tyty
OSCILLATIONS
Q # 1. What do you know about the oscillatory/vibratory motion?
Ans. To and fro motion of a body about a fixed point is called the vibratory or oscillatory motion.
Q # 2. State the Hook’s law.
Ans. For an object attached to an elastic spring, the displacement of the object from mean position is directly
proportional to the applied force. If an object of mass attached to an elastic spring, then the displacement produced in the object by the application of force is described as: Where k is the spring constant.
Q # 3. Define following
• Simple Harmonic Motion
• Restoring Force
Ans. Restoring Force
The force that brings back the oscillatory object towards its mean position is called the restoring force. This restoring force is equal and opposite to the applied force within the elastic limit of the spring. where is the spring constant and is the displacement of the oscillatory object from its mean position. Simple Harmonic Motion
It is a type of vibratory motion in which the acceleration of the body is proportional to displacement
and is directed towards its mean position. Mathematically it is described as: ∝ Where and are the acceleration and displacement of the oscillatory object from mean position. The –ve sign indicates that acceleration of the object is directed towards the mean position.
Q # 4. Show that an object attached to a horizontal mass spring system executes simple harmonic motion. Ans. Consider a body of mass is attached to a spring of spring constant is executing the oscillatory motion. The restoring force acting on the object can be find out by using expression: --------------- (1) The acceleration produced in the body of mass due to restoring force can be calculated using second law of motion. --------------- (2) Comparing equation (1) and (2), we get: or ∝ This shows that acceleration of the oscillating object is
directly proportional to displacement and also acceleration is directed
2 Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera
towards the mean position. Thus the object attached to a mass
spring system executes simple harmonic motion.
Q # 5. Define the terms for an object executing simple
harmonic motion.
• Vibration
• Instantaneous Displacement
• Amplitude
• Time Period
• Frequency
• Angular Frequency
Ans. Vibration
A vibration means one completer round trip of the body in motion.
Instantaneous Displacement
When a body is vibrating, its displacement from the mean position changes with time. The value of its
distance from mean position at any instant of time is known as instantaneous displacement.
Amplitude
The maximum value of displacement of vibratory body from its mean position is known as amplitude.
Time Period
It is the time required to complete one vibration.
Frequency
The number of vibrations executed by a body in one second is called frequency. It is measured in Hertz. The frequency and the time period of a vibrating body is related by the relation: 1 Angular Frequency If T is the time period of a body executing simple harmonic motion, its angular frequency will be: 2 2 Angular frequency is basically a characteristic of circular motion. Here it has been introduce in to compare SHM with circular motion.
Q # 6. Derive the expression of displacement, velocity and acceleration for an object executing SHM by
considering uniform circular motion. Ans. Consider a point P moves in a circle of radius , with uniform angular frequency . it can be visualized that when the point P moves along the circle of radius , its projection (point N) execute simple harmonic motion on the diameter DE of the circle.
Thus the expression of displacement, velocity and acceleration for the object executing SHM can be
derived using the analogy between the uniform circular motion of point P and SHM of point N on the diameter
of the circle.
Displacement
It is the distance of projection of point N from the mean position O and any instant.
Chapter 7 (Physics 1st Year) Oscillations
3 Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera
As from figure, it can be seen that ∠∠ If is the amplitude and is the displacement at any instant. Then from triangle NOP, we have sin """" """" sin ------------- (1) Velocity If the point P is moving in a circle of radius with uniform angular velocity , then the tangential velocity of point P will be: #$ We want to find out the expression of velocity for point N, which is executing SHM. The velocity of N is actually the component of velocity #$ in the direction parallel to the diameter DE. Thus we can write the velocity # of point N as: ##$sin%90(#$cos # cos ------------ (2) As from equation
(1), we have: sin+ +, cos√1sin.1+/ +, /.+, /0+/ +, / .+, /0+/ +, Putting this value in equation
(2), we get: # 1.+, /0+/ +, 23
This is the expression of velocity of the object executing simple
harmonic motion.
Acceleration When the point P moves in a circle of radius , the it will have an acceleration
$ that will be directed towards the center of the circle.
We want to find out the expression of acceleration of point N, that is executing SHM at the diameter of
the circle. It can be seen from the figure that the acceleration of point N is the vertical component of acceleration $ along the diameter DE. $sin sin --------- (3) As from equation
(1): sin+ +, Therefore the equation
(3) will become: 4+ +,5 Comparing the case of displacement and acceleration, it
Chapter 7 (Physics 1st Year) Oscillations
4 Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera
can be seen that the direction of displacement and acceleration are opposite to each other. Considering the direction of as reference, the acceleration will be represented by: ------- (4) Q # 7. What do know about the term phase? Ans. The angle which specifies the displacement as well as the direction of motion of the point executing SHM is known as phase.
Explanation The displacement and velocity # of the body executing SHM can be determined by using equations: sin and # cos, respectively. These equations indicates the displacement and velocity of simple harmonic oscillator are determine by the angle . This angle is obtained when the SHM is related with circular motion. It is the angle which the rotating radius OP makes with axis at any instant. If the body starts its motion from mean position, its phase at this point would be 0. Similarly at the extreme position, its phase would be . Initial Phase Concept In general at 60, the rotating radius can make any angle 7 with axis as shown in the figure. In time t, the radius would rotate by 6. So now the radius OP will make an angle %687( with . The displacement at instant t would be given by: sin%687( Now the phase angle is 687. At 60, 7. So 7 is the initial phase. Special Case If initial phase 7 , the expression of displacement will become: sin%6890˚(cos6 This equation describes the SHM for the object which starts its motion from extreme position.
Q # 8. Derive the expressions for angular frequency, time period, displacement and velocity for the case of
horizontal mass spring system.
Ans. Consider a body of mass m is attached to a spring as shown in the figure. The acceleration of the object is
described by the formula: : ; ----------- (1) The acceleration of the object executing SHM is described as: ----------- (2) Comparing equation (1) and (2), we have: < Time Period
The time period of simple harmonic oscillator (SHO) is
described as: 2
Chapter 7 (Physics 1st Year) Oscillations
5 Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera
⟹2 .
2. Instantaneous Displacement The instantaneous displacement of SHO is described by the formula: sin sin%6(sin1< 62 Instantaneous Velocity The instantaneous velocity # of SHO is described by the formula: # 3 ⟹# .: ;% ( ⟹# .: ;41+/ +, /5 ----------- (3) The velocity of SHO become maximum at mean position 0. If # is the velocity of SHO at mean position, then: # < Thus, equation (3) will become: ⟹# #<1 Q # 9. Find out the expression of time period of simple pendulum.
Ans. Consider a simple pendulum which consists of bob of mass m is suspended from a rigid support by a string
of length L as shown in the figure.
We want to find out the expression of time period for this
simple pendulum.
When the bob is displaced from its mean position through a small angle and released, it starts to and fro motion about mean position. The weight mg of the object can be resolved in two rectangular components; >cos and >sin. It is clear from the figure that the component of the weight >cos will balance the tension in the string. The restoring force acting on the object will be: >sin when the is very small, then sin?. Thus, > ----------- (1) According to Newton’s 2nd law of motion, ----------- (2) Comparing (1) and (2), we get: > ⟹> ----------- (3)
Please Visit us at: http://www.phycity.com
Chapter 7 (Physics 1st Year) Oscillations
6 Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera
The relationship between the arc length and angular displacement is described by the formula: @ where @ is the length of string. ⟹= @ Putting value of in eqation (3), we get: =−A B ----------- (4) The acceleration of the object executing SHM is described as: = − ----------- (5) Comparing equation (4) and (5), we have: =.> @ Time Period
The time period of simple harmonic oscillator (SHO) is described as: =2 ⟹=2 .> @ =2<@ > This expression shows that the time period of simple pendulum depends only on the length of the
pendulum and the acceleration due to gravity. It is independent of mass.
Q # 10. Prove that the law of conservation of energy is satisfied for an object executing SHM.
Statement
The total energy of the object executing SHM remains constant.
Proof
Consider a vibrating mass spring system. When the mass m is pulled slowly, the spring is stretched by an amount against the elastic restoring force. Derivation of Expression for P.E.
According to Hook’s law: C= When displacement = 0 force = 0 When displacement = force = Average force =D:+, = The work done in displacing the mass m through is: E=F=G1 2H%(=1 2 This work appears as elastic potential energy of spring. Hence .J=1 2 This equation gives the maximum P.E at the extreme position. Thus
Post a Comment