave. It is represented by “v”. its unit is m/sec. The speed of the
wave depends upon the nature of waves and medium of propagation.
Q14. Prove the relation between wave speed, wave
length and frequency of wave. Or Show that v = fλ
Ans. We know that the speed
is given
V=
------------- 1
Incase of waves, we know that
the distance covered by a wave in one time period “T” is equal to wavelength
“λ” so we get
V=
-------------------- 2
We also know that, T=
_________ (3)
Putting equation (3) in
equation (2), we get
V=
= λ×
= λ × f
= V = f λ ______________ (4)
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Equation (4) represents the
relation between wave speed, frequency and wavelength of a wave.
Q15.
Using ripple tank explain reflection, refraction and different of waves.
Ans. Ripple Tank: It
is a device with the help of which we can study the various characteristics of
water waves.
Construction: It consists of a rectangular
tray containing water fitted with a glass bottom. A light bulb is fixed above
the tray while a viewing screen is p[lace well below the tray.
Now when a disturbance is
produced in water with the help of uniform rod, then transverse waves are
produced. The patter of the waves can be obtained on the screen with the help
of parallel rays of light. The crest of the waves appear as bright bands while
the trough appear as dark bands.
Using ripple tank, we can study
the various properties of waves like reflection, refraction and diffraction.
(1) Reflection of waves: When an obstacle is placed in
the path of travelling waves, then the waves strike the obstacle and reflect
through various angles. Such phenomenon is known as reflection of waves. In
ripple tank, the reflection of waves can be studied by placing an upright
surface in water. The given indicates the reflection of waves from a flat
surface.
(2) Refraction of waves: When waves enters from a rare
medium into a denser medium then the waves bends away slightly from its
original path. Such phenomenon is known as refraction of waves.
In ripple
tank, refraction of waves can be studied by placing a plastic sheet at the
bottom of ripple tank. The waves will refract at the edges of plastic sheet.
The speed of waves of waves is greater in rare medium as compare to denser
medium.
(3) Diffraction of waves: when wave passes through a
narrow opening, then the waves spread out about the opening, such phenomenon is
called diffraction of waves.
In
ripple tank, the diffraction of waves can be studied by placing two obstacles
in such a way, that a small opening is left between them, as shown in the
figure. The waves will diffract around the corners of the opening after passing
through it.
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Solution
of Examples, Assignments
Example
1: what is the frequency of oscillation if the time period is 20ms?
Solution:
Given data: Time period = T = 20ms =
20×20×10-3s [ milli-10-3 ]
Required: Frequency of oscillation = f
=?
Calculation: We know that, f =
________ (1)
Putting the values in
equation (1), we get
f =
Hz
f = 0.05×103
Hz = f = 50Hz
Answer: F = 50Hz
Assignment
1: when an object oscillates with a frequency of 0.5 Hz, what is its time
period?
Solution:
Given data: Frequency= f = 0.5Hz
Required: Time period = T =?
Calculation: We know that, f =
=T
(1)
Putting the values in
equation (1), we get
T =
=
= 2 sec
Answer: T = 2sec
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Example 2:
A spring has a spring constant of 48.0N/m. this spring is pulled to a distance
of 55cm from equilibrium. What is the restoring force ?
Solution:
Given data: Displacement = x = 55cm =
0.55m,k = 48.0N/m
Required: restoring = F =?
Calculation: We know that, F= -kx ________
(1)
Putting the values of
equation (1), we get
= F = -48.0×0.55N = F = - 26.4 N
Answer: F - 26.4 N
Assignment
2: Determine the restoring force of a spring displaced 1.5m, with the spring
constant of 30.0N/m.?
Solution:
Given data: Displacement =x=1.5m, spring
constant = k = 30.0N/m
Required: Restoring force = F =?
Calculation: We know that, F = -Kx
(1)
Putting the values of
equation (1), we get
F = -Kx = -30.0×1.5N = F = -45N
Answer: F - 45 N
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Example
3: What is the mass of a vertical mass- spring system if it oscillates with a
period of 2.0sec and has a spring constant of 20.0N/m?
Solution:
Given data: Time period = T = 20 sec,
spring constant = k = 20.0N/m
Required: Mass of vertical mass- spring
system=m=?
Calculation: We know that,
T= 2Ï€
(1)
Squaring both sides of
equation (i), we get,
T2= 4Ï€2
= m=
_______ (2)
Putting the value in equation
(2), we get,
m =
= 2.02 kg
Answer: m 2kg
Assignment
3: A body of mass 0.2kg is attached to a spring placed on a frictionless
horizontal surface. The spring constant of spring is 4 N/m. find the time
period of oscillating mass-spring system.
Solution:
Given data: Mass = m= 0.2kg, spring
constant = k = 4N/m
Required: Time period = T =?
Calculation: We know that, T= 2Ï€
__________ (1)
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Putting the value of equation
(1), we get
T = 2×3.14
= 6.28×
T = 6.28×0.22sec = T =
1.38sec
Answer: T = 1.4sec
Example
4: Determine the magnitude of restoring force for a pendulum bob of mass
100.0gm that has been to an angle of 100 from the vertical.
Solution:
Given data: Mass = m= 100gm, Acceleration
due to gravity = g = 9.8m/sec2
Angle = Ó© = 100
Required: Restoring force = F =?
Calculation: We know that, in case of simple
pendulum, the component “mg sinÓ©” acts as restoring force, so we have, F= - mg
sinÓ© _________ (1)
Putting the values of
equation (1), we get
= F= -0.1×9.8× sin100
= - 0.98×0.17N
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Answer: F= -0.167N
Assignment
4: At what angle must a pendulum be displaced to create as restoring force of
4.00N on a bob with a mass of 500gm?
Solution:
Given data: Restoring force = F = 4.00N,
Mass = m= 500gm=0.5kg
Required: Angle =Ó©=?
Calculation: We know that, in case of
simple pendulum, the component “mg sinÓ©” acts as restoring force, so we get, F=
- mg sinÓ© = sinÓ© = F/mg _________ (1)
Putting the values of
equation (1),we get
= sinÓ© = 4.00/(0.5×9.8)=
4.00/4.9=0.816
= sinÓ© = 0.816 = Ó© = sin-1
(0.816)= 54.60
Answer: Ó© = 54.60
Example
5: What is the gravitational field strength on
planet mercury, if a 0.500 m pendulum swings with a solution of 2.30sec?
Solution:
Given data: Time Period = T = 2.3 sec,
Length = l = 0.500 m
Required: Gravitational field
strength = g=?
Calculation: we known that T
= 2Ï€
-------------------- (1)
Squaring both sides of
equation 1 we get
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Putting the values of
equation 2 we get
Answer: g = 3.73 N/Kg
Assignment
5: What is the gravitational field strength at the top of Mount Everest at an
altitude of 8954.0 m, if a pendulum with a length of 1.00m has a period of 2.01
sec?
Solution:
Given Data: Length = l= 1m
Time period = T = 2.01 sec
Required: Gravitational field
strength = g=?
We know that, T = 2Ï€
-------------------- (1)
Squaring both sides of
equation (1)
Putting the values of
equation 2 we get
g = 9.76 N/kg
Example
6: A student vibrates the end of a spring at 2.6 Hz. This produces a wave with
a wavelength of 0.37m. Calculate the speed of the wave.
Solution:
Given Data: Frequency = f =2.6 Hz
Wave length = λ = 0.37m
Required: Speed of the wave =
v =?
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Calculation: we know that v =
fλ---------------- 1
v = fλ = 2.6×0.37m/sec
v = 0.962 m/sec
Exercise
Q1. Choose the correct Answer.
1. A transverse wave on a
string has an amplitude A. A tiny spot on the string is colored red. As one
cycle of the wave passes by, what is the total distance traveled by the red
spot?
a.
A b. 2A c.
d.
2. Which of the following
does not affect the period of the mass spring system?
a.
Mass b. Spring Constant c. Amplitude of
vibration
d.
All of the above affect the period
3. An object of mass ‘m’
oscillates on the end of a spring. To double the period, replaced the object
with one of mass:
a.
2m b. m/2 c. 4m d. m/4
4. A car mounted on shock
absorbers is like a mass on a spring. If we ignore damping, will the frequency
of the oscillations change if passengers (or a heavy load) are added to the
car? The frequency will:
a.
Increase b.
Zero c. Stay
the same d. Zero
5. If the pendulum
completes exactly 12 cycles in 2.0min, what is frequency of the pendulum?
a.
0.10Hz b. 0.17 Hz c. 6.0 Hz d.10 Hz
6. A certain pendulum has
an iron bob. When the iron bob is replaced with lead bob of the same size, its
time period will:
a.
Stay
the same b. Decrease c. Increase d. Be zero
7. A wave transports:
a.
Energy
but not matter b. Matter but not
energy c. Both energy and
matter
d.
Air
8. The Bending of waves
around the edges of the obstacle is
a.
Reflection b.
Refraction c. Diffraction d.
Damping
Conceptual questions
Give a brief response to
the following questions.
1.
Is
every oscillatory motion simple harmonic? Give examples.
Ans. No, it is not necessary for
an oscillatory motion to be simple harmonic. In oscillatory motion a body moves
to and fro about a fixed point periodically. E.g. the motion of tuning fork,
swing etc. While for SHM, the following two conditions must be satisfied.
a. The acceleration of the
vibrating body is directly proportional to the displacement from the mean
position at any instant, and
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c. The acceleration is always
directed towards the mean position. For example
i.
Vibratory
motion of simple pendulum.
ii.
Vibratory
motion of a mass spring system
It is to be noted that every
SHM is oscillatory but every oscillatory motion is not necessary a SHM.
2.
For
a particle with simple harmonic motion, at what point of the motion does the
velocity attain maximum magnitude? Minimum magnitude?
Ans. For a particle executing SHM
its total energy at any instant of time is constant. That is the sum of K.E and
P.E remains the same all the time. Since K.E of the particle passing through
mean position is maximum (Equal to total energy) so at this position the
velocity of the particle will be maximum. When the particle is at either
position, the total energy of the particle is in the form of P.E and its K.E is
equal to zero. As K.E is zero at other extreme position, so the velocity of the
particle is also zero at these positions.
3.
Is
the restoring force on a mass attached to spring in simple harmonic motion ever
zero? If so, where?
Ans. Yes, the restoring force in
SHM becomes zero at the mean position. According to Hook’s law we have
F = - Kx -----------1
In equation 1 “x” represents
the displacement of vibrating body from mean position.
Now at the mean position, we
have x = 0
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So equation 1 becomes
F = - K (0) Ã F = 0 -------------- 2
Equation 2 shows that the
restoring force is zero at the mean position.
4.
If
we shorten the string of a pendulum to half to its original length, what is the
affect on its time period and frequency?
Ans. Effect on time period:
We know that time period of
simple pendulum is given by,
T = 2Ï€
------------------ 1
Now if the length becomes
half then we put l =
in equation 1 we get
T = 2Ï€
T =
[2Ï€
]
--------------- 2
Putting the equation 1
in equation 2 we get,
Equation 3 shows that
the time period will decrease by a factor is
when the length of the string becomes half.
Effect on frequency:
We know that
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Put
in equation 4
we get
Putting equation 4 in
equation 5 we get
Equation 6 shows that
frequency of simple pendulum will increase by a factor
when the length
of the string becomes half.
5.
Suppose
you stand on a swing instead of sitting on it. Will your frequency of
oscillation increase or decrease?
Ans. The swing may be
considered as a simple pendulum. In this case the frequency of oscillation is
given by,
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Where l = length the
pendulum and is equal to the distance from the point of suspension to the center
of gravity (Navel) of the person on the swing.
This length decreases as
the person stands up on the swing i.e. center of gravity of the person rises
up. Using equation 1 it is clear that frequency of oscillation is inversely
proportional to the square root of length of the pendulum. So, frequency of
oscillation increases as length decreases when the person stands up instead of
sitting.
6.
Explain the difference between the speed of a transverse
wave travelling along a cord and the speed of a tiny colored part of the cord.
Ans. Transverse waves are
those in which particles of the medium vibrate at right angle to the direction
of propagation of wave motion. Consider a cord having a coloured tiny part as
shown in the figure. Its one end is fixed and the other end is in our hand. If
we move our hand up and down transverse waves are produced moving in the
forward direction i.e. from left to right. As these are transverse waves, so
each part of the string moves up and down i.e. vibrating up and down, while the
transverse waves move in the forward direction while the coloured tiny part of
the string moves up and down executing SHM.
7.
Why waves refract at the boundary of shallow and deep water?
Ans. Refracting of waves
involves a change in the direction of waves as they pass from one medium to
another. In refraction both speed and wavelength of waves change. The speed of
wave depends upon the properties of a medium through which it travels. The speed
of waves is not the same in shallow water and deep water. Waves travel fast in
deep water. Thus if the waves are passing from deep water into shallow water,
they slow down. This decrease in speed is accompanied by decrease in
wavelength. So when waves are transmitted from deep water into shallow water,
its speed and wavelength decreases and waves change its direction. i.e.
refracted.
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8.
What is the effect on diffraction if the opening is made
small?
Ans. Diffraction is the
slight bending of a wave as it passes around the edge of an obstacle. The
amount of bending of a wave depends upon the relative size of the wavelength of
the wave and size of the opening. If the opening is much larger than the
wavelength, then very less bending occurs which is unnoticeable. How ever if
the two are comparable (or equal in size) then a considerable bending occurs
and can be seen easily with naked. Thus bends more and more if the opening is
made small i.e. making it comparable in size to the wavelength of the wave.
Numerical Questions
1.
A mass hung from a spring vibrates 15 times in 12s.
calculate the frequency and the period of the vibration.
2.
A spring requires a force of 100.0N to compress it to a
displacement of 4cm. what is its spring constant?
3.
A second pendulum is a pendulum with period of 2.0s. How
long must a second pendulum be on the earth (g=9.81 m/s2) and Moon
(Where g = 1.62m/s2)? What is the frequency of second pendulum at
earth and on moon?
4.
Calculate the period and frequency of a propeller on a plane
if it complete 250 cycles in 5.0s.
5.
What waves with wavelength 2.8m, produced in a ripple tank,
travel with a speed of 3.80 m/s. What is the frequency of the straight vibrator
that produced them?
6.
The distance between successive crests in a series of water
waves is 4.0m and the crest travel 9.0 m in 4.5s. what is the frequency of the
waves?
7.
A station broadcasts an AM radio wave whose frequency is
1230 x 103 Hz (1230 KHz on the dial) and an FM radio wave whose
frequency is 91.9 x 106Hz (91.6 MHz on the dial). Find the distance
between adjacent crests in each wave.
8.
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