All Exams >
Grade 12 >
Physics for Grade 12 >
All Questions
All questions of Oscillations for Grade 12 Exam
Find the amplitude of the S.H.M whose displacement y in cm is given by equation y= 3sin 157t +4cos157t where t is time in seconds.- a)20Hz
- b)25Hz
- c)50Hz
- d)40Hz
Correct answer is option 'B'. Can you explain this answer?
Find the amplitude of the S.H.M whose displacement y in cm is given by equation y= 3sin 157t +4cos157t where t is time in seconds.
a)
20Hz
b)
25Hz
c)
50Hz
d)
40Hz
|
Pioneer Academy answered |
When the displacement of a SHM is:
y=a sin wt+ b cos wt
y=a sin wt+ b cos wt
- Amplitude of the SHM will be:
A=√a2+b2
Here, a = 3, b = 4
Amplitude, A= √(32+42) = 5 cm
Amplitude, A= √(32+42) = 5 cm
Hence option B is correct.
Two identical spring, each of stiffness k are welded to each other at point P. The other two ends are fixed to the edge of a smooth horizontal tube as shown. A particle of mass m is welded at P. The entire system is horizontal. The period of oscillation of the particle in the direction of x is 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Two identical spring, each of stiffness k are welded to each other at point P. The other two ends are fixed to the edge of a smooth horizontal tube as shown. A particle of mass m is welded at P. The entire system is horizontal. The period of oscillation of the particle in the direction of x is
a)
b)
c)
d)
|
Mohit Rajpoot answered |
Let us assume a force dF is applied at P in positive x - direction. This will stretch each spring by dl inducing a spring force dFs in each spring.
Let the static deformation of the system is dx (along the x-direction). The particle is in equilibrium. So,
Using Pythagoras theorem,
Here y is constant.
A second pendulum is mounted in a space shuttle. Its period of oscillations will decrease when rocket is- a)moving in geostationary orbit
- b)ascending up with uniform acceleration
- c)descending down with uniform acceleration
- d)moving up with uniform velocity
Correct answer is option 'B'. Can you explain this answer?
A second pendulum is mounted in a space shuttle. Its period of oscillations will decrease when rocket is
a)
moving in geostationary orbit
b)
ascending up with uniform acceleration
c)
descending down with uniform acceleration
d)
moving up with uniform velocity
|
Top Rankers answered |
- Time Period, T = 2π √(l/g')where,
l = Length of seconds pendulum
g’ = Apparent Gravity - For the period of oscillations of Seconds Pendulum to decrease, the Apparent gravity (g’) has to increase because:
- Hence, Time Period of oscillations of Seconds Pendulum will decrease when the rocket is ascending up with uniform acceleration.
What is the maximum Kinetic energy and minimum potential energy of a harmonic oscillator with amplitude 0.03m, force constant 4×105 N/m and total mechanical energy of 230 J.- a)200J, 40J
- b)180J,50J
- c)190J, 60J
- d)210J,60J
Correct answer is option 'B'. Can you explain this answer?
What is the maximum Kinetic energy and minimum potential energy of a harmonic oscillator with amplitude 0.03m, force constant 4×105 N/m and total mechanical energy of 230 J.
a)
200J, 40J
b)
180J,50J
c)
190J, 60J
d)
210J,60J
|
Infinity Academy answered |
K. Σ=1/2 K(A2-x2)
Max of mean position,
K. Σ=1/2 KA2
=1/2 x4x105x(3x10-2)2
=180J
T.M. Σ=180+P.Σ
230=180+P.Σ
P.Σ=230-180
P.Σ=50J
Max of mean position,
K. Σ=1/2 KA2
=1/2 x4x105x(3x10-2)2
=180J
T.M. Σ=180+P.Σ
230=180+P.Σ
P.Σ=230-180
P.Σ=50J
If the length of a simple pendulum is increases by 1percent, then the new time-period
a) Decreases by 0.5 percent
b) Increases by 0.5 percent
c) Increases by 1 percent
d) Increases by 2.0 percent
Correct answer is option 'B'. Can you explain this answer?
|
Krishna Iyer answered |
Time period, T=2π√l/g
=>T∝√l
=>ΔT/T=1/2 Δl/l
Δl/l=1%
Therefore, ΔT/T=1/2x1%=0.5%
=>T∝√l
=>ΔT/T=1/2 Δl/l
Δl/l=1%
Therefore, ΔT/T=1/2x1%=0.5%
If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity isa)0.15 m/sb)0.1 m/sc)0.16 m/sd)0.8 m/sCorrect answer is option 'A'. Can you explain this answer?
|
Neha Joshi answered |
We know that in a simple harmonic motion the maximum velocity,
Vmax = A⍵
Here A = 50 mm
Vmax = A⍵
Here A = 50 mm
And ⍵ = 2π / T
= 2π / 2
= π
= 2π / 2
= π
Hence Vmax = 50 x 10-3.π
= 0.15 m/s
= 0.15 m/s
The displacement of a particle along the x-axis is given by x = a sin2ωt. The motion of the particle corresponds to: [2010]
- a) non simple harmonic motion
- b)simple harmonic motion of frequency 3ω /2π
- c)simple harmonic motion of frequency ω /π
- d)simple harmonic motion of frequency ω /2π
Correct answer is option 'A'. Can you explain this answer?
The displacement of a particle along the x-axis is given by x = a sin2ωt. The motion of the particle corresponds to: [2010]
a)
non simple harmonic motion
b)
simple harmonic motion of frequency 3ω /2π
c)
simple harmonic motion of frequency ω /π
d)
simple harmonic motion of frequency ω /2π
|
|
Athira Krishnana answered |
The work done by the string of a simple pendulum in S.H.M is- a)zero
- b)equal to total energy of system
- c)mg
- d)equal to kinetic energy of the system
Correct answer is option 'A'. Can you explain this answer?
The work done by the string of a simple pendulum in S.H.M is
a)
zero
b)
equal to total energy of system
c)
mg
d)
equal to kinetic energy of the system
|
Dilip Chaurasiya answered |
Zero becz. Tension and disp. Is at 90 degree.
What will be the phase difference between bigger pendulum (with time period 5T/4 )and smaller pendulum (with time period T) after one oscillation of bigger pendulum?- a)π/4
- b)π/2
- c)π/3
- d)π
Correct answer is option 'B'. Can you explain this answer?
What will be the phase difference between bigger pendulum (with time period 5T/4 )and smaller pendulum (with time period T) after one oscillation of bigger pendulum?
a)
π/4
b)
π/2
c)
π/3
d)
π
|
Preeti Iyer answered |
By the time bigger pendulum completes one vibration, the smaller pendulum would have completed 5/4 vibrations. That is smaller pendulum will be ahead by 1/4 vibration in phase. 1/4 vibration means λ/4 path or π/2 radians.
The dimensions and unit of phase constant Φ is- a)dinensionless,rad
- b)[T]-1 ,s-1
- c)dinensionless, no units
- d)[T] , s
Correct answer is option 'A'. Can you explain this answer?
The dimensions and unit of phase constant Φ is
a)
dinensionless,rad
b)
[T]-1 ,s-1
c)
dinensionless, no units
d)
[T] , s
|
Suresh Reddy answered |
Mathematical constants don't have any dimensions hence, it is dimensionless and phase constant is in angle therefore, the unit is radian.
A frequency of 1Hz corresponds to:- a)2 vibrations per second
- b)1 vibration per second
- c)10 vibrations per second
- d)a time period of ½ second
Correct answer is option 'B'. Can you explain this answer?
A frequency of 1Hz corresponds to:
a)
2 vibrations per second
b)
1 vibration per second
c)
10 vibrations per second
d)
a time period of ½ second
|
Alok Mehta answered |
Frequency used to be measured in cycles per second, but now we use the unit of frequency - the Hertz (abbreviated Hz). One Hertz (1Hz) is equal to one vibration per second. So the weight above is bouncing with a frequency of about 1Hz. The sound wave corresponding to Middle C on a piano is around 256Hz.
What determines the natural frequency of a body?- a)Position of the body with respect to force applied
- b)Mass and speed of the body
- c)Oscillations in the body
- d)Elastic properties and dimensions of the body
Correct answer is option 'D'. Can you explain this answer?
What determines the natural frequency of a body?
a)
Position of the body with respect to force applied
b)
Mass and speed of the body
c)
Oscillations in the body
d)
Elastic properties and dimensions of the body
|
Lavanya Menon answered |
Natural frequency is the frequency at which a body tends to oscillate in the absence of any driving or damping force.
Free vibrations of any elastic body are called natural vibration and happen at a frequency called natural frequency. Natural vibrations are different from forced vibration which happen at frequency of applied force .
Free vibrations of any elastic body are called natural vibration and happen at a frequency called natural frequency. Natural vibrations are different from forced vibration which happen at frequency of applied force .
If the period of oscillation of mass M suspended from a spring is one second, then the period of mass 4 M will be- a)1/ 4s
- b)4s
- c)1/ 2s
- d)2s
Correct answer is option 'D'. Can you explain this answer?
If the period of oscillation of mass M suspended from a spring is one second, then the period of mass 4 M will be
a)
1/ 4s
b)
4s
c)
1/ 2s
d)
2s
|
Geetika Shah answered |
T=2*3.14*√m/√k=1 s (given)
T’=2*3.14*√4m/√k=2*3.14*2√m/√k=2*1s
Therefore T’=2s
If 
is the natural frequency of the system and 
is the frequency of the external force that acts on an oscillating system then at resonance- a)ωd ≥ ω
- b)ωd = ω
- c)ωd ≤ ω
- d)ωd ≠ ω
Correct answer is option 'B'. Can you explain this answer?
If is the natural frequency of the system and is the frequency of the external force that acts on an oscillating system then at resonance
is the natural frequency of the system and is the frequency of the external force that acts on an oscillating system then at resonancea)
ωd ≥ ω
b)
ωd = ω
c)
ωd ≤ ω
d)
ωd ≠ ω
|
Om Desai answered |
When the natural frequency and external frequency of an object are equal the phenomenon is called resonance.
A mass m = 2.0 kg is attached to a spring having a force constant k = 290 N/m as in the figure. The mass is displaced from its equilibrium position and released. Its frequency of oscillation (in Hz) is approximately:
- a)12
- b)0.50
- c)1.9
- d)0.01
Correct answer is option 'C'. Can you explain this answer?
A mass m = 2.0 kg is attached to a spring having a force constant k = 290 N/m as in the figure. The mass is displaced from its equilibrium position and released. Its frequency of oscillation (in Hz) is approximately:
a)
12
b)
0.50
c)
1.9
d)
0.01
|
Riya Banerjee answered |
A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:- a)2π/√3
- b)2π√3
- c)√3/2π
- d)1/ 2π√3
Correct answer is option 'A'. Can you explain this answer?
A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:
a)
2π/√3
b)
2π√3
c)
√3/2π
d)
1/ 2π√3
|
Jyoti Kapoor answered |
velocity at distance x in shm is given as
v= w sqrt(A2-x2)
so A = 2 x =1
v = w sqrt(4-1) = wsqrt3
now acc at distance x is a= w2 x = w2
as mag a = mag v
w2 = wsqrt3
w= sqrt3
2pi/T= sqrt3
T = 2pi/sqrt3
A wave has S.H.M (Simple Harmonic Motion) whose period is 4 seconds while another wave which also possess SHM has its period 3 seconds. If both are combined, then the resultant wave will have the period equal to [1993]- a)4 seconds
- b)5 seconds
- c)12 seconds
- d)3 seconds
Correct answer is option 'C'. Can you explain this answer?
A wave has S.H.M (Simple Harmonic Motion) whose period is 4 seconds while another wave which also possess SHM has its period 3 seconds. If both are combined, then the resultant wave will have the period equal to [1993]
a)
4 seconds
b)
5 seconds
c)
12 seconds
d)
3 seconds
|
|
Siddheshwar Kale answered |
When wave in SHM combined beat will produce.
combine f= f2-f1= 1/T1 - 1/T2= 1/T=1/3 -1/4= 1/12.
since, T=12.
combine f= f2-f1= 1/T1 - 1/T2= 1/T=1/3 -1/4= 1/12.
since, T=12.
The restoring force in a simple harmonic motion is _________ in magnitude when the particle is instantaneously at rest.- a)zero
- b)maximum
- c)minimum
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The restoring force in a simple harmonic motion is _________ in magnitude when the particle is instantaneously at rest.
a)
zero
b)
maximum
c)
minimum
d)
none of these
|
Knowledge Hub answered |
The restoring force in a simple harmonic motion is maximum in magnitude when the particle is instantaneously at rest because in SHM object’s tendency is to return to mean position and here particle is instantaneously at rest after that instant restoring force will be max to bring particle to mean position.
A simple harmonic oscillator has an amplitude A and time period T. The time required by it to travel from x = A to x = A/2 is [1992]- a)T/6
- b)T/4
- c)T/3
- d)T/2
Correct answer is option 'A'. Can you explain this answer?
A simple harmonic oscillator has an amplitude A and time period T. The time required by it to travel from x = A to x = A/2 is [1992]
a)
T/6
b)
T/4
c)
T/3
d)
T/2
|
|
Geetika Shah answered |
For S.H.M.,
When x = A, 
When
or, 
Now, time taken to travel from x = A to x = A/2 is (T/4 – T/12) = T/6
The amplitude A of a simple harmonic oscillator (with period, T and energy, E) is tripled. What would happen to T and E?
- a)T and E triples
- b)T and E doubles
- c)T remains same and E becomes nine times
- d)T gets tripled and E becomes zero
Correct answer is option 'C'. Can you explain this answer?
The amplitude A of a simple harmonic oscillator (with period, T and energy, E) is tripled. What would happen to T and E?
a)
T and E triples
b)
T and E doubles
c)
T remains same and E becomes nine times
d)
T gets tripled and E becomes zero
|
|
Preeti Iyer answered |
E α A2 and hence the energy will become 9 times.
T is independent of A and hence no change will occur.
Hence C is the correct answer.
T is independent of A and hence no change will occur.
Hence C is the correct answer.
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately:- a)y = .10 cos (9.9t + .1)
- b)y = .10 sin 9.9t
- c)y = .10 cos 9.9t
- d)y = .05 cos 9.9t
Correct answer is option 'D'. Can you explain this answer?
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately:
a)
y = .10 cos (9.9t + .1)
b)
y = .10 sin 9.9t
c)
y = .10 cos 9.9t
d)
y = .05 cos 9.9t
|
|
Nandini Iyer answered |
a. y
= .10 sin 9.9t
b. y
= .10 cos 9.9t
c. y
= .10 cos (9.9t + .1)
d. y
= .10 sin (9.9t + 5)
e. y
= .05 cos 9.9t
Answer: e.
y = 0.5 . cos 9.9 t
If an external force with angular frequency ωd acts on an oscillating system with natural angular frequency ω, the system oscillates with angular frequency ωd. The amplitude of oscillations is the greatest when:- a)ωd > ω
- b)ωd < ω
- c)ωd ≥ ω
- d)ωd = ω
Correct answer is option 'D'. Can you explain this answer?
If an external force with angular frequency ωd acts on an oscillating system with natural angular frequency ω, the system oscillates with angular frequency ωd. The amplitude of oscillations is the greatest when:
a)
ωd > ω
b)
ωd < ω
c)
ωd ≥ ω
d)
ωd = ω
|
|
Jyoti Aiims Aspirant answered |
If the sign in equation F = -kx is changed what would happen to the motion of the oscillating body?- a)Motion would be linearly accelerated motion
- b)Body would come at rest
- c)Body would slow down
- d)Motion would be oscillating accelerated
Correct answer is option 'A'. Can you explain this answer?
If the sign in equation F = -kx is changed what would happen to the motion of the oscillating body?
a)
Motion would be linearly accelerated motion
b)
Body would come at rest
c)
Body would slow down
d)
Motion would be oscillating accelerated
|
Imk Pathsala answered |
If the sign is changed in F=-kx then the force and hence acceleration will not be opposite to the displacement. Due to this the particle will not oscillate and would accelerate in the direction of displacement. Hence the motion of the body will become linearly accelerated motion.
Under what condition angular frequency ω of the damped oscillator would be equivalent to angular velocity Ω of the undamped oscillator.- a)Velocity of oscillator is small
- b)Damping constant, b is small
- c)Damping constant ,b is large
- d)Force applied is small
Correct answer is option 'B'. Can you explain this answer?
Under what condition angular frequency ω of the damped oscillator would be equivalent to angular velocity Ω of the undamped oscillator.
a)
Velocity of oscillator is small
b)
Damping constant, b is small
c)
Damping constant ,b is large
d)
Force applied is small
|
|
Priya Patel answered |
The composition of two simple h ar monic motions of equal periods at right angle to each other and with a phase difference of π results in the displacement of the particle along [1990]- a)circle
- b)figures of eight
- c)straight line
- d)ellipse
Correct answer is option 'C'. Can you explain this answer?
The composition of two simple h ar monic motions of equal periods at right angle to each other and with a phase difference of π results in the displacement of the particle along [1990]
a)
circle
b)
figures of eight
c)
straight line
d)
ellipse
|
Aaksc Chemistry answered |
It is an equation of a straight line.
A simple pendulum with length l and bob of mass m is executing S.H.M of small amplitude a. The expression for maximum tension in the string will be- a)mg
- b)
- c)
- d)mg[1 + (a/l) ]
Correct answer is option 'B'. Can you explain this answer?
A simple pendulum with length l and bob of mass m is executing S.H.M of small amplitude a. The expression for maximum tension in the string will be
a)
mg
b)
c)
d)
mg[1 + (a/l) ]
|
|
Nandini Patel answered |
The tension in the string would be maximum when bob will be passing mean position.
Whose motion is an ideal case of SHM?- a)An electric dipole placed stably in a uniform electric field and then given a very small regular displacement
- b)A simple pendulum with very small angle of swing
- c)Shadow of a point mass having uniform circular motion, on the vertical diameter
- d)none
Correct answer is option 'A'. Can you explain this answer?
Whose motion is an ideal case of SHM?
a)
An electric dipole placed stably in a uniform electric field and then given a very small regular displacement
b)
A simple pendulum with very small angle of swing
c)
Shadow of a point mass having uniform circular motion, on the vertical diameter
d)
none
|
|
Raghav Bansal answered |
Simple pendulum is not a case of ideal SHM as it is mention in NCERT.
Shadow of a point mass having uniform circular motion on the horizontal diameter is a case of SHM but not on the vertical diameter.
Shadow of a point mass having uniform circular motion on the horizontal diameter is a case of SHM but not on the vertical diameter.
A rubber ball with water, having a small hole in its bottom is used as the bob of a simple pendulum. The time-period of such a pendulum:- a)Increases with time
- b)First increases and then decreases finally having same value as at the beginning
- c)Decreases with time
- d)Is a constant
Correct answer is option 'B'. Can you explain this answer?
A rubber ball with water, having a small hole in its bottom is used as the bob of a simple pendulum. The time-period of such a pendulum:
a)
Increases with time
b)
First increases and then decreases finally having same value as at the beginning
c)
Decreases with time
d)
Is a constant
|
|
Rajesh Gupta answered |
When rubber ball completely filled with water its centre of gravity will be at its centre, as water will fall through hole its COG will shift towards lower side leading to increase in length of pendulum and thus T, when very small amount of water will be left in rubber ball its COG will again shifts upward causing decrease in length and thus T, and finally when rubber ball becomes empty its COG will be at its centre and T will remains same as earlier.
What provides the restoring force for SHM in case of column of mercury in U-tube?- a)Temperature
- b)Weight of mercury
- c)Height of column
- d)Elasticity
Correct answer is option 'B'. Can you explain this answer?
What provides the restoring force for SHM in case of column of mercury in U-tube?
a)
Temperature
b)
Weight of mercury
c)
Height of column
d)
Elasticity
|
|
Riya Banerjee answered |
Weight of mercury provides the restoring force for SHM in case of column of mercury in U-tube.
The amplitude of a pendulum executing damped simple harmonic motion falls to 1/3 the original value after 100 oscillations. The amplitude falls to S times the original value after 200 oscillations, where S is [2002]- a)1/9
- b)1/2
- c)2/3
- d)1/6
Correct answer is option 'A'. Can you explain this answer?
The amplitude of a pendulum executing damped simple harmonic motion falls to 1/3 the original value after 100 oscillations. The amplitude falls to S times the original value after 200 oscillations, where S is [2002]
a)
1/9
b)
1/2
c)
2/3
d)
1/6
|
|
Santhana Priya answered |
Correct answer is option.A
Damping is due to- a)electrostatic forces
- b)resistive forces like air drag, friction etc.
- c)reaction forces
- d)conservative forces like gravity
Correct answer is option 'B'. Can you explain this answer?
Damping is due to
a)
electrostatic forces
b)
resistive forces like air drag, friction etc.
c)
reaction forces
d)
conservative forces like gravity
|
|
Om Desai answered |
Damping is due to resistive forces like air drag, friction etc.
In the figure shown pulley is massless. Initially the blocks are held at a height such that spring is in its natural length. The amplitude and velocity amplitude of block B1 respectively are (there is no slipping anywhere)
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
In the figure shown pulley is massless. Initially the blocks are held at a height such that spring is in its natural length. The amplitude and velocity amplitude of block B1 respectively are (there is no slipping anywhere)
a)
b)
c)
d)
|
Muskaan Mishra answered |
Decrease in GPE of B1 = Increase in Gravitational PE of B2 + Increase in elastic PE of spring.
Again applying law of conservation of mechanical energy at the mean position we get
The motion of simple pendulum is said to be S.H.M when its angle θ¸ through which bob is displaced from its equilibrium position is- a)θ is very small
- b)θ is zero
- c)θ is very large
- d)θ = cosθ
Correct answer is option 'A'. Can you explain this answer?
The motion of simple pendulum is said to be S.H.M when its angle θ¸ through which bob is displaced from its equilibrium position is
a)
θ is very small
b)
θ is zero
c)
θ is very large
d)
θ = cosθ
|
|
Priya Patel answered |
Consider a simple pendulum having mass 'm', length L and displaced by a small angle Θ with the vertical. Thus, it oscillates about its mean position. In the displaced position, two forces are acting on the bob, Gravitational force, which is the weight of the bob – 'mg' acting in the downward direction.
If the reference particle P moves in uniform circular motion, its projection along a diameter of the circle executes- a)motion along x axis
- b)motion along y axis
- c)circular motion
- d)Simple Harmonic motion
Correct answer is option 'D'. Can you explain this answer?
If the reference particle P moves in uniform circular motion, its projection along a diameter of the circle executes
a)
motion along x axis
b)
motion along y axis
c)
circular motion
d)
Simple Harmonic motion
|
|
Raghav Bansal answered |
SHM is a 1D projection of 2D UCM.
The ratio of maximum acceleration to maximum velocity of a particle performing S.H.M is equal to- a)Square of amplitude
- b)Square of angular velocity
- c)Amplitude
- d)Angular velocity
Correct answer is option 'D'. Can you explain this answer?
The ratio of maximum acceleration to maximum velocity of a particle performing S.H.M is equal to
a)
Square of amplitude
b)
Square of angular velocity
c)
Amplitude
d)
Angular velocity
|
|
Om Desai answered |
Maximum acceleration = w2A
Maximum velocity = wA
Ratio = w2A/wA = w
w = angular velocity
A = amplitude
Maximum velocity = wA
Ratio = w2A/wA = w
w = angular velocity
A = amplitude
A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is- a)A√3/2
- b)2A/√3
- c)A/2
- d)A/√2
Correct answer is option 'A'. Can you explain this answer?
A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is
a)
A√3/2
b)
2A/√3
c)
A/2
d)
A/√2
|
|
Rajesh Gupta answered |
The relation between angular frequency and displacement is given as
v=ω√A2−x2
Suppose
x=A sinω t
On differentiating the above equation w.r.t. time we get
dx/dt=Aωcosωt
The maximum value of velocity will be [{v{\max }} = A\omega \]
The displacement for the time when speed is half the maximum is given as
v=Aω/2
A2ω2=4ω(A2−x2)
By substituting the value in (1) we get the displacement as
x=A√3/2
v=ω√A2−x2
Suppose
x=A sinω t
On differentiating the above equation w.r.t. time we get
dx/dt=Aωcosωt
The maximum value of velocity will be [{v{\max }} = A\omega \]
The displacement for the time when speed is half the maximum is given as
v=Aω/2
A2ω2=4ω(A2−x2)
By substituting the value in (1) we get the displacement as
x=A√3/2
The potential energy of a long spring when stretched by 2cm is U. If the spring is stretched by 8 cm, the potential energy stored in it is- a)8 U
- b)16 U [2006]
- c)U/ 4
- d)4 U
Correct answer is option 'B'. Can you explain this answer?
The potential energy of a long spring when stretched by 2cm is U. If the spring is stretched by 8 cm, the potential energy stored in it is
a)
8 U
b)
16 U [2006]
c)
U/ 4
d)
4 U
|
|
Geetika Shah answered |
The potential energy of a spring
For x = 8 cm,
Energy stored
Which one of the following is a simple harmonic motion? [1994]- a)Ball bouncing between two rigid vertical walls
- b)Particle moving in a circle with uniform speed
- c)Wave moving through a string fixed at both ends
- d)Earth spinning about its own axis.
Correct answer is option 'C'. Can you explain this answer?
Which one of the following is a simple harmonic motion? [1994]
a)
Ball bouncing between two rigid vertical walls
b)
Particle moving in a circle with uniform speed
c)
Wave moving through a string fixed at both ends
d)
Earth spinning about its own axis.
|
Pankaj Kulkarni answered |
A wave moving through a string fixed at both ends, is a transverse wave formed as a result of simple harmonic motion of particles of the string.
Damped natural frequency is- a)same as natural frequency
- b)lower than natural frequency
- c)higher than natural frequency
- d)none of the above
Correct answer is option 'B'. Can you explain this answer?
Damped natural frequency is
a)
same as natural frequency
b)
lower than natural frequency
c)
higher than natural frequency
d)
none of the above
|
|
Geetika Shah answered |
If a resonant mechanical structure is set in motion and left to its own devices, it will continue to oscillate at a particular frequency known as its natural frequency, or "damped natural frequency". This will be a little lower in frequency than the resonant frequency, which is the frequency it would assume if there were no damping. The resonant frequency is also called the "undamped natural frequency”
If the length of a simple pendulum is increased by 2%, then the time period [1997]- a)increases by 2%
- b)decreases by 2%
- c)increases by 1%
- d)decreases by 1%
Correct answer is option 'C'. Can you explain this answer?
If the length of a simple pendulum is increased by 2%, then the time period [1997]
a)
increases by 2%
b)
decreases by 2%
c)
increases by 1%
d)
decreases by 1%
|
Ayesha Princess answered |
T=2π√l/g
here we take g=constant..
so T Will b proportional to √l....
for small change in length
ΔT/T=1/2Δl/l
=1/2(2%)
=1% increase (since it is directly proportional)
here we take g=constant..
so T Will b proportional to √l....
for small change in length
ΔT/T=1/2Δl/l
=1/2(2%)
=1% increase (since it is directly proportional)
At what distance from the mean position would the K.E of a particle in simple harmonic motion be equal to its potential energy?- a)a/√2
- b)a/2
- c)2√a
- d)a
Correct answer is option 'A'. Can you explain this answer?
At what distance from the mean position would the K.E of a particle in simple harmonic motion be equal to its potential energy?
a)
a/√2
b)
a/2
c)
2√a
d)
a
|
|
Raghav Bansal answered |
Let say from some distance x, the KE = PE and as total energy must be conserved and TE = -½ kA2
Thus we get 2PE = ½ kA2
Thus we get 2kx2 = kA2
We get x = A / √2
Thus we get 2PE = ½ kA2
Thus we get 2kx2 = kA2
We get x = A / √2
The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration is
- a)0.60 ms-2
- b)0.50 ms-2
- c)0.55 ms-2
- d)0.45 ms-2
Correct answer is option 'A'. Can you explain this answer?
The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration is
a)
0.60 ms-2
b)
0.50 ms-2
c)
0.55 ms-2
d)
0.45 ms-2
|
Nilotpal Singh answered |
+ 2) meters.
We can find the velocity and acceleration functions by taking the first and second derivatives of x with respect to time:
velocity v(t) = dx/dt = -3(0.45)sin(0.45t + 2)
acceleration a(t) = d^2x/dt^2 = -3(0.45)^2cos(0.45t + 2)
To find the maximum velocity, we set v(t) equal to zero and solve for t:
0 = -3(0.45)sin(0.45t + 2)
sin(0.45t + 2) = 0
0.45t + 2 = nπ, where n is an integer
t = (nπ - 2)/0.45
To find the maximum acceleration, we set a(t) equal to zero and solve for t:
0 = -3(0.45)^2cos(0.45t + 2)
cos(0.45t + 2) = 0
0.45t + 2 = (n + 0.5)π, where n is an integer
t = [(n + 0.5)π - 2]/0.45
Note that there are infinitely many solutions for both t_max_v and t_max_a, as there are infinitely many values of n.
To find the values of maximum velocity and maximum acceleration, we can substitute the values of t_max_v and t_max_a into the corresponding velocity and acceleration functions:
v_max = -3(0.45)sin(0.45t_max_v + 2)
a_max = -3(0.45)^2cos(0.45t_max_a + 2)
However, since there are infinitely many solutions for t_max_v and t_max_a, we cannot find a unique value for v_max and a_max without additional information.
We can find the velocity and acceleration functions by taking the first and second derivatives of x with respect to time:
velocity v(t) = dx/dt = -3(0.45)sin(0.45t + 2)
acceleration a(t) = d^2x/dt^2 = -3(0.45)^2cos(0.45t + 2)
To find the maximum velocity, we set v(t) equal to zero and solve for t:
0 = -3(0.45)sin(0.45t + 2)
sin(0.45t + 2) = 0
0.45t + 2 = nπ, where n is an integer
t = (nπ - 2)/0.45
To find the maximum acceleration, we set a(t) equal to zero and solve for t:
0 = -3(0.45)^2cos(0.45t + 2)
cos(0.45t + 2) = 0
0.45t + 2 = (n + 0.5)π, where n is an integer
t = [(n + 0.5)π - 2]/0.45
Note that there are infinitely many solutions for both t_max_v and t_max_a, as there are infinitely many values of n.
To find the values of maximum velocity and maximum acceleration, we can substitute the values of t_max_v and t_max_a into the corresponding velocity and acceleration functions:
v_max = -3(0.45)sin(0.45t_max_v + 2)
a_max = -3(0.45)^2cos(0.45t_max_a + 2)
However, since there are infinitely many solutions for t_max_v and t_max_a, we cannot find a unique value for v_max and a_max without additional information.
In a simple pendulum the restoring force is due to- a) The tangential component of the gravitational force
- b)The radial component of the gravitational force
- c)The tangential component of the tension in string
- d)The radial component of the tension in string
Correct answer is option 'A'. Can you explain this answer?
In a simple pendulum the restoring force is due to
a)
The tangential component of the gravitational force
b)
The radial component of the gravitational force
c)
The tangential component of the tension in string
d)
The radial component of the tension in string
|
|
Pooja Shah answered |
In a simple pendulum the restoring force is due to the tangential component of the gravitational force because on applying torque equation on the radial forces the effect is nullified due to it passing from centre & only tangential component remains.
The amplitude of S.H.M at resonance is _______ in the ideal case of zero damping.- a)Maximum
- b)Minimum
- c)Zero
- d)Infinite
Correct answer is option 'D'. Can you explain this answer?
The amplitude of S.H.M at resonance is _______ in the ideal case of zero damping.
a)
Maximum
b)
Minimum
c)
Zero
d)
Infinite
|
|
Neha Joshi answered |
In an ideal environment where there is no resistance to oscillation motion i.e. damping is zero, when we oscillate a system at its resonant frequency since there is no opposition to oscillation, the amplitude will go on increasing and reach infinity.
The instantaneous displacement of a simple harmonic oscillator is given by y = A cos (ωt + n/4). Its speed will be maximum at the time:). Its speed will be maximum at the time:- a)n/4ω
- b)ω/n
- c)ω/2n
- d)2n/ω
Correct answer is option 'A'. Can you explain this answer?
The instantaneous displacement of a simple harmonic oscillator is given by y = A cos (ωt + n/4). Its speed will be maximum at the time:). Its speed will be maximum at the time:
a)
n/4ω
b)
ω/n
c)
ω/2n
d)
2n/ω
|
|
Nandini Iyer answered |
The instantaneous displacement of a simple pendulum oscillator is given as y = acos(ωt + π/4)
differentiating with respect to time,
dy/dt = -ωasin(ωt + π/4)
here, dy/dt is the velocity of a particle executing SHM.
so, speed of particle = | dy/dt | = ωasin(ωt + π/4)
so, dy/dt will be maximum when sin(ωt + π/4) will be maximum i.e., 1
so, sin(ωt + π/4) = 1 = sin(π/2)
⇒ωt + π/4 = π/2
⇒ωt = π/4
⇒t = π/(4ω)
hence, at t = π/(4ω) , speed of the particle will be maximum.
differentiating with respect to time,
dy/dt = -ωasin(ωt + π/4)
here, dy/dt is the velocity of a particle executing SHM.
so, speed of particle = | dy/dt | = ωasin(ωt + π/4)
so, dy/dt will be maximum when sin(ωt + π/4) will be maximum i.e., 1
so, sin(ωt + π/4) = 1 = sin(π/2)
⇒ωt + π/4 = π/2
⇒ωt = π/4
⇒t = π/(4ω)
hence, at t = π/(4ω) , speed of the particle will be maximum.
The velocity of a particle moving with simple harmonic motion is . . . . at the mean position.- a)Zero
- b)Minimum
- c)Maximum
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
The velocity of a particle moving with simple harmonic motion is . . . . at the mean position.
a)
Zero
b)
Minimum
c)
Maximum
d)
None of the above
|
|
Niharika Nair answered |
V = ω√(A2 – x2)
So the velocity is maximum at mean position
So the velocity is maximum at mean position
The kinetic energy of a body executing S.H.M. is 1/3 of the potential energy. Then, the displacement of the body is x percent of the amplitude, where x is- a)33
- b)50
- c)67
- d)87
Correct answer is option 'A'. Can you explain this answer?
The kinetic energy of a body executing S.H.M. is 1/3 of the potential energy. Then, the displacement of the body is x percent of the amplitude, where x is
a)
33
b)
50
c)
67
d)
87
|
|
Pooja Shah answered |
We know that PE + KE = TE = constant
Hence at the extremis TE = PE = ½ kz2
Where z is amplitude and k is shm constant.
Thus when KE = ⅓ PE
We get PE = ¾ TE = ½ kz2
Hence we get ½ kx2 = ¾ ½ kz2
We get x/z = √3/4
= 1.73 / 2
= .87
Thus we get x is 87 percent of the amplitude.
Hence at the extremis TE = PE = ½ kz2
Where z is amplitude and k is shm constant.
Thus when KE = ⅓ PE
We get PE = ¾ TE = ½ kz2
Hence we get ½ kx2 = ¾ ½ kz2
We get x/z = √3/4
= 1.73 / 2
= .87
Thus we get x is 87 percent of the amplitude.
Chapter doubts & questions for Oscillations - Physics for Grade 12 2025 is part of Grade 12 exam preparation. The chapters have been prepared according to the Grade 12 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Grade 12 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Oscillations - Physics for Grade 12 in English & Hindi are available as part of Grade 12 exam.
Download more important topics, notes, lectures and mock test series for Grade 12 Exam by signing up for free.
Physics for Grade 12
161 videos|453 docs|190 tests
|
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup