956 + Interview Questions in OSCILLATIONS IN GENERAL AWARENESS Page 9 InterviewSolution

401.	If the length of a simple pendulum is increased to 4 times its initial length, its frequency of oscillation will (A) reduce to half its initial frequency (B) increase to twice its initial frequency (C) reduce to \(\frac14\)th its initial frequency (D) increase to 4 times its initial frequency.
Answer» (A) reduce to half its initial frequency

Discussion

402.	A pendulum clock shows correct time at `0^(@)C`. On a summer dayA. it runs slow and gain timeB. it runs fast and loses timeC. it runs slow and loses timeD. it runs fast and gains time
Answer» Correct Answer - C

Discussion

403.	If the bob of a simple pendulum is made to oscillate in some fluid of density greater than the density of air (density of the bob > density of the fluid), then time period of the pendulum increased or decrease.
Answer» Time period of the pendulum Increased.

Discussion

404.	Pendulum A and B having periodic times 4 s and 4.2 s, are made to oscillate simultaneously. At time t = 0, they are in the same phase. After how many complete oscillations of A, they will be again in the same phase?A. 7B. 14C. 21D. 28
Answer» Correct Answer - C `(T_(2))/(T_(1))=sqrt((g_(1))/(g_(2)))=sqrt((M_(2))/(M_(1))xx((R_(1))/(R_(2)))^(2))=sqrt((1)/(2)xx4)` `T_(2)=T_(1)sqrt(2)" "2sqrt(2)s`.

Discussion

405.	During summer, the time period of the pendulum of a clock changes to 2.02 s from 2 s. The clock runsA. slow by 14.4 minutes per dayB. fast by 14.4 minutes per dayC. slow by 28.8 minutes per dayD. fast by 28.8 minutes per day
Answer» Correct Answer - A In one oscillation pendulum loses time 0.02 s. `therefore` In day `((86400)/(2))`oscillation loses time is, `((86400)/(2))xx0.02=864s` `=(864)/(60)=14.4min`.

Discussion

406.	Identify correct statement among the followingA. the greater the mass of a pendulum bob, the shorter is its frequency of oscillationB. a simple pendulum with a bob of mass m swings with an anguluar amplitude of `40^(@)` its angular amplitude is `20^(@)`, the tension in the string earlier is less than the tension in the string laterC. as the length of a simple pendulum is increased, the maximum velocity of its bob during its oscillations will also increasesD. the fractional change in the time period of a pendulum on changing the temperature is independent of the length of the pendulum
Answer» Correct Answer - B

Discussion

407.	The time period of a simple pendulum is T. When the length is increased by 10 cm, its period is `T_(1)`. When the length is decreased by 10 cm, its period is `T_(2)`. Then, the relation between T, `T_(1)` and `T_(2)` isA. `(2)/(T^(2))=(1)/(T_(1)^(2))+(1)/(T_(2)^(2))`B. `(2)/(T^(2))=(1)/(T_(1)^(2))-(1)/(T_(2)^(2))`C. `2T^(2)=T_(1)^(2)+T_(2)^(2)`D. `2T^(2)=T_(1)^(2)-T_(2)^(2)`
Answer» Correct Answer - C `Tpropsqrt(l)` `T_(1)propsqrt(l+10),T_(2)propsqrt(l-10)` `(T_(1)^(2))/(T^(2))=(l+10)/(l),(T_(2)^(2))/(T^(2))=(l-10)/(l)` `(T_(1)^(2))/(T^(2))+(T_(2)^(2))/(T^(2))=(l+10)/(l)+(l-10)/(l)=2` `(T_(1)^(2))/(T^(2))+(T_(2)^(2))/(T^(2))=2T^(2) therefore T_(1)^(2)+T_(2)^(2)=2T^(2)`

Discussion

408.	If A is amplitude of a particle in SHM, its displacement from the mean position when its kinetic energy is thrice that to its potential energyA. AB. `A//4`C. `A//2`D. `3A//4`
Answer» Correct Answer - C `K.E.=3P.E.` `(1)/(2)momega^(2)(A^(2)-x^(2))=3xx(1)/(2)momega^(2)x^(2)` `A^(2)-x^(2)=3x^(2)` ` A^(2)=Ax^(2)` `A=2x` `x=(A)/(2)`

Discussion

409.	If the length of a simple pendulum is equal to the radius of the earth, its time period will beA. 84 minB. 60 minC. 75 minD. 48 min
Answer» Correct Answer - B `T=2pisqrt((R)/(2g))=(84.6)/(1.414)=59.40` min

Discussion

410.	A simple pendulum whose length is 20 cm is suspended from the ceiling of a lift which is rising up with an acceleration of `3.0ms^(-2)`. Calculate the time-period of the pendulum.
Answer» Correct Answer - 0.79s

Discussion

411.	Simple pendulum is executing simple harmonic motion with time period `T`. If the length of the pendulum is increased by `21%`, then the increase in the time period of the pendulum of the increased length is:A. 0.22B. 0.2C. 0.33D. 0.44
Answer» Correct Answer - B `(T_(2))/(T_(1))=sqrt((l_(2))/(l_(1)))=sqrt(1.44)=1.2` `(T_(2)-T_(1))/(T_(1))=1.2-1=0.2=20%`.

Discussion

412.	A particle oscillates simple harmonically along a straight line with period 8 seconds and amplitude `8 sqrt(2)`m. It starts from the mean position, then the ratio of the distances travelled by it in the second second and first second of its motion isA. `sqrt(2)`B. 2C. `(sqrt(2)-1)`D. `sqrt(3)`
Answer» Correct Answer - C Given, `8 sqrt(2)`m and T = 8 s and `omega = (2pi)/(8)=(pi)/(4)` The particle starts from the mean position. `therefore" "x = A sin omega t` `therefore` The distance travelled by the particle in one second is `therefore" "x_(1)=8 sqrt(2)sin((pi)/(4)xx1)` `= 8 sqrt(2)xx(1)/(sqrt(2))=8m = s_(1)` and the distance travelled by the particle in 2s is `x_(2)=8 sqrt(2)sin((pi)/(4)xx2)=8 sqrt(2)m` `therefore` Distance travelled in second second `(s_(2))` `=x_(2)-x_(1)=8 sqrt(2)-8 =(sqrt(2)-1)m` `therefore" "(s_(2))/(s_(1))=(8 sqrt(2)-1)/(8)=sqrt(2)-1`

Discussion

413.	The bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change if the water begins to drain out of the hollow sphere?
Answer» When the sphere is filled with water, its C.G. at the centre of the sphere. So, let its time period be 'T'. As the water begins to drain out, C.G. of the sphere will be lowered and hence length l will increases. Therefore, time-period increases. When whole of the water is drained out, C.G. again shifts to the centre of the sphere and hence the period will be the same (i.e, T) as and when the sphere was full of water.

Discussion

414.	STATEMENT-1`:` Time period of the liquid executing S.H.M.in a U-tube depends on the area of cross section of U-tube. and STATEMENT-2 `:` The restoring force acting on liquid displaced from equilibrium position of U -tube depends on the difference in levels of liquid in the two limbs of U -tube.
Answer» Correct Answer - 4

Discussion

415.	If the period of oscillation of mass m suspended from a spring is 2s, then the period of mass 4m will beA) 1 s B) 2 s C) 3 s D) 4 s
Answer» B) 2 s Explanations: \(T=2\pi\sqrt{\frac{m}{k}}\) T'= 2sec

415.

If the period of oscillation of mass m suspended from a spring is 2s, then the period of mass 4m will beA) 1 s B) 2 s C) 3 s D) 4 s

Answer»

B) 2 s

Explanations:

\(T=2\pi\sqrt{\frac{m}{k}}\)

T'= 2sec

Discussion

416.	Define Second’s pendulum.
Answer» A second’s pendulum is a pendulum whose time period is two seconds. Its length is 99.3 cm.

Discussion

417.	A particle executes `SHM` with a amplitude `0.5 cm` and frequency `100s^(-1)`. The maximum speed of the particle is (in m/s)A. `pi`B. `0.5`C. `5pi xx 10^(-5)`D. `100 pi`
Answer» Correct Answer - A `V_(max) = A omega = 2pi fA`

Discussion

418.	What is meant by Simple pendulum ?
Answer» A simple pendulum is a heavy point mass suspended by a Weightless, inextensible and a perfectly flexible string from a rigid support about which it can vibrate freely. The distance between the point of suspension and the point of oscillation is called length of the pendulum.

Discussion

419.	When a particle performing uniform circular motion of radius 10 cm undergoes the SHM, what will be its amplitude?A. 10 cmB. 5 cmC. 2.5 cmD. 20 cm
Answer» Correct Answer - A Radius = Amplitude = 10 cm.

Discussion

420.	For a body in `SHM` the velocity is given by the relation `V = sqrt(144-16x^(2))ms^(-1)`. The maximum acceleration isA. `12 m//s^(2)`B. `16 m//s^(2)`C. `36 m//s^(2)`D. `48 m//s^(2)`
Answer» Correct Answer - D `a_(max) = omega^(2)A, V = omega sqrt(A^(2) -x^(2))`

Discussion

421.	The frequency of the seconds pendulum isA. 1 HzB. 2 HzC. 0.5 HzD. 3 Hz
Answer» Correct Answer - C

Discussion

422.	A particle of mass `m` is allowed to oscillate near the minimum of a vertical parabolic path having the equaiton `x^(2) =4ay`. The angular frequency of small oscillation is given by A. `sqrt(ga)`B. `sqrt((g)/(a))`C. `sqrt((g)/(2a))`D. `sqrt((2g)/(a))`
Answer» Correct Answer - 2

Discussion

423.	A body of mass 1 kg suspended by a light vertical spring of spring constant 100 N/m executes SHM. Its angular frequency and frequency areA. `(5)/(pi)` rad/s, 100 HzB. 10 rad/s, `(pi)/(5)` HzC. 10 rad/s, `(10)/(pi)` HzD. 10 rad/s, `(5)/(pi)` Hz
Answer» Correct Answer - D `omega=sqrt((K)/(m))=sqrt((100)/(1))=10` rad/s `n=(omega)/(2pi)=(10)/(2pi)=(5)/(pi)Hz`

Discussion

424.	Two particles executes S.H.M. of the same amplitude and frequency along the same straight line. They pass one another when going in opposite directions, each time their displacement is root three by two times amplitude. The phase difference between them isA. `(2pi)/(3)` radB. `(pi)/(2)` radC. `(pi)/(6)` radD. `(pi)/(3)` rad
Answer» Correct Answer - D `x_(1)=Asinomegat` `omegat=sin^(-1)((x)/(A))=sin^(-1)((sqrt(3))/(2))` `delta_(1)=omegat=(pi)/(3)` `deltatheta=pi-2delta=pi-(2pi)/(3)=(pi)/(3)`

Discussion

425.	In S.H.M., the phase difference between the displacement and velocity of a particle, at any isntant isA. `pi`B. `2pi`C. `pi//2`D. 0
Answer» Correct Answer - C Phase different between displacement and velocity is `(pi)/(2)` rad.

Discussion

426.	A small spherical steel ball is placed a little away from the centre of a large concave mirror of radius of curvature 2.5 m, if the ball is released, then the period of the motion will be `(g=10m//s^(2))`A. `pi//4 s`B. `pi//2 s`C. `pi//s`D. `2pi s`
Answer» Correct Answer - C `T=2pisqrt((R)/(g))=2pisqrt((2.5)/(10))=(2pixx5)/(10)=pis`

Discussion

427.	The phase of a particle executing simple harmonic motion is `pi/2` when it hasA. maximum velocityB. minimum accelerationC. maximum kinetic energyD. maximum displacement
Answer» Correct Answer - D

Discussion

428.	The phase quantity depends uponA. time and displacementB. time, direction and displacementC. displacement and directionD. displacement
Answer» Correct Answer - B

Discussion

429.	The amplitude of particle performing S.H.M. isA. tensorB. vectorC. scalarD. depending upon magnitude
Answer» Correct Answer - B

Discussion

430.	Acceleration amplitude of a particle performing S.H.M. is the product ofA. amplitude and velocityB. amplitude and accelerationC. amplitude and square of angular velocityD. square of amplitude and angular velocity
Answer» Correct Answer - C

Discussion

431.	A particle performing S.H.M. about equilibrium position. Then the velocity of the particle isA. slower from mean to extreme positionB. faster from mean to extreme positionC. slower initial from mean position and faster lateral and stop at the endD. faster initil from mean position and later on falls off, suddenly
Answer» Correct Answer - D

Discussion

432.	The acceleration of a particle performing S.H.M. at mean position isA. minimum or zeroB. constantC. maximumD. half of the maximum
Answer» Correct Answer - A

Discussion

433.	The acceleration of a particle performing S.H.M. at extreme position isA. minimumB. constantC. maximumD. in between maximum and minimum
Answer» Correct Answer - C

Discussion

434.	A particle performing S.H.M., its velocity when the particle moves from mean to extreme position isA. slower initially and faster laterallyB. uniformly movesC. faster initially and momentraily falls to zeroD. fast moves and stop at extreme position
Answer» Correct Answer - C

Discussion

435.	If a watch with a wound spring is taken on to the moon, itA. runs fasterB. runs slowerC. does not workD. shows no change
Answer» Correct Answer - D

Discussion

436.	Particle moves from extreme position to mean position, itsA. kinetic energy increases, potential energy decreasesB. kinetic energy decreases, potential increasesC. both remains constantD. potential energy becomes zero and kinetic energy remains contant
Answer» Correct Answer - A

Discussion

437.	A magnet of magnetic moment M oscillates in magnetic field B with time period 2 sec. If now the magnet is cut into two half pieces parallel to the axis, then what is new time period is only one part oscillate in field?A. 2 sB. `2 sqrt(2) s`C. `(1)/(sqrt(2)) s`D. `2.4 s`
Answer» Correct Answer - A

Discussion

438.	The phase angle between the two projections of uniform circular motion on any two mutually perpendicular diameter isA. zeroB. `pi//2`C. `3pi//4`D. `pi`
Answer» Correct Answer - B `x=Asinomegat" ... (i)"` `y=Asin(omegat+(pi)/(2))` `=Acosomegat" ... (ii)"` Squaring and adding equation (i) and (ii), `x^(2)+y^(2)=A^(2)(sin^(2)omegat+cos^(2)omegat)` `x^(2)+y^(2)=A^(2)` This is standard equation of circle.

Discussion

439.	1 kg weight is suspended to a weightless spring and it has time period T. If now 4 kg weight is suspended from the same spring the time period will beA. TB. `T//2`C. `2T`D. `4T`
Answer» Correct Answer - C `T_(1)=2pisqrt((m_(1))/(k))andT_(2)=2pisqrt((m_(2))/(k))` `(T_(2))/(T_(1))=sqrt((m_(2))/(m_(1)))=sqrt((4)/(1))` `T_(2)=2T_(1)`

Discussion

440.	Two simple harmonic motions given by, `x = a sin (omega t+delta)` and `y = a sin (omega t + delta + (pi)/(2))` act on a particle will beA. circular anti-clockwiseB. circular clockwiseC. elliptical anti-clockwiseD. elliptical clockwise
Answer» Correct Answer - B

Discussion

441.	A pendulum is displaced to an angle `theta` from its equilibrium position, then it will pass through its mean position with a velocity v equal toA. `sqrt(2g l)`B. `sqrt(2g l sin theta)`C. `sqrt(2 g l cos theta)`D. `sqrt(2gl(1-cos theta))`
Answer» Correct Answer - D

Discussion

442.	The velocity of sound in a tube containing air at 27°C and pressure of 76 cm of Hg is 330 ms–1. What will be its velocity, when pressure is increased to 152 cm of mercury and temperature is kept constant ?
Answer» At a given temperature, the velocity of sound is independent of pressure, so velocity of sound in tube will remain 330 ms^–1.

Discussion

443.	Which of the following functions of time represents (a) simple harmonic motion and (b) periodic motion? Given the period for each case. i. `Sin omega t - cos omegat` ii. `Sin^(2)omegat`
Answer» i. `sin omegat - cos omegat = sqrt(2) [(1)/(sqrt(2))Sin omegat-(1)/(sqrt(2))Cos omegat]` `= sqrt(2) [Sin omegat cos.(pi)/(4) -cos omega t "sin"(pi)/(4)]` `= sqrt(2) sin (omegat - pi//4]` This fuctions represent a simple harmonic motion having a period `T = (2pi)/(omega)` and a phase angle `(-pi//4)` or `(7pi//4)`. ii) `Sin^(2) omegat =(1-cos.(2omegat))/(2) = (1)/(2) -(1)/(2) cos (2 omegat)` The function is periodic having a period `T = pi//omega` It is not `SHM` Note: If displacement of a particle moving along straight line is given by `y = A sin^(2)omegat` then it represents `SHM`.

Discussion

444.	For a particle executing `S.H.M.,` the kinetic energy `K` is given `K = K_(0) cos ^(2)omega t`. The maximum value of potential energy is:A. `K_(o)`B. `2K_(o)`C. `K_(o)//2`D. `4K_(o)`
Answer» Correct Answer - A

Discussion

445.	If the length of the seconds pendulum is increases by 4%, how many seconds it will lose per day?A. 3456 sB. 864 sC. 432 sD. 1728 s
Answer» Correct Answer - D `l_(2)=1.02l_(1)(T_(2))/(T_(1))=sqrt(1.02)` `=(1+0.02)^(1//2)` `(T_(2)-T_(1))/(T_(1))=0.01` `T_(2)-T_(1)=0.02` In one oscillation it looses 0.02 `therefore 43200.......? 0.04xx43200=1728s`

Discussion

446.	A particle executes `SHM` with a time period `T`. The time period with which its potential energy changes isA. `2T`B. `T`C. `T//2`D. `3T//2`
Answer» Correct Answer - C

Discussion

447.	If the length of seconds pendulum is decreased by 1%, the gain or lose time per day by the pendulum will beA. gain 432 sB. lose 443 sC. gain 4.4 sD. lose 0.44 s
Answer» Correct Answer - A `l_(2)=(l_(1)-0.01)` `T_(1)=2pisqrt((l)/(g))andT_(2)=2pisqrt((0.99)/(g))` `(T_(2))/(T_(1))=sqrt((0.99)/(1))` `therefore T_(2)=2sqrt(99xx10^(-2))=2xx9.95xx10^(-1)` `T_(2)=1.990s` In 1 osci. Gain of time is, `2-1.990=0.01` s `therefore` In one day i.e. 43200 oscillations The time gain is 43200 `xx` 0.01 = 432

Discussion

448.	A simple pendulum simple harmonic motion about `x = 0` with an amplitude a and time period `T` speed of the pendulum at `x = a//2` will beA. `pi A sqrt(3) / T`B. `pi A/T`C. `pi A sqrt(3) / 2T`D. `3pi^(2) A/T`
Answer» Correct Answer - A `v=omegasqrt(A^(2)-x^(2))=omegasqrt(A^(2)-(A^(2))/(4))` `=omegasqrt((4A^(2)-A^(2))/(2))=(Aomega)/(2)sqrt(3)` `=(sqrt(3))/(2)A(2pi)/(T)=piAsqrt(3)//T`

Discussion

449.	A simple pendulum, suspended from the coiling of a lift, has a period of oscillation T, when the lift is at rest. If the lift starts moving upwards with an acceleration a =3g, then the new period will beA. T/2B. 2TC. T/3D. 3T
Answer» Correct Answer - A The period of pendulum when lift is stationary `T=2pisqrt((L)/(g))` When the lift is moving upwards, the acceleration of the lift increases. `therefore T_(1)=2pisqrt((L)/(g+a))=2pisqrt((L)/(g+3g))=2pisqrt((L)/(4g))` `(T_(1))/(T)=sqrt((g)/(4g))=(1)/(2) therefore T_(1)=(T)/(2)`

Discussion

450.	A seconds pendulum is suspended form the roof of a lift. If the lift is moving up with an acceleration `9.8 m//s^(2)`, its time period isA. `1s`B. `sqrt(2)`C. `(1)/(sqrt(2))s`D. `2sqrt(2)s`
Answer» Correct Answer - B `T = 2pi sqrt((l)/(g+a))`

Discussion

Explore topic-wise InterviewSolutions in .

If the length of a simple pendulum is increased to 4 times its initial length, its frequency of oscillation will (A) reduce to half its initial frequency (B) increase to twice its initial frequency (C) reduce to \(\frac14\)th its initial frequency (D) increase to 4 times its initial frequency.

A pendulum clock shows correct time at `0^(@)C`. On a summer dayA. it runs slow and gain timeB. it runs fast and loses timeC. it runs slow and loses timeD. it runs fast and gains time

If the bob of a simple pendulum is made to oscillate in some fluid of density greater than the density of air (density of the bob &gt; density of the fluid), then time period of the pendulum increased or decrease.

Pendulum A and B having periodic times 4 s and 4.2 s, are made to oscillate simultaneously. At time t = 0, they are in the same phase. After how many complete oscillations of A, they will be again in the same phase?A. 7B. 14C. 21D. 28

During summer, the time period of the pendulum of a clock changes to 2.02 s from 2 s. The clock runsA. slow by 14.4 minutes per dayB. fast by 14.4 minutes per dayC. slow by 28.8 minutes per dayD. fast by 28.8 minutes per day

If A is amplitude of a particle in SHM, its displacement from the mean position when its kinetic energy is thrice that to its potential energyA. AB. `A//4`C. `A//2`D. `3A//4`

If the length of a simple pendulum is equal to the radius of the earth, its time period will beA. 84 minB. 60 minC. 75 minD. 48 min

A simple pendulum whose length is 20 cm is suspended from the ceiling of a lift which is rising up with an acceleration of `3.0ms^(-2)`. Calculate the time-period of the pendulum.

Simple pendulum is executing simple harmonic motion with time period `T`. If the length of the pendulum is increased by `21%`, then the increase in the time period of the pendulum of the increased length is:A. 0.22B. 0.2C. 0.33D. 0.44

A particle oscillates simple harmonically along a straight line with period 8 seconds and amplitude `8 sqrt(2)`m. It starts from the mean position, then the ratio of the distances travelled by it in the second second and first second of its motion isA. `sqrt(2)`B. 2C. `(sqrt(2)-1)`D. `sqrt(3)`

The bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change if the water begins to drain out of the hollow sphere?

STATEMENT-1`:` Time period of the liquid executing S.H.M.in a U-tube depends on the area of cross section of U-tube. and STATEMENT-2 `:` The restoring force acting on liquid displaced from equilibrium position of U -tube depends on the difference in levels of liquid in the two limbs of U -tube.

If the period of oscillation of mass m suspended from a spring is 2s, then the period of mass 4m will beA) 1 s B) 2 s C) 3 s D) 4 s

Define Second’s pendulum.

A particle executes `SHM` with a amplitude `0.5 cm` and frequency `100s^(-1)`. The maximum speed of the particle is (in m/s)A. `pi`B. `0.5`C. `5pi xx 10^(-5)`D. `100 pi`

What is meant by Simple pendulum ?

When a particle performing uniform circular motion of radius 10 cm undergoes the SHM, what will be its amplitude?A. 10 cmB. 5 cmC. 2.5 cmD. 20 cm

For a body in `SHM` the velocity is given by the relation `V = sqrt(144-16x^(2))ms^(-1)`. The maximum acceleration isA. `12 m//s^(2)`B. `16 m//s^(2)`C. `36 m//s^(2)`D. `48 m//s^(2)`

The frequency of the seconds pendulum isA. 1 HzB. 2 HzC. 0.5 HzD. 3 Hz

A particle of mass `m` is allowed to oscillate near the minimum of a vertical parabolic path having the equaiton `x^(2) =4ay`. The angular frequency of small oscillation is given by A. `sqrt(ga)`B. `sqrt((g)/(a))`C. `sqrt((g)/(2a))`D. `sqrt((2g)/(a))`

A body of mass 1 kg suspended by a light vertical spring of spring constant 100 N/m executes SHM. Its angular frequency and frequency areA. `(5)/(pi)` rad/s, 100 HzB. 10 rad/s, `(pi)/(5)` HzC. 10 rad/s, `(10)/(pi)` HzD. 10 rad/s, `(5)/(pi)` Hz

In S.H.M., the phase difference between the displacement and velocity of a particle, at any isntant isA. `pi`B. `2pi`C. `pi//2`D. 0

A small spherical steel ball is placed a little away from the centre of a large concave mirror of radius of curvature 2.5 m, if the ball is released, then the period of the motion will be `(g=10m//s^(2))`A. `pi//4 s`B. `pi//2 s`C. `pi//s`D. `2pi s`

The phase of a particle executing simple harmonic motion is `pi/2` when it hasA. maximum velocityB. minimum accelerationC. maximum kinetic energyD. maximum displacement

The phase quantity depends uponA. time and displacementB. time, direction and displacementC. displacement and directionD. displacement

The amplitude of particle performing S.H.M. isA. tensorB. vectorC. scalarD. depending upon magnitude

Acceleration amplitude of a particle performing S.H.M. is the product ofA. amplitude and velocityB. amplitude and accelerationC. amplitude and square of angular velocityD. square of amplitude and angular velocity

The acceleration of a particle performing S.H.M. at mean position isA. minimum or zeroB. constantC. maximumD. half of the maximum

The acceleration of a particle performing S.H.M. at extreme position isA. minimumB. constantC. maximumD. in between maximum and minimum

A particle performing S.H.M., its velocity when the particle moves from mean to extreme position isA. slower initially and faster laterallyB. uniformly movesC. faster initially and momentraily falls to zeroD. fast moves and stop at extreme position

If a watch with a wound spring is taken on to the moon, itA. runs fasterB. runs slowerC. does not workD. shows no change

Particle moves from extreme position to mean position, itsA. kinetic energy increases, potential energy decreasesB. kinetic energy decreases, potential increasesC. both remains constantD. potential energy becomes zero and kinetic energy remains contant

A magnet of magnetic moment M oscillates in magnetic field B with time period 2 sec. If now the magnet is cut into two half pieces parallel to the axis, then what is new time period is only one part oscillate in field?A. 2 sB. `2 sqrt(2) s`C. `(1)/(sqrt(2)) s`D. `2.4 s`

The phase angle between the two projections of uniform circular motion on any two mutually perpendicular diameter isA. zeroB. `pi//2`C. `3pi//4`D. `pi`

1 kg weight is suspended to a weightless spring and it has time period T. If now 4 kg weight is suspended from the same spring the time period will beA. TB. `T//2`C. `2T`D. `4T`

Two simple harmonic motions given by, `x = a sin (omega t+delta)` and `y = a sin (omega t + delta + (pi)/(2))` act on a particle will beA. circular anti-clockwiseB. circular clockwiseC. elliptical anti-clockwiseD. elliptical clockwise

A pendulum is displaced to an angle `theta` from its equilibrium position, then it will pass through its mean position with a velocity v equal toA. `sqrt(2g l)`B. `sqrt(2g l sin theta)`C. `sqrt(2 g l cos theta)`D. `sqrt(2gl(1-cos theta))`

The velocity of sound in a tube containing air at 27°C and pressure of 76 cm of Hg is 330 ms–1. What will be its velocity, when pressure is increased to 152 cm of mercury and temperature is kept constant ?

Which of the following functions of time represents (a) simple harmonic motion and (b) periodic motion? Given the period for each case. i. `Sin omega t - cos omegat` ii. `Sin^(2)omegat`

For a particle executing `S.H.M.,` the kinetic energy `K` is given `K = K_(0) cos ^(2)omega t`. The maximum value of potential energy is:A. `K_(o)`B. `2K_(o)`C. `K_(o)//2`D. `4K_(o)`

If the length of the seconds pendulum is increases by 4%, how many seconds it will lose per day?A. 3456 sB. 864 sC. 432 sD. 1728 s

A particle executes `SHM` with a time period `T`. The time period with which its potential energy changes isA. `2T`B. `T`C. `T//2`D. `3T//2`

If the length of seconds pendulum is decreased by 1%, the gain or lose time per day by the pendulum will beA. gain 432 sB. lose 443 sC. gain 4.4 sD. lose 0.44 s

A simple pendulum simple harmonic motion about `x = 0` with an amplitude a and time period `T` speed of the pendulum at `x = a//2` will beA. `pi A sqrt(3) / T`B. `pi A/T`C. `pi A sqrt(3) / 2T`D. `3pi^(2) A/T`

A simple pendulum, suspended from the coiling of a lift, has a period of oscillation T, when the lift is at rest. If the lift starts moving upwards with an acceleration a =3g, then the new period will beA. T/2B. 2TC. T/3D. 3T

A seconds pendulum is suspended form the roof of a lift. If the lift is moving up with an acceleration `9.8 m//s^(2)`, its time period isA. `1s`B. `sqrt(2)`C. `(1)/(sqrt(2))s`D. `2sqrt(2)s`

If the bob of a simple pendulum is made to oscillate in some fluid of density greater than the density of air (density of the bob > density of the fluid), then time period of the pendulum increased or decrease.