PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
CHAPTER 1 (PART 3 SUBJECTIVE
Y
QUESTIONS)
EM
Question 1
D
A man with a wrist watch on his hands fall from the
CA
top of a tower. Does the watch give correct time?
A
Answer:
E
The wrist watch works on the principle of simple harmonic
N
oscillation. Notice that the time period of simple harmonic
LI
oscillation does not depend on the acceleration due to gravity it
N
means if a man falls from the tower his watch will not be
O
affected by it and will sure the correct time during the free fall.
R
A
ST
G
N
SI
RI
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
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Question 2
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A vibrating simple pendulum of period is
D
placed in a lift is accelerating downwards. What will
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be the effect on the time period?
A
Answer:
E
When the lift accelerates downwards, so net acceleration
N
decreases so time period increases.
LI
N
O
R
A
ST
G
N
SI
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
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Question 3
EM
A girl is sitting on a swing. Another girl
D
sits by her side. What will be the effect on the periodic
CA
time of the swing?
A
Answer:
E
There will be no change in the periodic time because the
N
periodic time is independent of mass but depends upon length of
LI
pendulum and acceleration due to gravity at a place, which are
N
not affected by sitting another girl on a swing.
O
R
A
ST
G
N
SI
RI
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
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Question 4
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Can a motion be oscillatory but not SHM?
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If your answer Yes, give an example and if not explain
CA
why?
A
Answer:
E
Yes, when a ball is dropped from a height on a perfectly elastic
N
plane surface, the motion of ball is oscillatory but not simple
LI
harmonic as restoring force F = mg = constant and not F ∝ − y.
N
O
R
A
ST
G
N
SI
RI
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
Question 5
If you double the period of a pendulum,
Y
what happens to its length?
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Answer:
D
We know that period of pendulum is given by
CA
√
A
Squaring both sides, we get
E
N( )
LI
N
If we double the period of the pendulum, let and length is .
O
R
Now,
A
ST
√
G
( )
N
SI
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It is clear that, if we double the period of a pendulum, then its length
becomes 4 times the original.
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
Question 6
Y
A 2kg body suspended from a spring of
EM
negligible mass is found to stretch the
D
spring by 10cm. If the body is pulled down and then
CA
released. What is the frequency of the oscillation?
Solution:
A
Given mass = m = 2kg
E
N
LI
Extension = x = 10 cm =
N
O
R
A
√ √
ST
G
N
SI
RI
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
Question 7
Y
A body undergoes SHM of amplitude 3cm
EM
and frequency 20Hz. What is the velocity
D
of the body at the maximum acceleration of the
CA
object?
Solution:
A
Amplitude = A = 3cm = 0.03m
E
N
Frequency = f = 20Hz
LI
N
Velocity = v = ?
O
Acceleration = a = ?
R
A
ST
G
N
SI
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
Question 8
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An object of mass 0.35kg oscillates in SHM with an
EM
amplitude of 140 mm and a frequency of 0.60 Hz.
Calculate or find:
D
CA
(a) Maximum Kinetic Energy of the object
(b) Maximum Potential Energy of the object
A
(c) Potential and Kinetic energy at the midway point
between the centre and the extremity of the motion.
E
N
Solution:
LI
Given m = 0.35 kg,
N
Amplitude = A = 140 mm = 0.140 m
O
Frequency = f = 0.60 Hz
R
A
(a) As we know that Kinetic Energy in SHM is given as
…….(1)
ST
At x = 0, Kinetic Energy is Maximum
G
N
SI
RI
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
b) The P.E. of SHM if given as
Y
EM
At , potential energy is maximum,
D
CA
A
c) At the midway point P.E. and K.E. is equal
i.e.
E
N
LI
N
O
R
A
ST
After Simplification, we get
G
N
Taking Square root on both sides,
SI
RI
√
From Equation (1), we get
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
Y
EM
D
CA
From Equation (2), we get
A
E
N
LI
N
O
R
A
ST
G
N
SI
RI
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
Question 9
Find the equation for the acceleration of
Y
SHM using differentiation.
EM
Acceleration of SHM
Again we know that Acceleration of a particle is given by
D
,
CA
where v is the velocity of particle executing motion.
A
In SHM velocity of particle is give by,
E
N
LI
differentiating this we get,
N
O
----
R
This equation gives acceleration of particle executing SHM and quantity
A
ω2 is called acceleration amplitude and the acceleration of oscillating
ST
particle varies betwen the limits ±ω2A.
G
Putting Equation in Equation
N
we get =>
SI
which shows that acceleration is proportional to the displacement but in
opposite direction.
RI
Thus from above equation we can see that when x is maximum
(+A or -A), the acceleration is also maximum (-ω2A or +ω2A)but is
directed in direction opposite to that of displacement.
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
Question 10
At a particular frequency, the velocity of the
Y
cap in m/s is given by the expression
EM
(a) Calculate the frequency of the sound emitted.
D
CA
(b) Calculate the amplitude of the vibration.
Solution:
A
(a)
E
N
LI
N
O
R
(b)
A
ST
G
N
SI
RI
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
Question 11
A set of spring all have initial length of 10cm.
Each …. Which of the oscillating systems has
Y
EM
the highest frequency.
Solution:
D
CA
For spring A,
A
So, √ √ √ ; √
E
N
For spring B,
LI
N
So, √ √ √ ; √
O
R
For spring C,
A
ST
So, √ √ √ ; √
G
For spring D,
N
SI
So, √ √ √ ; √
RI
So, spring C has the lowest time period and maximum
frequency, so answer C is correct
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
Question 12
Four different masses are hung from four
Y
springs with an unstretched length of 10cm,
causing the spring to stretch ……. Which of the
EM
oscillating system has the highest frequency?
D
Solution:
CA
We know that
A
E
N
Frequency of oscillation,
LI
N
√
O
R
A
√
ST
G
N
Less the value of , more is the frequency, which is possible is
SI
A. So, Mass A is the answer.
RI
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
Question 13
A mass oscillating up and down on a spring; the
Y
motion is illustrated at right.
EM
D
CA
A
E
1. At which time or times shown in the acceleration zero?
N
2. At which time or times shown is the kinetic energy a
LI
maximum?
N
3. At which time or times shown is the potential energy is
O
maximum?
R
Answer:
A
ST
1. Acceleration is zero at A, C , E
2. Maximum K.E is at A, C, E
G
3. Maximum P.E. is at B
N
SI
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PHYSICS EXERCSIES CHAPTER 1 EXERCISE
SIMPLE HARMONIC MOTION
Question 14
A pendulum is pulled to the side and released.
Y
The mass swings to the right as shown. The
EM
diagram shows positions for the half of a
complete oscillation.
D
CA
A
E
N
LI
N
1. At which point or points is the speed the highest?
O
2. At which point or points is the acceleration the
greatest?
R
A
3. At which point or points is the restoring force the
ST
greatest?
Answers:
G
N
1. Highest speed at point C
SI
2. Greatest acceleration at point A and E
3. Greatest restoring force at point A and E
RI
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SI
N
G
SOLUTION:
ST
A
QUESTION # 16
R
O
N
LI
N
E
A
CA
D
EM
Y
RI
SI
N
G
ST
A
R
O
N
LI
N
E
A
CA
D
EM
Y
(b) When student releases the mass,
(i) Force acting on mass, will be elastic force
which is proportional to elongation in
spring,
Y
𝐹𝑠𝑝𝑟𝑖𝑛𝑔 𝛼 (𝑦 − 𝑦𝑜 )
EM
𝐹𝑠𝑝𝑟𝑖𝑛𝑔 = −𝑘 (𝑦 − 𝑦𝑜 )
D
(𝑦 − 𝑦𝑜 ) = displacement of mass
CA
(ii) To find acceleration,
𝐹𝑦 = 𝑚 × 𝑎𝑦
A
𝑘 (𝑦 − 𝑦𝑜 ) = 𝑚 × 𝑎𝑦
E
N
𝒌 (𝑦 − 𝑦𝑜 )
𝑎𝑦 =
LI
𝑚
N
𝟑𝟓 × (60𝑚𝑚)
𝑎𝑦 =
O
0.5
R
𝑎𝑦 = 4.2 𝑚/𝑠𝑒𝑐 2
A
ST
Calculation of Spring Constant:
G
T ′ = 𝑘𝑦 => 2.1 = 𝑘 × (60𝑚𝑚)
N
2.1
SI
=> 𝒌 = = 𝟑𝟓
0.06
RI
(iii) state the initial amplitude of the Oscillation
𝑉 = 𝜔 √ 𝐴2 − 𝑦 2
Y
Where,
EM
𝑉 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦;
D
𝜔 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
CA
𝐴 = 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒
A
𝑦 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
E
N
Now, making A as subject,
LI
N
𝑉 2
𝐴 = √( ) + 𝑦 2
O
𝜔
R
A
As, velocity will be zero, at max displacement from
ST
mean position, so
G
𝐴 = √𝑦 2
N
SI
𝐴 = √(60 𝑚𝑚)2
RI
𝐴 = 60 𝑚𝑚
Question # 17
The piston moves up and down between points A and B as shown in the diagram
which are separated 0.10 m apart.
The frequency and oscillation of that piston's motion is
Y
EM
Mass of piston
Since the piston is moving between points A and B separated by 0.10 m, its mean
point will be at its centre point i.e. 0.05 m from any of the one end points.
D
CA
Thus, the amplitude of oscillation of this piston will be A = 0.05 m
The angular frequency of the piston will be given by;
A
E
Substituting all thw values; N
LI
N
The general equation of displacement for a particle performing simple harmonic
O
motion is given by;
R
A
Here, y is the displacement of the particle / object
ST
A is the amplitude of the object performing SHM
is its angular frequency
G
is its initial phase
N
Considering the piston to be staring from the mean position, we can say that
SI
its initial phase is zero.
Hence, subsituting all these values in the general equaition,
RI
The magnitude of maximum acceleration of any particle / body performing SHM is
given by;
Now, substituting all the values in the above equation;
Y
EM
Hence, the maximum acceleratio of this piston will be 1917.2 m/s2
D
The force applied due to this maximum acceleration will also be maximum.
CA
Substituting the values;
A
E
N
LI
Hence, the maximum force on this piston will be 9465.216 N
Maximum kinetic energy for a particle / object performing SHM is given by;
N
O
Here, k is the force constant that is given by
R
A
ST
Substituting all the values,
G
N
SI
Hence, maximum kinetic energy of this piston will be 473.261 J.
RI
