1003 + Interview Questions in SIMPLE-HARMONIC-MOTION IN GENERAL AWARENESS Page 12 InterviewSolution

551.	In the arrangement shown in figure, pulleys are light and spring are ideal. `K_(1)`, `k_(2)`, `k_(3)`and `k_(4)` are force constant of the spring. Calculate period of small vertical oscillations of block of mass `m`.
Answer» Correct Answer - A::B::C::D Let block be displacement by x than displacement in springs be `x_(1),x_(2),x_(3)` and `x_(4)` Such that `x = 2x_(1) + 2x_(2) + 2x_(3) + 2x_(4)` Now let restoring force on m be `F = kx` then `2f = k_(1)x_(1) = k_(2)x_(2) = k_(3)x_(3) = k_(4)x_(4)` `rArr F/k = (4F)/(k_(1)) + (4F)/(k_(2)) + (4F)/(k_(3)) + (4F)/(k_(4))` `rArr 1/k = 4(1/(k_(1)) + 1/(k_(2)) + 1/(k_(3) ) + 1/(k_(4)))` `T = 2pisqrt((m)/(k)) = 2pisqrt(4m(1/(k_(1))+(1)/(k_(2)) + 1/(k_(3)) + 1/(k_(4))))`

Discussion

552.	The two torsion pendula differ only by the addition of cylindrical masses as shown in the figure. The radius of each additional mass is `1//4` the radius fo the disc. Each cylinder and disc have equal mass. The ratio of time periods of the two torsion pendula is `p`. Then `27 p^(2) ` is.
Answer» `I_(1) = (MR^(2))/(2)` `I_(2) = [M/2(R/4)^(2) + M((3R)/4)^(2)] + [M/2(R/4)^(2) + M((3R)/4)^(2)] + (MR^(2))/(2)` `T prop sqrt(1)` `(T_(1))/(T_(2)) prop sqrt((I_(1))/(I_(2)))`

Discussion

553.	Two springs, each of spring constant `k`, are attached to a block of mass `m` as shown in the figure. The block can slide smoothly alongs a horizontal platform clamped to the oppsite walls of the trolley of mass `M`. If the block is displaced by `x cm` and released, the period of oscillaion is : A. `T = 2pisqrt((Mm)/(2k))`B. `T = 2pisqrt(((M + m))/(kmM))`C. `T = 2pisqrt((mM)/(2k(M + m)))`D. `T = 2pi((M + m))/(k)`
Answer» Correct Answer - C

Discussion

554.	A particle moves along the x - axis according to `x = A[1 + sin omega t]`. What distance does is travel in time interval from `t = 0` to `t = 2.5pi//omega` ?A. `4A`B. `6A`C. `5A`D. None
Answer» Correct Answer - C

Discussion

555.	A particle of mass m oscillates with simple harmonic motion between points `x_(1)` and `x_(2)`, the equilibrium position being O. Its potential energy is plotted. It will be as given below in the graphA. B. C. D.
Answer» Correct Answer - C (c) Potential energy graph is parabolic with its minimum value at mean position.

Discussion

556.	A particle moves with simple harmonic motion in a straight line. When the distances of the particle from the equilibrium position are `x_(1)` and `x_(2)`, the corresponding velocities are `u_(1)` and `u_(2)`. Find the period of the moton.
Answer» Correct Answer - 2pi`((x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2)))^(1/2)`

Discussion

557.	The maximum velocity of a body undergoing SHM is 0.004 m`s^(-1)` and its acceleration at 0.002 m from the mean position is 0.06 m`s^(-2)`. Find its amplitude and period of vibration. [Hint: Use `v_(m) = aomega^(2)x`numerically.[
Answer» Correct Answer - `2.31xx10^(-2)m,3.63s`

Discussion

558.	What are undamped and damped oscillations ?
Answer» Correct Answer - Oscillation of a system is said to be undamped it its amplitude does not change with time.

Discussion

559.	Define forced and resonant vibrations
Answer» Correct Answer - When a body maintained in a state of oscillation by a strong periodic force of frequency other than the natural frequency of the body, the oscillations are called forced vibrations.

Discussion

560.	The time period of a simple pendulum of infinite length is (R=radius of earth).A. infiniteB. `2pisqrt((R)/(g))`C. `2pisqrt((g)/(R))`D. `(1)/(2pi)sqrt((R)/(g))`
Answer» Correct Answer - B (b) `T=2pisqrt((R_(e))/(g))=84.6` min [If l is infinte]

Discussion

561.	If the length of a simple pendulum is equal to the radius of the earth, its time period will beA. `2pisqrt(R//g)`B. `2pi sqrt (R//2g)`C. `2 pi sqrt (2R//g)`D. infinite
Answer» Correct Answer - B `T = 2pi sqrt((1)/(g((1)/(l) + (1)/(R))` put `l = R`

Discussion

562.	Two sources of sound are in resonance whenA. they look likeB. they are situated at a particular distance from each otherC. they produce the sound of same frequencyD. they are excited by the same excittng device
Answer» Correct Answer - C (c) Two sources of sound are in resonance when they produce the sound of same frequency.

Discussion

563.	Define resonance.
Answer» Correct Answer - The phenomenon of setting a body into oscillations of large amplitude by the influence of another virbrating body having the same netural frequency is called resonance.

Discussion

564.	During the phenonmenon of resonanceA. the amplitude of oscillation becomes largeB. the frequency of oscillation become largeC. the time period of oscillation becomes largeD. All of the above
Answer» Correct Answer - A (a) During the phenomenon of resonance , the amplitude of oscillation becomes large . Because applied frequency is equal to natural frequency.

Discussion

565.	A block of mass M executes SHM with amplitude a and time period T. When it passes through the mean position, a lump of putty of mass m is dropped on it. Find the new amplitude and time period.
Answer» Correct Answer - `sqrt(M/(M+m)).asqrt((M+m)/(M)).T

Discussion

566.	A plank with a body of mass `m` places on it starts moving straight up according to the law `y=a(1-cos omegat t )`, whee `y` is the displacement from the initial position, `omega =11 s^(-1)`. Find : (a) the time dependence of the force that the body exerts on the plank if `a=4.0cm,` plot this dependence, `(b)` the minimum amplitude of oscillation of the plank at which the body starts falling behind the plank, (c) the amplitude of oscillation of the plank at which the body springs up to a height `h=50cm` relative to the initial position (at the moment `t=0)`.
Answer» Correct Answer - `(g)/(Omega^(2))(Omega.sqrt((2h)/(g))-1)`

Discussion

567.	A particle of mass 0.10 kg executes SHM with an amplitude 0.05 m and frequency 20 vib/s. Its energy os oscillation isA. 2 JB. 4 JC. 1 JD. zero
Answer» Correct Answer - A (a) `E=(1)/(2)momega^(2)A^(2)=(1)/(2)m(2pif)^(2)A^(2)=2pi^(2)f^(2)mA^(2)` `=2pi^(2)(20)^(2)(0.1)(0.05)^(2)=2 J" "` (approximately)

Discussion

568.	Energy of particle executing SHM depends uponA. amplitude onlyB. Amplitude and frequencyC. velocity onlyD. frequency only
Answer» Correct Answer - B (b) Energy in SHM , `E=U_("mean")+(1)/(2)momega^(2)A^(2)` i.e., E depend on `omega` and A both.

Discussion

569.	The equation of a damped simple harmonic motion is `m(d^2x)/(dt^2)+b(dx)/(dt)+kx=0`. Then the angular frequency of oscillation isA. `omega=((k)/(m)-(b^(2))/(4m^(2)))^(1//2)`B. `omega=((k)/(m)-(b)/(4m))^(1//2)`C. `omega=((k)/(m)-(b^(2))/(4m))^(1//2)`D. `omega=((k)/(m)-(b^(2))/(4m^(2)))^(-1//2)`
Answer» Correct Answer - A (a)`omega=sqrt((k)/(m)-((b)/(2m))^(2))`

Discussion

570.	The equation of a damped simple harmonic motion is `m(d^2x)/(dt^2)+b(dx)/(dt)+kx=0`. Then the angular frequency of oscillation isA. `omega=(k/m-(b^2)/(4m^2))^(1//2)`B. `omega=(k/m-(b)/(4m))^(1//2)`C. `omega=(k/m-(b^2)/(4m))^(1//2)`D. `omega=(k/m-(b^2)/(4m^2))^(1//2)`
Answer» Correct Answer - A `omega=sqrt(k/m-((b)/(2m))^2)`

Discussion

571.	The total energy of a particle having a displacement x, executing simple harmonic motion isA. `prop x`B. `prop x^(2)`C. independent of xD. `prop x^(1//2)`
Answer» Correct Answer - C (c) Total energy, E`=(1)/(2)momega^(2)A^(2)` `implies` It is independent of x

Discussion

572.	In case of damped oscillation frequency of oscillation isA. greater than natural frequencyB. less than natural frequencyC. equal to nature frequencyD. Both (a) and (b)
Answer» Correct Answer - B (b) Frequency is less than with dmaping .

Discussion

573.	The total energy of a harmonic oscillator of mass 2 kg is 9 J. If its potential energy at mean position is 5 J , its KE at the mean position will beA. 9 JB. 14 JC. 4 JD. 11 J
Answer» Correct Answer - C (c)From law of conservation of energy, KE=E-U=4 J

Discussion

574.	The bob of a simple pendulum it displaced position `O` to a equilibrium position `Q` which is at height h above `O` and the bob to then mass released Assuming the mass of the bob is m and time period `2.0` sec of oscillation to be string when the bob passes through `O` is A. `m(g = pi sqrt(2 gh))`B. `m(g = sqrt(pi^(2) gh))`C. `m (g + sqrt(pi^(2)/(2)gh))`D. `m (g + sqrt(pi^(2)/(3)gh))`
Answer» Correct Answer - A

Discussion

575.	The energy at the mean position of a pendulum will beA. zeroB. partial PE and partial KEC. totally KED. totally PE
Answer» Correct Answer - C At mean positions .KE is maximum and PE is zero.

Discussion

576.	The figure shows the displacement-time graph of a particle executing `SHM`. If the time period of oscillation is `2s`, then the equation of motion is given by
Answer» Correct Answer - `X=10 sin (pit+pi//6)`

Discussion

577.	Frequency of oscillation is proportional to A. `sqrt((3k)/(m))`B. `sqrt((k)/(m))`C. `sqrt((2k)/(m))`D. `sqrt((m)/(3k))`
Answer» Correct Answer - A (a)`f=(1)/(2pi)sqrt((3k)/(m))` or `sqrt((k)/(m))` Here value of `k_(e)` will become 3k `" "`(`because` Springs are connected in parallel) `therefore f prop sqrt((3k)/(m))`

Discussion

578.	Assertion : If amplitude of SHM is increased , time period of SHM will increase. Reason: If amplitude is increased, body have to travel more distance in one complete oscillation.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true.
Answer» Correct Answer - D (d)Time period does not depend on amplitude of oscillation.

Discussion

579.	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 isA. `(1)/(2pisqrt(3))`B. `2pisqrt(3)`C. `(2pi)/(sqrt(3))`D. `(sqrt(3))/(2pi)`
Answer» Correct Answer - C (c) Velocity , `v=omegasqrt(A^(2)-x^(2))` and acceleration `=omega^(2)x` Now given, `omega^(2)x=omegasqrt(A^(2)-x^(2))` `impliesomega^(2).1=omegasqrt(2^(2)-1^(2))` `implies omega=sqrt(3)` `therefore T=(2pi)/(omega)=(2pi)/(sqrt(3))`

Discussion

580.	The system shown in figure is in equilibrium . The mass of the container with Liquid is M, density of liquid in the container is `rho` and the volume of the block is V. If the container is now displaced downwards through a distance `x_(0)` and released such that the block remains well inside the liquid then during subsequent motion A. time period of SHM of the container will be `2pisqrt((M)/(k))`B. time period of SHM of the container will be `2pisqrt((M+rhoV)/(k))`C. amplitude of SHM of the container is `x_(0)`D. amplitude of SHM of the container is `2x_(0)`
Answer» Correct Answer - B (b) `M_(e)`= Mass of container + `("Upthrust")/(g)` `M+(rhoVg)/(g)=M+rhoV`

Discussion

581.	Assertion : Simple harmonic motion is an example on one dimensional motion with non-uniform acceleration . Reason : In simple harmonic motion, acceleration varies with displacement linearly.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true.
Answer» Correct Answer - D (d) Simple harmonic motion is not always one -dimensional . In case of angular simple harmonic motion , it is 2-D also.

Discussion

582.	The shortest distance travelled by a particle executing SHM from mean position in 2 s is equal to `(sqrt(3)//2)` times its amplitude. Determine its time period.
Answer» `t=2 s, x=(sqrt(3))/(2) A,T =?` Using x=A sin `omegat` , we get `(sqrt(3))/(2)A =A "sin"(2pi)/(T)xx2` `implies "sin"(4pi)/(T)=(sqrt(3))/(2)="sin"(pi)/(3) implies (4pi)/(T)=(pi)/(3)` `therefore T=12 s`

Discussion

583.	A ring of mass m and radius r rolls without slipping on a fixed hemispherical surface of radius R as shown. The time period of small oscillations of ring is `2pisqrt((n(R-r))/(3g))` then find the value of n.
Answer» Correct Answer - 6

Discussion

584.	A vertical U-tube has one of the its arms bent at `theta` from the vertical and then filled with a liquid of density `rho` up to a height h in the vertical arm. Calculate the period of oscillations of the liquid when disturbed a little and released.
Answer» Correct Answer - `T=2pisqrt(h/(g cos theta))`

Discussion

585.	An object of mass `0.2 kg` executes simple harmonic oscillation along the `x - axis`with a frequency of `(25//pi) Hz`. At the position `x = 0.04`, the object has Kinetic energy of `0.5 J` and potential energy `0.4 J. The amplitude of oscillations is……m.A. 0.05B. 0.06C. 0.01D. none of these
Answer» Correct Answer - B

Discussion

586.	An object of mass `0.2 kg` executes simple harmonic oscillation along the `x - axis`with a frequency of `(25//pi) Hz`. At the position `x = 0.04`, the object has Kinetic energy of `0.5 J` and potential energy `0.4 J. The amplitude of oscillations is……m.
Answer» `x = 0.04 m, K.E. = 0.5 J and P.E. = 0.4 J` `T.E. = (0.5 + 0.4) J = 0.9 J` Now, `T.E. = (1)/(2)m omega^2 a^2 = (1)/(2) m xx 4 pi^2 v^2 a^2` rArr `0.9 = (1)/(2) xx 0.1 xx 4 pi^2 xx (25)/(pi) xx (25)/(pi) xx a^2 rArr a = 0.06 m`.

Discussion

587.	Assertion : Time period of a spring-block is T. If length of spring is decreased, time period will decrease. Reason If length is decreased, then the block will have to travel less distance and it will take less time.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true.
Answer» Correct Answer - C (c) Force constant of a spring , `k prop (1)/(l)` and `T prop (1)/(sqrt(k))` `therefore T prop sqrt(l)`

Discussion

588.	Assertion : Time period of a spring-block equation of a particle moving along X-axis is x=4+6`"sin"omegat` . Under this situation, motion of particle is not simple harmonic. Reason : `(d^(2)x)/(dt^(2))` for the given equation is proportional to -x.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true.
Answer» Correct Answer - D (d) Motion of particle is simple harmonic, but mean position is at x=4. Amplitude is 6. The particle will oscillation between x=10 and x=-2.

Discussion

589.	A spring mass system has time period of `2` second. What should be the spring constant of spring if the mass of the block is `10 grams` ?
Answer» Correct Answer - `0.1 N//m` `T = 2pisqrt((m)/(K)) rArr 2 = 2pisqrt((10//1000)/(K)) rArr K = 0.1 N//m`

Discussion

590.	An object of mass 0.2 kg executes simple harmonic oscillation along the x-axis with a frequency `(25)/pi`. At the position x = 0.04m, the object has kinetic amplitude of osciallation is (potential energy is zero mean position).
Answer» Correct Answer - `A = 0.06 m`

Discussion

591.	An object of mass 0.2 kg executes simple harmonic oscillation along the x-axis with a frequency `(25)/pi`. At the position x = 0.04m, the object has kinetic amplitude of osciallation is (potential energy is zero mean position).A. 6cmB. 4cmC. 8 cmD. 2cm
Answer» Correct Answer - A

Discussion

592.	An object of mass `0.2 kg` exectes simple harmonic oscillations alongs `x-`axis with a freqency of `(25//pi) Hz`. At the positions `x = 0.04m`, the object has kinetic energy of `0.5 J` and potential energy `0.4 J`. Find the amplitude of oscillaties
Answer» Correct Answer - `A = 0.06 m`

Discussion

593.	A particle is attached to a vertical spring and is pulled down a distance 0.04m below its equilibrium position and is released from rest. The initial upward acceleration of the particle is `0.30 ms^(-2)`. The period of the oscillation isA. 4.08 sB. 1.92 sC. 3.90 sD. 2.29 s
Answer» Correct Answer - D (d)`A=0.04m,omega^(2)A=0.3 ms^(-2)` `therefore omega=2.74 "rad"s^(-1)` Now, `T=(2pi)/(omega)=2.29 s`

Discussion

594.	The instantaneous displacement of a simple pendulum oscillator is given by `x= A cos (omegat+ pi / 4 )`Its speed will be maximum at timeA. `pi/(4 omega)`B. `pi/(2 omega)`C. `(pi)/(omega)`D. `(2pi)/(omega)`
Answer» Correct Answer - A

Discussion

595.	A box B of mass M hangs from an ideal spring of force constant k. A small particle, also of mass M, is stuck to the ceiling of the box and the system is in equilibrium. The particle gets detached from the ceiling and falls to strike the floor of the box. It takes time t for the particle to hit the floor after it gets detached from the ceiling. The particle, on hitting the floor, sticks to it and the system thereafter oscillates with a time period T. Find the height H of the box if it is given that `t=(T)/(6sqrt(2))`. Assume that the floor and ceiling of the box always remian horizontal.
Answer» Correct Answer - `H=(Mg)/(2k)(1+(pi^(2))/(9))`

Discussion

596.	A body executing `SHM` passes through its equilibrium. At this instant, it hasA. maximum potential energyB. maximum kinetic energyC. minimum kinetic energyD. maximum acceleration
Answer» Correct Answer - B At equilibrium position `K.E.` is maximum.

Discussion

597.	Velocity at mean position of a particle executing S.H.M. is v , they velocity of the particle at a distance equal to half of the amplitudeA. 4 vB. 2 vC. `sqrt3/2v`D. `sqrt3/4 v`
Answer» Correct Answer - C

Discussion

598.	(i) In the system shown in figure, find the time period of vertical oscillations of the block A. Both the blocks A and B have equal mass of m and the force constant of the ideal spring is k. Pulley and threads are massless (ii) In the arrangement shown in the figure the spring, string and the pulley are mass less. The force constant of the spring is k. A rope of mass per unit length equal to `lamda (kg m^(-1))` hangs from the string as shown. In equilibrium a length L of the rope is in air and its bottom part lies in a heap on the floor. The rope is very thin and size of the heap is negligible though the heap contains a fairly long length of the rope. The rope is raised by a very small distance and released. Show that motion will be simple harmonic and calculate the time period. Assume that the hanging part of the rope does not experience any force from the heap or the floor (For example there is no impact force while the rope hits the floor while moving downward and there is no impulsive pull when the vertical part jerks a small element of heap into motion).
Answer» Correct Answer - (i) ` T=pi sqrt((5m)/(k)), (ii) T=2pi sqrt((lamda L)/(k+lmadag))`

Discussion

599.	For a particle executing SHM, x = displacement from mean position, v = velocity and a = acceleration at any instant, thenA. `v-X` graph is a circleB. `v-X` graph is an ellipseC. `a-X` graph is a straight lineD. `a-v` ggraph is an ellipse
Answer» Correct Answer - B::C::D

Discussion

600.	For a particle executing SHM, x = displacement from mean position, v = velocity and a = acceleration at any instant, thenA. `v - x` graph is a circleB. `v - x` graph is an ellipseC. `a - x` graph is a straight lineD. `a - x` graph is a circle
Answer» Correct Answer - B::C `v = omega sqrt (A^(2) - x^(2))` `:. (v^(2))/(omega^(2)) + (x^(2))/((1)^(2)) = A^(2)` i.e. `v - x` graph is an ellipse. `a = - omega^(2)x` i.e. `a - x` graph is a straight line passing through origin with negative slope.

Discussion

Explore topic-wise InterviewSolutions in .

In the arrangement shown in figure, pulleys are light and spring are ideal. `K_(1)`, `k_(2)`, `k_(3)`and `k_(4)` are force constant of the spring. Calculate period of small vertical oscillations of block of mass `m`.

The two torsion pendula differ only by the addition of cylindrical masses as shown in the figure. The radius of each additional mass is `1//4` the radius fo the disc. Each cylinder and disc have equal mass. The ratio of time periods of the two torsion pendula is `p`. Then `27 p^(2) ` is.

A particle moves along the x - axis according to `x = A[1 + sin omega t]`. What distance does is travel in time interval from `t = 0` to `t = 2.5pi//omega` ?A. `4A`B. `6A`C. `5A`D. None

A particle of mass m oscillates with simple harmonic motion between points `x_(1)` and `x_(2)`, the equilibrium position being O. Its potential energy is plotted. It will be as given below in the graphA. B. C. D.

A particle moves with simple harmonic motion in a straight line. When the distances of the particle from the equilibrium position are `x_(1)` and `x_(2)`, the corresponding velocities are `u_(1)` and `u_(2)`. Find the period of the moton.

The maximum velocity of a body undergoing SHM is 0.004 m`s^(-1)` and its acceleration at 0.002 m from the mean position is 0.06 m`s^(-2)`. Find its amplitude and period of vibration. [Hint: Use `v_(m) = aomega^(2)x`numerically.[

What are undamped and damped oscillations ?

Define forced and resonant vibrations

The time period of a simple pendulum of infinite length is (R=radius of earth).A. infiniteB. `2pisqrt((R)/(g))`C. `2pisqrt((g)/(R))`D. `(1)/(2pi)sqrt((R)/(g))`

If the length of a simple pendulum is equal to the radius of the earth, its time period will beA. `2pisqrt(R//g)`B. `2pi sqrt (R//2g)`C. `2 pi sqrt (2R//g)`D. infinite

Two sources of sound are in resonance whenA. they look likeB. they are situated at a particular distance from each otherC. they produce the sound of same frequencyD. they are excited by the same excittng device

Define resonance.

During the phenonmenon of resonanceA. the amplitude of oscillation becomes largeB. the frequency of oscillation become largeC. the time period of oscillation becomes largeD. All of the above

A block of mass M executes SHM with amplitude a and time period T. When it passes through the mean position, a lump of putty of mass m is dropped on it. Find the new amplitude and time period.

A particle of mass 0.10 kg executes SHM with an amplitude 0.05 m and frequency 20 vib/s. Its energy os oscillation isA. 2 JB. 4 JC. 1 JD. zero

Energy of particle executing SHM depends uponA. amplitude onlyB. Amplitude and frequencyC. velocity onlyD. frequency only

The equation of a damped simple harmonic motion is `m(d^2x)/(dt^2)+b(dx)/(dt)+kx=0`. Then the angular frequency of oscillation isA. `omega=((k)/(m)-(b^(2))/(4m^(2)))^(1//2)`B. `omega=((k)/(m)-(b)/(4m))^(1//2)`C. `omega=((k)/(m)-(b^(2))/(4m))^(1//2)`D. `omega=((k)/(m)-(b^(2))/(4m^(2)))^(-1//2)`

The equation of a damped simple harmonic motion is `m(d^2x)/(dt^2)+b(dx)/(dt)+kx=0`. Then the angular frequency of oscillation isA. `omega=(k/m-(b^2)/(4m^2))^(1//2)`B. `omega=(k/m-(b)/(4m))^(1//2)`C. `omega=(k/m-(b^2)/(4m))^(1//2)`D. `omega=(k/m-(b^2)/(4m^2))^(1//2)`

The total energy of a particle having a displacement x, executing simple harmonic motion isA. `prop x`B. `prop x^(2)`C. independent of xD. `prop x^(1//2)`

In case of damped oscillation frequency of oscillation isA. greater than natural frequencyB. less than natural frequencyC. equal to nature frequencyD. Both (a) and (b)

The total energy of a harmonic oscillator of mass 2 kg is 9 J. If its potential energy at mean position is 5 J , its KE at the mean position will beA. 9 JB. 14 JC. 4 JD. 11 J

The energy at the mean position of a pendulum will beA. zeroB. partial PE and partial KEC. totally KED. totally PE

The figure shows the displacement-time graph of a particle executing `SHM`. If the time period of oscillation is `2s`, then the equation of motion is given by

Frequency of oscillation is proportional to A. `sqrt((3k)/(m))`B. `sqrt((k)/(m))`C. `sqrt((2k)/(m))`D. `sqrt((m)/(3k))`

The shortest distance travelled by a particle executing SHM from mean position in 2 s is equal to `(sqrt(3)//2)` times its amplitude. Determine its time period.

A ring of mass m and radius r rolls without slipping on a fixed hemispherical surface of radius R as shown. The time period of small oscillations of ring is `2pisqrt((n(R-r))/(3g))` then find the value of n.

A vertical U-tube has one of the its arms bent at `theta` from the vertical and then filled with a liquid of density `rho` up to a height h in the vertical arm. Calculate the period of oscillations of the liquid when disturbed a little and released.

An object of mass `0.2 kg` executes simple harmonic oscillation along the `x - axis`with a frequency of `(25//pi) Hz`. At the position `x = 0.04`, the object has Kinetic energy of `0.5 J` and potential energy `0.4 J. The amplitude of oscillations is……m.A. 0.05B. 0.06C. 0.01D. none of these

An object of mass `0.2 kg` executes simple harmonic oscillation along the `x - axis`with a frequency of `(25//pi) Hz`. At the position `x = 0.04`, the object has Kinetic energy of `0.5 J` and potential energy `0.4 J. The amplitude of oscillations is……m.

A spring mass system has time period of `2` second. What should be the spring constant of spring if the mass of the block is `10 grams` ?

An object of mass 0.2 kg executes simple harmonic oscillation along the x-axis with a frequency `(25)/pi`. At the position x = 0.04m, the object has kinetic amplitude of osciallation is (potential energy is zero mean position).

An object of mass 0.2 kg executes simple harmonic oscillation along the x-axis with a frequency `(25)/pi`. At the position x = 0.04m, the object has kinetic amplitude of osciallation is (potential energy is zero mean position).A. 6cmB. 4cmC. 8 cmD. 2cm

An object of mass `0.2 kg` exectes simple harmonic oscillations alongs `x-`axis with a freqency of `(25//pi) Hz`. At the positions `x = 0.04m`, the object has kinetic energy of `0.5 J` and potential energy `0.4 J`. Find the amplitude of oscillaties

A particle is attached to a vertical spring and is pulled down a distance 0.04m below its equilibrium position and is released from rest. The initial upward acceleration of the particle is `0.30 ms^(-2)`. The period of the oscillation isA. 4.08 sB. 1.92 sC. 3.90 sD. 2.29 s

The instantaneous displacement of a simple pendulum oscillator is given by `x= A cos (omegat+ pi / 4 )`Its speed will be maximum at timeA. `pi/(4 omega)`B. `pi/(2 omega)`C. `(pi)/(omega)`D. `(2pi)/(omega)`

A body executing `SHM` passes through its equilibrium. At this instant, it hasA. maximum potential energyB. maximum kinetic energyC. minimum kinetic energyD. maximum acceleration

Velocity at mean position of a particle executing S.H.M. is v , they velocity of the particle at a distance equal to half of the amplitudeA. 4 vB. 2 vC. `sqrt3/2v`D. `sqrt3/4 v`

For a particle executing SHM, x = displacement from mean position, v = velocity and a = acceleration at any instant, thenA. `v-X` graph is a circleB. `v-X` graph is an ellipseC. `a-X` graph is a straight lineD. `a-v` ggraph is an ellipse

For a particle executing SHM, x = displacement from mean position, v = velocity and a = acceleration at any instant, thenA. `v - x` graph is a circleB. `v - x` graph is an ellipseC. `a - x` graph is a straight lineD. `a - x` graph is a circle