Q1. A mass attached to a spring is free to oscillate , with angular velocity ω , in a frictionless horizontal plane. The mass is displaced from it's equilibrium position by a distance x0 towards the center by pushing it with velocity v0 at time t=0. Find the amplitude of resulting oscillations in terms ofω,x0 and v0 .
Ans. A=√[x02+(v02/ω2)]
Q2. A uniform cylinder of length l and mass m having crosssectional area Ais suspended , with length vertical , from a fixed point by a massless spring , such that it is half submearged in a liquid of density σ at equilibrium position.. When the cylinder is given a small downward push and released , it starts oscillating verticallywith small amplitude . Calculate the frequency of oscillations of cylinder.
(IIT 1990)
Ans. f=[(k+(σAg)/m]1/2
Q3. What should be the percentage change of length of pendulum in order that clock have same time period when moved from place where g=9.8 m/s2 to another where g=9.81 m/s2.
Ans. .102%
Q4. A 4 kg particle is moving along x axis under the action of the force F=-(π2/16)x N
when t=2 s the particle passes through origin.If x0 is the amplitude of oscillating particle find the equation of elongation.
Ans. x=x0 cos( πt/8 + π/4)
Q5. Two blocks of masses m1 and m2 are connected by a spring and these masses are free to oscillate along the axis of the spring. Find the angular frequency of oscillation.
Ans.
ω=√[k(m1+m2)/m1m2]
Wednesday, 9 April 2008
Tuesday, 8 April 2008
Objective Question for SHM
Q 1. Total energy of mass spring system in harmonic motion is E=1/2(mω2A2). Consider another system executing SHM with same amplitude having value of spring constant as half the previous one and mass twice as that of previous one. The energy of second oscillator will be
(a) E
(b) 2E
(c) √ 2E
(d) E/2
Q 2. A particle is executing linear SHM of amplitude A. What fraction of total energy is potential when the displacement is 1/4 times amplitude.
(a) 3/2
(b) 1/16
(c) 1/4
(d) 1/2√ 2
Q 3. Fig below shows two spring mass systems. All the springs are identical having spring constant k and are of negligible mass. If m is the mass of block attached to the spring then the ratio of time period of oscillations of both systems is
(a) 1:2√ 2
(b) 1:1/2√ 2
(c) 1:√ 2
(d) √ 2:1
Q 4. Fig below shows two equal masses of mass m joined by a rope passing over a light pully. First mass is attached to a spring and another end of spring is attached to a rigid support. Neglacting frictional forces total energy of the system when spring is extended by a distance x is
(a) mv2+1/2(Kx2)+mgx
(b) mv2-1/2(Kx2)+mgx
(c) mv2-1/2(Kx2)-mgx
(d) mv2+1/2(Kx2)-mgx
where v = dx/dt , the velocity of mass
Q 5. A spring of force constant k is cut into two pieces such that one piece is four times the length of the other. the longer piece will have force constant equal to
(a) 4k/5
(b) 5k/4
(c) 3k/2
(d) 4k
Q 6. In the system shown below frequency of oscillation when mass is displaced slightely is
(a) f=1/2π(k1k2/(k1+k2)m)1/2
(b) f=1/2π((k1+k2)/m)1/2
(c) f=1/2π(m/(k1k2))1/2
(d) f=1/2π((k1+k2)/(k1k2)m)1/2
Q 7. A simple pendulum is displaced from it's mean position o to a position A such that hight of A above O is 0.05m. It is then released it's velocity when it passes mean position is
(a) .1m/s
(b) 5.0m/s
(c) 1m/s
(d) 1.5m/s
Q 8. A particle is executing SHM at mid point of mean position and extreme position . What is it's KE in terms of total energy E.
(a) E/2
(b) 4E/3
(c) √ 2E
(d) 3E/4
Q 9. A solid cylinder of radius r and mass m is connected to a spring of spring constant k and it slips on a frictionless surface without rolling with angular frequency
(a) √(k/mr)
(b) √(kr/m)
(c) √(k/m)
(d) √(2k/m)
Solutions
(a) E
(b) 2E
(c) √ 2E
(d) E/2
Q 2. A particle is executing linear SHM of amplitude A. What fraction of total energy is potential when the displacement is 1/4 times amplitude.
(a) 3/2
(b) 1/16
(c) 1/4
(d) 1/2√ 2
Q 3. Fig below shows two spring mass systems. All the springs are identical having spring constant k and are of negligible mass. If m is the mass of block attached to the spring then the ratio of time period of oscillations of both systems is
(a) 1:2√ 2
(b) 1:1/2√ 2
(c) 1:√ 2
(d) √ 2:1
Q 4. Fig below shows two equal masses of mass m joined by a rope passing over a light pully. First mass is attached to a spring and another end of spring is attached to a rigid support. Neglacting frictional forces total energy of the system when spring is extended by a distance x is
(a) mv2+1/2(Kx2)+mgx
(b) mv2-1/2(Kx2)+mgx
(c) mv2-1/2(Kx2)-mgx
(d) mv2+1/2(Kx2)-mgx
where v = dx/dt , the velocity of mass
Q 5. A spring of force constant k is cut into two pieces such that one piece is four times the length of the other. the longer piece will have force constant equal to
(a) 4k/5
(b) 5k/4
(c) 3k/2
(d) 4k
Q 6. In the system shown below frequency of oscillation when mass is displaced slightely is
(a) f=1/2π(k1k2/(k1+k2)m)1/2
(b) f=1/2π((k1+k2)/m)1/2
(c) f=1/2π(m/(k1k2))1/2
(d) f=1/2π((k1+k2)/(k1k2)m)1/2
Q 7. A simple pendulum is displaced from it's mean position o to a position A such that hight of A above O is 0.05m. It is then released it's velocity when it passes mean position is
(a) .1m/s
(b) 5.0m/s
(c) 1m/s
(d) 1.5m/s
Q 8. A particle is executing SHM at mid point of mean position and extreme position . What is it's KE in terms of total energy E.
(a) E/2
(b) 4E/3
(c) √ 2E
(d) 3E/4
Q 9. A solid cylinder of radius r and mass m is connected to a spring of spring constant k and it slips on a frictionless surface without rolling with angular frequency
(a) √(k/mr)
(b) √(kr/m)
(c) √(k/m)
(d) √(2k/m)
Solutions
Saturday, 5 April 2008
Friday, 4 April 2008
Conceptual Question of SHM
1.To execute SHM system must have
a. Elasticity
b. Moment of Inertia
c. Inertia
d. all the above
2. Angular frequency of system executing SHM depends on
a. mass
b. total energy
c.force constant
d. Amplitude
3.A particle of mass m is attached to a massless string of lenght L and is oscillating in vertical plane with one end of string fixed to rigid support.Tension in the string at a certain instant is T=kmg.Then
a. K can never be equal to 1
b. K can never be greater than 1
c. K can never be greater than 3
d K can never be less than 1
4.The bob A of a simple pendulum is released when the string makes an angle 45 with the verical.Its hit another bob B of the same mateial and same mass kept at rest on the table.If the collsion is elastic
a. B moves first and A follows it with half of its intial velocity
b.A comes to rest and B moves with the velocity of A
c Both A and B moves with same velocity of A
d Both A and B comes to rest at B
5.For a particle executing SHM
a.Acceleration is proportional to the displacement in the direction of the motion
b.Acceleration is proportional to the displacement but in opposite direction of the motion
c. Total energy of particle remains constant
d KE and PE of particle remains constant
6. which one of the following statement is true
a. Maximum value of velocity in SHM is A2ω
b.In SHM velocity of the particle is maximum when displacment is maximum
c.Velocity of the particle is zero in SHM when displacement attains its maximum on either side
d.Velocity in SHM vary periodically with time
7. which one of the following statement is true
a. Amplitude and intial displacement of particle in SHM are always equal
b.Amplitude and intial displacement of particle in SHM are never equal
c. Amplitnude of a particle in SHM can be equal to its initial displacement
d. Amplitnude of a particle in SHM can be greater to its initial displacement
8.The amplitutde and phase of a particle executing SHM depends on
a.The displacemnt of particle at t=0
b.The velocity of particle at t=0
c Both Velocity and displacement at t=0
d Neither velocity and displacemnt at t=0
Solutions
a. Elasticity
b. Moment of Inertia
c. Inertia
d. all the above
2. Angular frequency of system executing SHM depends on
a. mass
b. total energy
c.force constant
d. Amplitude
3.A particle of mass m is attached to a massless string of lenght L and is oscillating in vertical plane with one end of string fixed to rigid support.Tension in the string at a certain instant is T=kmg.Then
a. K can never be equal to 1
b. K can never be greater than 1
c. K can never be greater than 3
d K can never be less than 1
4.The bob A of a simple pendulum is released when the string makes an angle 45 with the verical.Its hit another bob B of the same mateial and same mass kept at rest on the table.If the collsion is elastic
a. B moves first and A follows it with half of its intial velocity
b.A comes to rest and B moves with the velocity of A
c Both A and B moves with same velocity of A
d Both A and B comes to rest at B
5.For a particle executing SHM
a.Acceleration is proportional to the displacement in the direction of the motion
b.Acceleration is proportional to the displacement but in opposite direction of the motion
c. Total energy of particle remains constant
d KE and PE of particle remains constant
6. which one of the following statement is true
a. Maximum value of velocity in SHM is A2ω
b.In SHM velocity of the particle is maximum when displacment is maximum
c.Velocity of the particle is zero in SHM when displacement attains its maximum on either side
d.Velocity in SHM vary periodically with time
7. which one of the following statement is true
a. Amplitude and intial displacement of particle in SHM are always equal
b.Amplitude and intial displacement of particle in SHM are never equal
c. Amplitnude of a particle in SHM can be equal to its initial displacement
d. Amplitnude of a particle in SHM can be greater to its initial displacement
8.The amplitutde and phase of a particle executing SHM depends on
a.The displacemnt of particle at t=0
b.The velocity of particle at t=0
c Both Velocity and displacement at t=0
d Neither velocity and displacemnt at t=0
Solutions
Thursday, 3 April 2008
Oscillations
PART 2
(1) Some system Executing SHM
a)Oscillations of a Spring mass system
-In this case particle of mass m oscillates under the influence of hooke's law restoring force given by F=-Kx where K is the spring constant
Angular Frequency ω=√(K/m)
Time period T=2π√(m/K)
And frequency is =(1/2π)√(K/m)
Time period of both horizontal ans vertical oscillation are same but spring constant have diffrent value for horizontal and vertical motion
b) Simple pendulum
-Motion of simple pendulum oscillating through small angles is a case of SHM with angular frequency given by
ω=√(g/L)
and Timeperiod
T=2π√(L/g)
Where L is the length of the string.
-Here we notice that period of oscillation is independent of the mass m of the pendulum
c) Compound Pendulum
- Compound pendulum is a rigid body of any shape,capable of oscillating about the horizontal axis passing through it.
-Such a pendulum swinging with small angle executes SHM with the timeperiod
T=2π√(I/mgL)
Where I =Moment of inertia of pendulum about the axis of suspension
L is the lenght of the pendulum
(2) Damped Oscillation
-SHM which continues indefinitely without the loss of the amplitude are called free oscillation or undamped and it is not a real case
- In real physical systems energy of the oscillator gradually decreases with time and oscillator will eventually come to rest.This happens because in acutal physical systems,friction(or damping ) is always present
-The reduction in amplitude or energy of the oscilaltor is called damping and oscillation are call damped
(3) Forced Oscillations and Resonance.
- Oscillations of a system under the influence of an external periodic force are called forced oscillations
- If frequency of externally applied driving force is equal to the natural frequency of the oscillator resonance is said to occur
(1) Some system Executing SHM
a)Oscillations of a Spring mass system
-In this case particle of mass m oscillates under the influence of hooke's law restoring force given by F=-Kx where K is the spring constant
Angular Frequency ω=√(K/m)
Time period T=2π√(m/K)
And frequency is =(1/2π)√(K/m)
Time period of both horizontal ans vertical oscillation are same but spring constant have diffrent value for horizontal and vertical motion
b) Simple pendulum
-Motion of simple pendulum oscillating through small angles is a case of SHM with angular frequency given by
ω=√(g/L)
and Timeperiod
T=2π√(L/g)
Where L is the length of the string.
-Here we notice that period of oscillation is independent of the mass m of the pendulum
c) Compound Pendulum
- Compound pendulum is a rigid body of any shape,capable of oscillating about the horizontal axis passing through it.
-Such a pendulum swinging with small angle executes SHM with the timeperiod
T=2π√(I/mgL)
Where I =Moment of inertia of pendulum about the axis of suspension
L is the lenght of the pendulum
(2) Damped Oscillation
-SHM which continues indefinitely without the loss of the amplitude are called free oscillation or undamped and it is not a real case
- In real physical systems energy of the oscillator gradually decreases with time and oscillator will eventually come to rest.This happens because in acutal physical systems,friction(or damping ) is always present
-The reduction in amplitude or energy of the oscilaltor is called damping and oscillation are call damped
(3) Forced Oscillations and Resonance.
- Oscillations of a system under the influence of an external periodic force are called forced oscillations
- If frequency of externally applied driving force is equal to the natural frequency of the oscillator resonance is said to occur
Wednesday, 2 April 2008
Oscillations
PART I
-If a particle moves such that it retraces its path regularly after regular interval of time,its motion is said to be periodic Ex-Motion of earth around Sun
-If a body in periodic motion moves back and forth over the same path then the motion is said to be oscillatory motion
-Simple harmonic motion is simplest form of oscillatory motion
-SHM is a kind of motion in which the restoring force is propotional to the displacement from the mean position and opposes its increase.Mathematically restoring force is
F=-Kx
Where K=Force constant
x=displacement of the system from its mean or equilibrium position
Diffrential Equation of SHM is
d2x/dt2 + ω2x=0
Solutions of this equation can both be sine or cosine functions .We conveniently choose
x=Acos(ωt+φ) where A,ω and φ all are constants
-Quantity A is known as amplitude of SHM which is the magnitude of maximum value of displacement on either sides from the equilibrium position
-Time period (T) of SHM the time during which oscillation repeats itself i.e, repeats its one cycle of motion and it is given by
T=2π/ω where ω is the angular frequency
-Frequency of the SHM is the number of the complete oscillation per unit time i.e, frequency is reciprocal of the time period
f=1/T
Thus angular frequncy
ω=2πf
-Velocity of a system executing SHM as a function of time is
v=-ωAsin(ωt+φ)
-Acceleration of particle executing SHM is
a=-ω2Acos(ωt+φ)
So a=-ω2x
This shows that acceleration is proportional to the displacement but in opposite direction
-At any time t KE of system in SHM is
KE=(1/2)mv2
=(1/2)mω2A2sin2(ωt+φ)
which is a function varying periodically in time
-PE of system in SHM at any time t is
PE=(1/2)Kx2
=(1/2)mω2A2cos2(ωt+φ)
-Total Energy in SHM
E=KE+PE
=(1/2)mω2A2
and it remain constant in absense of dissapative forces like frictional forces
-If a particle moves such that it retraces its path regularly after regular interval of time,its motion is said to be periodic Ex-Motion of earth around Sun
-If a body in periodic motion moves back and forth over the same path then the motion is said to be oscillatory motion
-Simple harmonic motion is simplest form of oscillatory motion
-SHM is a kind of motion in which the restoring force is propotional to the displacement from the mean position and opposes its increase.Mathematically restoring force is
F=-Kx
Where K=Force constant
x=displacement of the system from its mean or equilibrium position
Diffrential Equation of SHM is
d2x/dt2 + ω2x=0
Solutions of this equation can both be sine or cosine functions .We conveniently choose
x=Acos(ωt+φ) where A,ω and φ all are constants
-Quantity A is known as amplitude of SHM which is the magnitude of maximum value of displacement on either sides from the equilibrium position
-Time period (T) of SHM the time during which oscillation repeats itself i.e, repeats its one cycle of motion and it is given by
T=2π/ω where ω is the angular frequency
-Frequency of the SHM is the number of the complete oscillation per unit time i.e, frequency is reciprocal of the time period
f=1/T
Thus angular frequncy
ω=2πf
-Velocity of a system executing SHM as a function of time is
v=-ωAsin(ωt+φ)
-Acceleration of particle executing SHM is
a=-ω2Acos(ωt+φ)
So a=-ω2x
This shows that acceleration is proportional to the displacement but in opposite direction
-At any time t KE of system in SHM is
KE=(1/2)mv2
=(1/2)mω2A2sin2(ωt+φ)
which is a function varying periodically in time
-PE of system in SHM at any time t is
PE=(1/2)Kx2
=(1/2)mω2A2cos2(ωt+φ)
-Total Energy in SHM
E=KE+PE
=(1/2)mω2A2
and it remain constant in absense of dissapative forces like frictional forces
Tuesday, 1 April 2008
Solutions for Kinematics objective
1 a,c,d
Hint:Eliminating t from both the equation,you got trajotory and diffrentiating gives velocity y
x=2t y=2t2 So eliminating gives y=x2/2
2.d
3.d
4.c
5.b,c
6.a,b
7.b
8.b
9.a
10.d
Hint:Eliminating t from both the equation,you got trajotory and diffrentiating gives velocity y
x=2t y=2t2 So eliminating gives y=x2/2
2.d
3.d
4.c
5.b,c
6.a,b
7.b
8.b
9.a
10.d
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