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OSCILLATIONS
CONCEPTUAL PROBLEMS
Q-01
Ans
Q-02
Ans
A girl is sitting on a swing. Another girl sits by her side. What will be the effect on the periodic-time of the
swing ?
There will be no change in the periodic time because the periodic time is independent of mass but
depends upon length of pendulum and acceleration due to gravity at a place, which are not affected by
sitting another girl on a swing.
The girl sitting in a swing stands up. What will be the effect on the periodic time of the swing ?
The periodic time, T will decrease because in the standing position, the location of centre of mass of the
girl shifts upwards. Due to which the effective length of the pendulum. ‘ ’ (i.e. of swing) becomes less. As
T∝
Q-03
Ans
Q-04
Ans
, therefore T decreases.
The bob of a simple pendulum of length is negatively charged. A positively-charged metal plate is
placed just below the bob and the pendulum is made to oscillate. What will be the effect on the time
period of the pendulum ?
Due to force of attraction between positively charged metal plate and negatively charged bob of
pendulum, the effective value of ‘g’ increases and so the time period will decrease.
If a hollow pipe passes across the diameter of earth, then what changes take place in the velocity and
acceleration of a ball dropped in the pipe ?
The ball dropped in the pipe will execute S.H.M. with centre of earth as mean position. The velocity of the
ball is maximum at the centre of earth and minimum on the surface of earth and acceleration of the ball is
maximum on earth’s surface and minimum at the centre of earth.
Q-05
Ans
Explain that sin θ and cos θ are periodic function.
A periodic function is one whose value repeats after a definite interval of time. Sin θ and cos θ both are
the periodic functions because sin (θ + 2 π n) = sin θ and cos (θ + 2 π n) = cos θ, where n = 1,2,3 …
Q-06
Ans
Can the function
The function,
Or
= sin ω t + cos ω , represent the simple harmonic function ? Explain.
= sin ω t + cos ω
ω
ω
ω
ω
This represents a simple harmonic motion with period 2π/ω.
Q-07
If a pendulum clock is taken to a mountain top, does it lose or gain time, assuming it is correct at a lower
elevation ?
1
Ans
Time period of a simple pendulum is inversely proportional to the square root of g i.e. T ∝ 1/ . At the
mountain top, the value of g decreases. Hence the value of T increases i.e. the pendulum will take longer
time to complete one vibration . This shows that the pendulum clock will become slow. Hence the
pendulum clock will lose time on the mountain top.
Q-08
Ans
The soldiers marching on a suspended bridge are advised to go out of steps. Why ?
If the soldiers while crossing a suspended bridge march in steps, the frequency of marching steps of
soldiers may match with the natural frequency of oscillations of the suspended bridge. In that situation
resonace will take place, then the amplitude of oscillations of the suspended bridge will increase
enormously, which may cause the collapsing of the bridge. To avoid this situation, the soldiers are
advised to go out of steps on suspended bridge.
Q-09
For an oscillating simple pendulum, is the tension in the string constant throughout the oscillation ? If not
, when it is (a) the least, (b) the greatest ?
In simple pendulum, when bob is in deflection position, the tension in the string is T = mg cos θ. Since the
value of θ is different at different positions, hence tension in the string is not constant throughout the
oscillations. (a) At end points, θ is maximum : the value of cos θ is least, hence the so the value of tension
is greatest.
Ans
Q-10
Ans
At what distance from the mean position is the K.E. in simple harmonic oscillator equal to P.E ?
When the displacement of a particle executing S.H.M is y, then its
K.E. =
If
Q-11
Ans
K.E = P.E,
and
P.E. =
then
or
2
or y =
The amplitude of a simple harmonic oscillator is doubled. How does this affect (i) the maximum velocity
(ii) the total energy and (iii) the period of the oscillator ?
(i) max = ω. If amplitude a is doubled, the value of maximum velocity becomes double.
(ii) Total energy, E =
. If a is doubled, then E becomes four times.
(iii) T = 2
, since and do not change with the change in amplitude of oscillation, hence period of
oscillator remains unchanged with change in amplitude of oscillation.
Q-12
Ans
What provides the restoring force for simple harmonic oscillations in the following cases ? (i) simple
pendulum (ii) spring (iii) column of mercury in U tube.
The source of restoring force is as follows :
(i) Simple pendulum : gravity (ii) spring : Elasticity (iii) column mercury in U tube : weight
Q-13
What is the frequency of a second pendulum in an elevator moving up with an acceleration of g/2 ?
Ans
For second pendulum, frequency
. When elevator is moving upwards with accelerations a, the
effective acceleration due to gravity is g1 = g + a = g + g/2 = 3g/2.
As
so
∝ .
2
or
∴
Q-14
What is the percentage change in the time period, if the length of simple pendulum increases by 3%.
Ans
Time period of simple pendulum of length is T = 2π
…(1)
When length of simple pendulum increases by 3% then
Taking log of (i) and differentiating it we have
∴
Q-15
Ans
.
% change in time period =
A shelf moves vertically with simple harmonic motion whose time period of motion is T seconds. Find the
maximum amplitude that it can have so that objects resting on the shelf may always remain in contact
with it.
When the objects are resting on the shelf, then accelerations of the object will be the same as that of the
shelf. If R is the reaction of the shelf on the object of mass m placed on the shelf, then the equation of
motion of a mass on the shelf during upward motion of the shelf is
R – mg = mass × acceleration = m ω2 y
Or
R = mg – m ω2 y = m (g
ω2 y) = m
.
The object will just leave the shelf if R = 0 i.e.
g=
or
This is the greatest amplitude the shelf should have during its vibration. For amplitude greater than
,
the object will leave the shelf.
Q-16
A particle is subjected to two simple harmonic motions in the same direction having equal amplitude and
equal frequency. If the resultant amplitude is equal to the amplitude of individual motions, what is the
phase difference between the motions.
Ans
Here,
Q-17
and
As
R2 =
∴
r2 = r2 + r2 + 2r.r cos θ
Or
cos θ =
Or
θ = 1200 = 2π/3 radian
R=r;
θ=?
r2 (1+cos θ)
or
1+cos θ =
A pendulum is mounted on a cart rolling without friction down on an inclined surface of inclination θ
with the horizontal. The period of the pendulum on an immobile cart is T. What will be the period of the
pendulum on the cart when the cart rolls down the surface.
3
Ans
When the cart is at rest, the effective acceleration due to gravity is g and when the cart is rolling down the
inclined surface, the effective acceleration due to gravity involved perpendicular to plane is g cos θ. Time
period of pendulum when cart is immobile, T = 2 π
.
Time period of pendulum when cart is moving down the plane, T’ = 2 π
.
Q-18
A particle moves with S.H.M. in a straight line. In the first second after starting from rest, it travels a
distance
cm and in the next second it travels a distance
cm in the same direction. Prove that the
amplitude of oscillation is
.
Ans
As the particle starts from rest, it must start from the extreme position. Hence (i) when t = 0,
is the required amplitude. Using the relation, = r cos ωt, we have
rand
r–(
= r cos ω × 1 = r cos ω
) = r cos ω × 2 = r cos 2 ω
= r where r
…(i)
or
r-
(2 cos2 ω - 1)
…(ii)
Solving (i) and (ii), we get r =
Q-19
Ans
Two springs have force constants
and
respectively. They are attached to a mass m and two fixed
supports as shown in Fig. 10(a).37. If the surface is frictionless, find the time period of oscillations. What is
the spring factor of this combination.
If the mass m is displaced a little through distance towards
right hand side, the spring gets compressed and spring
gets stretched. Due to which, the restoring forces and
developed in two springs will be towards left i.e. in the same direction. Since and are the spring
constants of the two springs, hence,
and
Total restoring force, F = F1 + F2 = ( K1 )+ ( K2 ) =
(K1+K2)
…(i)
i.e. F ∝ . This force F is directed towards equilibrium position of the body. If body is left free, it will
execute linear S.H.M. If k is the force constant of a spring which is equivalent to the combination of the
two springs as given above, then
F = ky
…(ii)
From (i) and (ii),
Here,
k = k1 + k2
spring factor = k1 + k2
Inertia factor = mass of the body = m
∴ Time period, T = 2π
π
Q-20
All simple harmonic motions are periodic motions but all periodic motions are not simple harmonic
motions. Explain.
Ans
Periodic motion is that motion which is repeated identically after a fixed interval of times e.g. (i) the
revolution of earth around the sun, (ii) the rotation of earth about its own axis etc. Simple harmonic
motion is a special case of periodic motion in which the body moves to and fro about its equilibrium
4
Q-21
Ans
position. The force acting on the body at an instant is directed towards equilibrium position and is
proportional to the displacement of the body form equilibrium position i.e.
F = − ky
A simple
harmonic motion is represented by a single harmonic function (i.e. sine or cosine function) and of constant
amplitude.
For example, the oscillations of the bob of simple pendulum is simple harmonic motion which is periodic
also. But the revolution of earth around the sun is only periodic and not simple harmonic one, as it is not
to and fro motion about a fixed point.
Equations for two waves is given as
If amplitude and time period
of resultant wave dose not change, the calculate (
).
The equation of resultant wave is
The amplitude of the resultant wave is
Given, A = , then
Or
=2
cos
or
cos
or
Q-22
Two simple pendulums of length 1 metre and 16 metre respectively are both given small displacements in
the same direction at the same instant. After how many oscillations the shorter pendulum has completed,
the two pendulum will be in the same phase again.
Ans
Time period, T = 2π
or T ∝
∴
or T2 = 4 T1
It means, when the pendulum of smaller length will complete 4 oscillations, the pendulum of larger
length will complete 1 oscillation. It means, the two pendulum will be in the same phase, when shorter
pendulum has completed 4 oscillations.
Q-23
A block is resting on a piston which is moving vertically with simple harmonic motion of period 1.0
second. At what amplitude of motion will the black and piston separate ? What is the maximum velocity
of the piston at this amplitude ?
Ans
Restoring force = m ω2 ; weight of block = mg
For just separation of block and piston ; m ω2
= mg
Or
Max. velocity of block, Vmax =
ω
π T
× 3.14/1 = 1.57 ms 1.
5
VERY SHORT ANSWER QUESTIONS
Q-01
Ans
Which of the following conditions is not sufficient for S.H.M. and why ?
(i) acceleration ∝ displacement, (ii) restoring force ∝ displacement.
Conidtion (i) is not sufficient because it gives no reference of the direction of acceleration, where as in
S.H.M the acceleration is always in a direction opposite to that of the displacement
Q-02
The particle is executing S.H.M. The amplitude of motion is . State those positions of the particle in terms
of when (i) the K.E. of the particles is zero (ii) P.E. is zero, (iii) P.E. is one-fourth of the total energy (iv)
P.E and K.E are equal.
Ans
(i) ±
Q-03
When a body of mass 2.0 kg is suspended by a spring, the spring is stretched. If the body is pulled down
slightly and released, it oscillates up and down. What force is applied on the body by the spring when it
passes through the mean position ? (g = 9.8 N kg−1)
Since there is no acceleration in the body at the mean position, hence the resultant force on it will be zero
i.e. the force applied by the spring will be exactly equal to the weight of the body.
Ans
(ii) 0
(iii) ± /2
(iv) ±
.
Q-04
How will the period of a simple pendulum change when its length is doubled ?
Ans
The time period becomes
Q-05
Ans
Why the amplitude of the vibrating pendulum should be small ?
When amplitude of the vibrating pendulum is small then angular displacement of the bob used in simple
pendulum is small. Here the restoring force F = mg sin θ = mg θ = mg / . Where is the displacement of
the bob and is the length of pendulum. Hence F ∝ . Since F is directed towards mean position, therefore
the motion of the bob of simple pendulum will be S.H.M. if θ is small.
Q-06
Ans
Why a point on a rotating wheel can not be considered as executing S.H.M ?
The motion of rotating points on the wheel is not oscillatory but is only periodic.
Q-07
Ans
A man with a wrist watch on his hands fall from the top of a tower. Does the watch give correct time ?
Yes ; because the working of wrist (spring wound) watch is independent of ‘g’ but depends upon the P.E.
stored in the spring.
Q-08
How would the period of spring mass system change, when it is made to oscillate horizontally and then
vertically ?
The time period remains same in both the cases.
Ans
times the original value. ∵ T ∝
.
Q-09
A vibrating simple pendulum of period T is placed in a lift which is accelerating upwards. What will be
the effect on the time period ?
Ans
Time period decreases as effective value of acc. due to gravity increases (i.e. g’ = g + ) and (T ∝ 1/
Q-10
A vibrating simple pendulum of period T is placed in a lift which is accelerating downwards. What will
be the effect on the time period ?
6
)
Ans
Time period increases as effective value of acceleration due to gravity decreases (∵ g’=g−a) and (T ∝
1/
).
Q-11
Ans
Two simple pendulum of equal length cross each other at mean position. What is their phase difference ?
1800 or π radians
Q-12
Two simple pendulums of unequal length meet each other at mean position while oscillating. What is
their phase difference ?
00 if both are moving in the same direction and 1800 or π radians if moving in opposite directions.
Ans
Q-13
Ans
What is the phase relationship between displacement, velocity and acceleration in S.H.M ?
In S.H.M, the velocity leads the displacement by a phase π/2 radians and acceleration leads the velocity
by a phase π/2 radians.
Q-14
Ans
Can the motion of an artificial satellite around earth be taken as S.H.M ?
No, it is a circular and periodic motion but not to and fro about a mean position which is essential for
S.H.M.
Q-15
Ans
When is the tension maximum in the string of a simple pendulum ?
The tension in the string is maximum at the mean position because tension in the string = mg cos θ and at
mean position θ = 00 or cos θ = 1.
Q-16
Ans
How many times in one vibration, K.E. and P.E. becomes maximum ?
two times.
Q-17
Ans
Is the damping force, constant on a system executing S.H.M ?
No, because damping force depends upon velocity and is more when the system moves fast and is less
when system moves slow.
Q-18
Ans
Will the time period of loaded spring change when taken to moon ?
No, because time period of loaded spring does not depend upon the acceleration due to gravity but
depends upon the mass attached and spring constant of the spring.
Q-19
Ans
What determine the natural frequency of a body ?
Natural frequency of a body depends upon (i) elastic properties of the material of the body and (ii)
dimensions of the body.
Q-20
Can a motion be periodic but not oscillatory ? If your answer is yes give an example and if not explain
why ?
Yes, uniform circular motion is the example of it.
Ans
Q-21
Can a motion be oscillatory but not simple harmonic ? If your answer is yes give an example and if not
explain why ?
7
Ans
Yes, when a ball is dropped from a height on a perfectly elastic plane surface, the motion of ball is
oscillatory but not simple harmonic as restoring force F = mg = constant and not F ∝ − y
Q-22
Ans
What is the basic condition for the motion of a particle to be S.H.M. ?
The necessary and sufficient condition for motion to be simple harmonic is that the restoring force must
be linear i.e. F = − ky or Torque, τ = − c θ.
Q-23
What is the ( ) distance moved by (b) displacement of, a body executing S.H.M. in a time equal to its
period if its amplitude is A.
A body while executing S.H.M. completes one vibration in a time equal to its period, so the body reaches
its initial position after a time equal to its period. Thus the total distance travelled is 4A and displacement
is zero.
Ans
Q-24
Ans
Determine whether or not the following quantities can be in the same direction for a simple harmonic
motion ; ( ) displacement and velocity ( ) velocity and acceleration (c) displacement and acceleration.
( ) Yes, when the particle is moving from equilibrium position to extreme position (b) Yes, when the
particle is moving from extreme position to mean position to mean position (c) No, because is S.H.M the
displacement is always opposite to acceleration.
Q-25
Ans
What is a second’s pendulum ?
A pendulum, whose time period is 2 second is called a second’s pendulum.
Q-26
Ans
Why does a simple pendulum eventually stop ?
Due to frictional resistance between air and bob, the amplitude of oscillations of the pendulum gradually
decreases and it finally stops.
Q-27
What will be the time period of oscillation, if the length of a second pendulum is one third ?
Ans
or
or
s
Q-28
Ans
When a pendulum clock gains time, what adjustment should be made ?
When a pendulum clock gains time, it means it has gone fast i.e. it makes more vibrations per day than
required. This shows that the time period of oscillation has decreased. Therefore, to correct it, the length of
pendulum should be properly increased.
Q-29
Ans
During the oscillations of the bob of a simple pendulum ; what is the quantity that remains constant ?
Total energy of bob in simple pendulum remains constant at all instants.
Q-30
Ans
What is an epoch ? Name the unit in which it is measured ?
The initial phase of the oscillating particle is called epoch. It is expressed in radians.
Q-31
Ans
What do you understand by the term ‘phase’ of an oscillating particle ?
The phase of an oscillating particle at an instant is the argument of the sine or cosine function involved to
represent the given oscillatory motion. In an oscillatory motion
is the phase
of the oscillating particle at the time t. Here
is the initial phase or epoch of the oscillating particle.
8
Q-32
Ans
What forces keep the simple pendulum in simple harmonic motion ?
The component of weight due to gravitational force (i.e. mg sin θ), keeps the simple pendulum in simple
harmonic motion.
Q-33
On an average a human heart is found to beat 75 times in a minute. Calculate its beat frequency and
period.
Ans
The beat frequency of heart,
The time period,
Q-34
Ans
A particle starts oscillating from half the amplitude position. What is its initial phase ?
y = sin (ωt + 0). At
t = 0, y = /2
or
∴
or
.
Q-35
Ans
Can the motion of a satellite round a planet be taken as S.H.M. ? Explain.
The motion of a satellite round a planet is a periodic motion only and not oscillatory. Hence it is not a
S.H.M. because for that the motion has to be periodic and oscillatory.
Q-36
Ans
Can the oscillation of a mass suspended by a spring be taken as S.H.M ?
Yes, because it will be periodic as well as oscillatory.
Q-37
Ans
When will the motion of a simple pendulum be a simple harmonic ?
The motion of a simple pendulum will be S.H.M., if the angle θ through which the bob is displaced from
its equilibrium position is small, so that sin θ ≈ θ.
Q-38
Ans
When is the P.E. and K.E. of a harmonic oscillator maximum and what are these maximum values ?
The P.E. of a harmonic oscillator is maximum at extreme positions and K.E. of the harmonic oscillator is
maximum at the mean position. The maximum value of P.E. or K.E. is equal to total energy
Q-39
Ans
A S.H.M. has an amplitude and time period T. What is the time taken to travel from
Here displacement ( ) from the mean position for the motion
to
is =
=
to
=
/2 ?
i.e.
. Since the time is to be noted from the extreme position, hence using
We have,
or
or
or
Q-40
The total energy of simple harmonic motion is E. What will be the kinetic energy of the particle when
displacement is half of the amplitude ?
Ans
Here, total energy,
When
Q-41
, then Kinetic energy, K =
A body of mass 1 kg is executing S.H.M. Its displacement x (in cm) at time t (in second) is given by x = 6
sin (100 t + π/4) What is the maximum kinetic energy of the body ?
9
Ans
Comparing the given equation with,
we have,
Max. K.E. =
cm = 6 × 10−2 m ; ω = 100 rad/s.
J
SHORT ANSWER QUESTIONS
Q-01
Ans
What will be the effect on the time period, if the amplitude of a simple pendulum increases ?
We know that time period of a simple pendulum is independent of amplitude of vibration so long as its
motion is simple harmonic i.e. restoring force is linear. If amplitude of a simple pendulum increases, then
angle θ is large. Now sin θ ≠ θ. In this situation the motion of simple pendulum will be oscillatory but not
simple harmonic.
Q-02
Ans
Define force constant of a spring. Give its S.I. unit and dimensional formula.
Force constant of a spring is defined as the force required to produce unit extension or compression in the
spring i.e. k = F/y. The S.I. unit of k is Nm-1. Its dimensional formula = MLT-2/L = [M1 L0 T 2].
Q-03
Ans
Show that in S.H.M., the acceleration is directly proportional to its displacement at the given instant.
In S.H.M., the displacement of the particle at an instant is given by
y = r sin ωt.
velocity,
V
.
…(i)
Acceleration, A =
…(ii)
[From (i)] From (ii), we note that A ∝ y i.e. acceleration in S.H.M at an instant is directly proportional to
the displacement of the particle from the mean position at that instant.
Q-04
The length of a second’s pendulum on the surface of earth is 100 cm. What will be length of a second’s
pendulum on the surface of moon ?
Ans
For a simple pendulum, time period (T) is given by ; T = 2 π
Q-05
At what displacement (i) the P.E. of a simple harmonic oscillator is maximum (ii) the K.E. is maximum ?
Ans
The P.E. of a particle executing S.H.M. is given by ; U
. On moon surface, g’ = g/6, so
.
U is maximum when y = r = amplitude of vibration i.e. the particle is passing from the extreme position
and is minimum when y = 0 i.e. the particle is passing from the mean position :
The K.E. of a particle executing S.H.M. is given by K =
K is maximum when y = 0 i.e. the particle is passing form the mean position and K is minimum when y =
r i.e. the particle is passing from the extreme position.
Q-06
Ans
The maximum acceleration of a simple harmonic oscillator is A0 while the maximum velocity is
is the displacement amplitude ?
Here,
or
.
Thus
.
10
. What
Q-07
Ans
Q-08
Ans
The bob of a vibration simple pendulum is made of ice. How will the period of swing will change when
the ice starts melting ?
The period of swing of simple pendulum will remain unchanged till the location of centre of gravity of the
bob left after melting the ice remains at a fixed distance from the point of suspension. If the centre of
gravity of ice bob after melting is raised upwards, then the effective length of pendulum decreases and
hence the time period of swing decreases. If the centre of gravity of ice shifts on lower side, the time
period of swing increases.
Two identical springs have the same force constant of 147 Nm-1. What elongation will be produced in each
spring in each case shown in Fig – 10 (a).38 ? g = 9.8 ms 2.
Here k = 14 Nm 1 ; In Fig. 10(a).38 (a), the effective spring constant,
K = k + k = 2k = 2 × 147 = 294 Nm-1
∴
elongation in the spring,
In Fig.10(a).38 (b), the effective spring constant,
K=
∴
Total elongation in the spring,
Thus elongation in each spring 1/3 m
In Fig. 10(a).38 (c), the effective spring constant,
∴
Q-09
Ans
K=147 Nm
1
elongation in the spring,
The frequency of oscillations of a mass m suspended by a spring is . If the length of the spring is cut to
one-half, the same mass oscillates with frequency . Determine the value of / .
Let the full spring be the combination of two springs in series, each of force constant k. Then in case I, the
effective spring constant (K1) is given by K1 =
Frequency of oscillation,
…(i)
In case (ii) ; When the spring is cut to one-half, the effective spring constant
Frequency of oscillation,
Q-10
Ans
…(ii)
The periodic time of a mass suspended by a spring (force constant K) is T. If the spring is cut in three
equal pieces, what will be the force constant of each part ? If the same mass be suspended from one piece,
what will be the periodic time ?
Consider the spring the made of combination of three springs in series each of spring constant k. The
effective spring constant K is given by
∴
or
K=
or
k=3K
Time period of vibration of a body attached to the end of this spring,
T=
…(i)
When the spring is cut into three pieces, the spring constant = k
11
Time period of vibration of a body attached to the end of this spring, T1 = 2 π
From (i) and (ii),
Q-11
Ans
…(ii)
or
The bob of a simple pendulum is a ball full of water. If a fine hole is made in the bottom of the ball, what
will be its effect on the time period of the pendulum ?
When a hole is made in the bottom of a hollow sphere full of water, the water flows out slowly through
the hole. As the water flows out of the sphere, the centre of mass of the system will first move down and
then will come up. Due to this the equivalent length of pendulum L and hence time period T first
increases, reaches a maximum value and then decreases till it becomes equal to its initial value.
Q-12
Two exactly identical pendulums are oscillating with amplitudes 2 cm and 6 cm. Calculate the ratio of
their energies of oscillation.
Ans
Total energy of the bob of simple pendulum is given by E =
i.e.
E ∝ r2 ;
So
Q-13
Ans
A block rests on a horizontal table which is excuting S.H.M. in the horizontal plane with an amplitude A.
What will be the frequency of oscillation, the block will just start to slip ? Coefficient of friction = μ.
Restoring force on the block =
∴
Acceleration in the block = μ mg.
Frequency of oscillation,
Q-14
Ans
A mass m is dropped in a tunnel along the diameter of earth from a height h (<< R) above the surface of
earth. Find the time period of motion. Is the motion simple harmonic ?
When a ball is dropped from a height , it gains velocity due to gravity pull. The body will entre the
tunnel of earth with velocity,
; after a time,
. The body will go out of earth on the
other side through the same distance before coming back towards the earth. When the body is outside the
earth, the restoring force F ∝ ( 1/r2) and not ( r) so the motion does not remain S.H.M. but becomes
oscillatory. The period of oscillation of the body will be,
T=2π
Q-15
.
Ans
What would happen to the motion of an oscillating system if the sign of the force term in equation
F=−
is changed ?
When the sign in the force equation is changed, the force and hence acceleration will not be opposite to
displacement. Due to which the particle will not oscillate but will be accelerated in the direction of
displacement. As a result of it, the motion will become a linearly accelerated motion.
Q-16
Two simple harmonic motion are represented by the following equations :
and
Ans
, what is the ratio of their amplitudes ?
Here ;
…(i)
12
= 10 sin (3 π t + π/3)
…(ii)
Thus the amplitude ratio of motion (i) and (ii)
= 10/10 = 1/1 = 1 : 1.
Q-17
A simple pendulum performs S.H.M. about
speed of the pendulum at
?
= 0 with an amplitude
Ans
Speed of the pendulum,
Q-18
A pendulum clock is placed on the moon, where object weighs only one-sixth as much as on earth. How
many times the clock tick out in an actual time of 1 minute, the clock keeps good time on earth.
Ans
T’=2 π
As
T.
T = 1 sec, so T’ =
second.
Thus the clock will tick in one minute =
Q-19
Ans
and time period T. What is the
24.5 times.
A ball of radius r is made to oscillate in a bowl of radius R, find its time period of oscillation.
Here, equivalent length of simple pendulum = distance between centre of ball to centre of bowl i.e
∴
time period of oscillation of ball = 2 π
Q-20
The angular velocity and amplitude of simple pendulum is ω and r respective. At a displacement
the mean position, if its kinetic energy is T and potential energy is V, find the ratio of T to V.
Ans
Kinetic energy,
Potential energy,
from
so
Q-21
Which of the following functions of time represent (a) periodic and (b) non-periodic motion ? Give the
period for each of periodic motion. (ω is any positive constant)
(i) sin ωt + cos ωt
(ii) sin ω t + cos ω t + sin 4 ω t
(iii)
(iv) log ω t.
Ans
(i) sin ω t + cos ω t =
As
So the given function is a periodic one and its periodic is 2 π/ω.
(ii) sin ω t + cos 2 ω t + sin 4 ω t, it represents the periodic function with different angular frequency. Since
the period is the least time interval after which a function is repeated in its value. Here sin ω t, has a
period T =
; cos 2 ω t has a period
ω
and sin 4 ω t has a period
ω
ω
. The last two terms
repeat after any integral multiple of their period. Therefore, each term in the function repeats itself after
time interval T. That is why, the given function is a periodic function with a period T = 2 π/ ω.
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(iii) The value of function
, decreases with increasing time t and as t → ∞, it tends to zero. Therefore,
the function is non-periodic.
(iv) The value of function log ω increases with time t. As t → ∞, log ω t approaches to ∞. Therefore, the
value of this function never repeats. Hence, it represents non-periodic function.
Q-22
What is the length of a simple pendulum which ticks seconds ?
Ans
A simple pendulum which ticks seconds is a second pendulum, Its time period T=2 s. If is the length of
this pendulum, then
or
Q-23
Ans
Fig. 10(a).39 shows four different springs arrangements. If the mass m in each arrangement is displaced
from its equilibrium position and released, what is the released, what is the resulting frequency of
vibration in each case ? Neglect the mass of the spring. Figs. 10(a).41 (a) and (b) represent an arrangement
of springs in parallel and (c) and (d) represent springs in series ?
In the arrangements shown in Figs. 10(a).39 (a) and (b), the two springs are in parallel. The effective
spring constant of each of these arrangements is,
K = k1 + k2
Frequency of oscillation,
In the arrangements shown in Figs. 10(a).39 (c) and (d), the
two springs are in series. The effective spring constant K of
each of these arrangements is
or
Frequency of oscillation,
=
14
.
Q-24
Which of the following functions of time represent
(a) simple harmonic motion and (b) periodic but not simple harmonic motion ? Give the period for each
case. (i) sin ω t cos ω t
(ii) sin2 ω t.
Ans
(i) sin ω t
cos ω t =
ω
Thus, this function represents S.H.M., having period T = 2 π/ω and initial phase =
(ii) sin2 ω t =
ω
ω
The function is periodic, having a period T =
Q-25
Ans
π/4 rad.
but does not represent S.H.M.
The displacement of a harmonic oscillator is given by
sin ωt + β cos ωt. What is the amplitude of the
Oscillation ?
Given,
sin ωt + β cos ωt. Let = r cos θ and β = r sin θ. Then
cos θ sin ωt + r sin θ cos ωt = r sin (ωt + θ)
It is an equation of simple harmonic motion, whose amplitude is r.
From above,
or r =
.
Q-26
A simple harmonic motion of amplitude A, has a time period T. What will be the acceleration of the
oscillator, when its displacement is half of the amplitude
Ans
Acceleration,
Q-27
What will happen to the motion of simple pendulum if the amplitude of motion is large ? How does
period of oscillation change with amplitude ?
If amplitude of motion of simple pendulum is large, then θ is large. In that case, sin θ ≠ θ and restoring
torque will not be linear, hence motion will not remain simple harmonic but will become oscillatory. If θ0
is the angular amplitude of motion in this situation, then time period woule be given by
Ans
.
≈2π
Q-28
Amplitude of a wave is represented by A =
Ans
The resonance will occur when amplitude A becomes infinite. It will be so if a+b–c = 0. Which will be
possible if (i) a = c – b or (ii) b=0 and a = c.
Q-29
A particle moves according to the law
cos (π t/2). What is the distance covered by it in time interval
t = 0 to t = 3 second.
When t = 0,
cos (π × 0/2) = cos 0 = . It means the origin for time measurement lies at the extreme
position. When t = 3 s ;
cos 0 (π × 3/2) = cos (3 π/2)=0. It means the particle has covered a phase
3 π/2 from its starting point, while passing from mean position. So the total distance travelled is
=2
.
Ans
Q-30
Ans
Under what conditions, the resonance will occur.
The amplitude of an oscillating simple pendulum is 10 cm and its period is 4 second. What will be its
speed, one second after it passes its equilibrium position ?
Here, r = 10 cm, T = 4 s, t = 1 s ; = ?
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.
Q-31
Explain periodic motion and oscillatory motion with illustrations.
Q-32
Distinguish clearly Harmonic oscillations and Non- Harmonic oscillations
Q-33
What do you understand by S.H.M ? Explain its geometrical interpretation.
Q-34
Find the total energy of the particle executing S.H.M and show graphically the variation of P.E. and K.E.
with time in S.H.M. What is the frequency of these energies w.r.t. the frequency of the particle executing
S.H.M.
Q-35
What is a simple pendulum ? Find an expression for the time period and frequency of a simple pendulum.
Q-36
Explain the oscillations of a loaded spring and find the relations for the time period and frequency in case
of (i) horizontal spring (ii) vertical spring
Q-37
What is a spring factor ? Find its value in case of two springs connected in (i) series and (ii) parallel.
Q-38
Distinguish clearly with an illustration between free, forced and resonant oscillations.
Q-39
A periodic motion may not be an oscillatory motion. Explain.
Q-40
Derive an expression for the time period of a S.H.M. in terms of inertia factor and spring factor.
Q-41
A cylindrical piece of cork of base area A density ρ and height floats in a liquid of density ρL. The cork is
depressed slightly and then released. Show that cork oscillates up and down simple harmonically and
find its time period of oscillations.
LONG ANSWER QUESTIONS
Q-01
Explain periodic Functions with illustrations and show that combination of periodic functions is also
periodic one.
Q-02
Explain phase and phase difference, angular frequency, displacement in periodic motion with
illustrations.
Q-03
Explain displacement, velocity, acceleration and time period in S.H.Ms. Find the relation between them.
Q-04
Explain the relation in phase between displacement, velocity and acceleration in SHM, graphically as well
as theortically.
Q-05
Find the expressions for time period and frequency
(i) in the oscillations of a liquid in a U tube
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(ii) in the oscillations of a floating cylinder
Q-06
Explain (i) Undamped and damped oscillations and (ii) free, forced and resonant oscillations with
illustrations.
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