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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
A spring mass system preforms `S.H.M` if the mass is doubled keeping amplitude same, then the total energy of `S.H.M` will become :A. doubleB. halfC. inchangedD. 4 times |
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Answer» Correct Answer - C `E=(1)/(2)mA^2omega^2` `=(1)/(2)mA^2((k)/(m))` `=(1)/(2)A^2k` (independent of mass) |
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| 52. |
A particle executing SHM of amplitude 4 cm and `T=4 s` .The time taken by it to move from position to half the amplitude isA. `1 sec `B. `1// sec`C. `2//3sec `D. `sqrt(3//2) sec` |
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Answer» Correct Answer - C |
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| 53. |
Two particles execute `SHM` of same amplitude of `20 cm` with same period along the same line about the same equilibrium position. If phase difference is `pi//3` then the maximum distance between these two will beA. `10 cm `B. `20 cm`C. `10sqrt2 cm `D. `20 sqrt2 cm ` |
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Answer» Correct Answer - A |
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| 54. |
A point particle if mass `0.1 kg` is executing SHM of amplitude `0.1 m`. When the particle passes through the mean position, its kinetic energy is `8 xx 10^(-3)J`. Write down the equation of motion of this particle when the initial phase of oscillation is `45^(@)`. |
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Answer» Given that amplitude `a = 0.1 m, m = 0.1 kg phi = 45^(@) = (pi//4)` rad , so the wquatude of SHM will be `y = 0.1 sin [(omega t + (pi//4)]` (i) Now as in SHM, `KE` is given by `K = (1)/(2) m omega^(2) (A^(2) - y^(2))` which according to the given problem is `8 xx 10^(-3) = (1)/(2) xx 0.1 xx omega^(2) (0.1^(2) - 0^(2))`, i.e. `omega = 4 ras//s` (ii) Sumstituting the value of `omega` from Eq (ii) in Eq (i), we get `y = 0.1 sin [4 t + (pi//4)]` |
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| 55. |
A block of mass `1kg` hangs without vibrating at the end of a spring whose force constant is `200(N)/(m)` and which is attached to the ceiling of an elevator. The elevator is rising with an upward acceleration of `(g)/(3)` when the acceleration suddenly ceases. The angular frequency of the block after the acceleration ceases isA. `13(rad)/(s)`B. `14(rad)/(s)`C. `15(rad)/(s)`D. none of these |
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Answer» Correct Answer - B The angular frequency is `omega=sqrt((k)/(m))=sqrt((200)/(1))approx14(rad)/(s)` |
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| 56. |
A uniform of radius `5.0 cm` and mass `200 g` is fixed at its centre to a matel wire , the other end of which is fixed to a celling. The hanging dise is rotate obout the wire through an angle and is released. If the dise makes torsional oscillations with time period `0.20 s` , find the torsional constant of the wire. |
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Answer» The situation is shown in figure . The moment of inertia of the about the wire is `l = mr^((2)/(2)) = ((0.200 kg) (5.0 xx 10^(-2) m)^(2))/(2)` `= 2.5 xx 10^(-4) kg-m^(2)` The time period is given by `T = 2 pi sqrt((l)/(k)) or k = sqrt((4 pi^(2) l)/(T^(2))` `= (4 pi^(2) (2.5 xx 10^(-4) lg-m^(2)))/((0.20s)^(2) = 0.25 kg-m^(2)//s^(2)` |
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| 57. |
A vertical spring carries a `5kg` body and is hanging in equilibrium an additional force is applied so that the spring is further stretched. When released from this position. It performs 50 complete oscillation in 25 s, with an amplitude of 5 cm. The additional force applied isA. 80 NB. `80pi^2N`C. `4pi^2N`D. `4N` |
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Answer» Correct Answer - C `T=(25)/(50)=(1)/(2)simpliesomega=(2pi)/(T)=4pi(rad)/(s)` Spring constant `k=momega^2=5xx(4pi)^2=80pi^2(N)/(m)` Force required to stretch the spring by 5 cm is `F=kx=80pi^2xx0.05N=4pi^2` |
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| 58. |
Two blocks connected by a spring rest on a smooth horizontal plane as shown in Fig. A constant force `F` start acting on block `m_2` as shown in the figure. Which of the following statements are not correct?A. Length of the spring increases continuously if `m_1gtm_2`.B. Blocks start performing SHM about centre of mass of the system, which moves rectilinearly with constant acceleration.C. Blocks start performing oscillation about centre of mass of the system with incresing amplitude.D. Acceleration of `m_2` is maximum at initial moment of time only. |
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Answer» Correct Answer - A::C::D The only external horizontal force acting on the system of the two blocks and the spring is `F`. Therefore, acceleration of the centre of mass of the system is equal to `(F)/(m_1)+m^2`. Hence, centre of mass of the system moves with a constant acceleration Initially, there is no tenstion in the spring. Therefore at initial moment `m_2` has an acceleration `(F)/(m_2)` and it starts to move to the right. Due to its motion, the spring elongates and a tension is developed. Therefore, acceleration of `m_2` decreases while that of `m_1` increases from zero initial value. The blocks start to perform SHM about their centre of mass and the centre of mass moves with the acceleration calculated above. Hence Option (b) is correct. Since the block perform SHM about centre of mass, therefore the length of the spring varies periodically. Hence option (a) is wrong. Since magnitude of the force F remains contant therefore amplitude of oscillation also remains constant. So, option (c ) is also wrong. Acceleration of `m_2` is maximum at the instant when the spring in its manimum possible length, which is equal to its natural length. Hence, at initial moments, acceleration of `m_2` is maximum possible. The spring is in its natural length not only at initial moment but at time `t=T`, `T=2`, `3T` .. also, where T is the period of oscillation, hence option (d) is wrong. |
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| 59. |
A body of mass `m` is released from a height h to a scale pan hung from a spring. The spring constant of the spring is `k`, the mass of the scale pan is negligible and the body does not bounce relative to the pan, then the amplitude of vibration isA. `(mg)/(k)sqrt(1-(2hk)/(mg))`B. `(mg)/(k)`C. `(mg)/(k)+(mg)/(k)sqrt(1+(2hk)/(mg))`D. `(mg)/(k)-(mg)/(k)sqrt(1-(2hk)/(mg))` |
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Answer» Correct Answer - C Decreases in potential energy of the mass when the pan gets lowered by distance `y` (due to mass hitting on the pan) `=mg(h+y)`, where `h` is the height through which the mass falls on the pan. Increases in elastic potential of the spring `=(1)/(2)ky^2` (according to low of conservation of energy) or `mg(h+y)=(1)/(2)ky^2` or `ky^2-2mgy-2mgh=0` `y=(2mg+-sqrt(4m^2g^2+8mghk))/(2k)` `=(mg)/(k)+-(mg)/(k)sqrt((1+(2hk)/(mg))` Velocity of the pan will be maximum at the time of collision and will be zero at the lowet position. Hence y should be the amplitude of oscillation. So, amplitude of vibration`=[(mg)/(k)+(mg)/(k)sqrt((1+(2hk)/(mg)))]` |
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| 60. |
An object of mass 4 kg is attached to a spring having spring constant `100(N)/(m)`. It performs simple harmonic motion on a smooth horizontal surface with an amplitude of 2 m. A 6 kg object is dropped vertically onto the 4 kg object when it crosses the mean position, and sticks to it. the change in amplitude of oscillation due to collision isA. `1m`B. `zero`C. `2[1-sqrt((2)/(5))]`D. `2[1-(1)/(sqrt5)]` |
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Answer» Correct Answer - C Time period of the system (object of mass 4 kg) before collision is `T_1=2pisqrt((4)/(100))` After collision, time period of the combined mass is `T_2=2pisqrt((10)/(100))` We can apply momentum conservation for just before the collision and just after the collision in the horizontal direction. `4A_1omega_1=10A_2omega_2impliesA(4xx2xxsqrt((100)/(4)))/(10xxsqrt((100)/(10)))=2sqrt((2)/(5))`m So change in amplitude, `triangleA=A_1-A_2=2[1-sqrt((2)/(5))]`m |
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| 61. |
A particle is perfroming simple harmoic motion along ` x-"axis"` with amplitude 4 cm and time period `1.2 sec`.The minimum time taken by the particle to move from x=2 cm to `x= + 4 cm` and back again is given byA. `0.6 sec`B. `0.4 sec`C. `0.3 sec`D. `0.2 sec` |
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Answer» Correct Answer - B |
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| 62. |
If a partical moves in a potential energy held `U = U_(0) - ax + bx^(2)`, where are a and b partical constents obtian an expression for the force acting on if as a function of position. At what point does the force vanish? Is this a point of stable equilibriun ? Calculate the force constant and friquency of the partical. |
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Answer» `F = - (dU)/(dx) = a - 2b x` `F = 0, at x = a//2b` If stable equilibrum is present at this position, `(d^(2(U)/(dx^(2))) gt 0)` Hence, `(d^(2(U)/(dx^(2))) = 2b gt 0)` i.e., `x = (a)/(2 b)` is a point of minimum potential energy Hence, the equilibrium is stable. The partical will oscillate about `x = (a)/(2 b)` The effective force constant of oscillation `K_(eff) = ((d^(2)U)/(dx^(2))|_(x= (a)/(2b))) = 2b` As `K_(eff) = m omeg^(2) = 2b = m omeg^(2)` or `omega^(2) = (2 b)/(m) implies omega = 2 pi f = sqrt((2b)/(m)) implies f = (1)/(2pi) sqrt((k)/(m))` |
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| 63. |
A particle of mass m moves in the potential energy U shoen above. The period of the motion when the particle has total energy E isA. `2pisqrt((m)/(k))+4sqrt((2E)/(mg^2))`B. `2pisqrt((m)/(k))`C. `pisqrt((m)/(k))+2sqrt((2E)/(mg^2))`D. `2sqrt((2E)/(mg^2))` |
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Answer» Correct Answer - C For `xlt0` `F=-(dU)/(dx)=-kx` `ma=-kx` or `a=-(k)/(m)x` `-omega_1^2x=-(k)/(m)x` `omega_1=sqrt((k)/(m))` `T_1=2pisqrt((m)/(k))` For `xgt0` `U=mgx` `F=-(dv)/(dx)=-mg` But `E=(1)/(2)mv_0^2` `v_0=sqrt((2E)/(m))` It is speed at lowest point `T_2=(2v_0)/(g)=(2)/(g)sqrt((2E)/(m))` `T=(T_1)/(2)+T_2=pisqrt((m)/(k))+(2)/(g)sqrt((2E)/(m))` |
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| 64. |
The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by `U=5x(x-4)J`, where x is in meter. It can be concluded thatA. the particle is acted upon by a constant forceB. the speed of the particle is maximum at `x=2m`C. the particle executes SHMD. the period of oscillation of the particle `((pi)/(5))` s |
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Answer» Correct Answer - B::C::D `U=5x(x-4)=5x^2-20x` `F=-(dU)/(dx)=-10x+20` i.e., force is not contant. KE or speed of the particle will be maximum at the mean position where force becomes zero. `F=0` or `x=2m` Acceleration experienced by the particle is `a=(F)/(m)=(-10x+20)/(0.1)=-(100x-200)` i.e., particle executes SHM. As `omega^2=100` `omega=10` Hence `T=(2pi)/(omega)=(pi)/(5)s` |
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| 65. |
The potential energy U of a body of unit mass moving in one dimensional conservative force field is given by `U=x^2-4x+3`. All units are is SI. For this situation mark out the correct statement (s).A. The body will perform simple harmonic motion about `x=2` units.B. The body will perform oscillatory motion but not simple harmonic motion.C. The body will perform simple harmonic motion with time period `sqrt2pis`.D. If speed of the body at equilibrium position is `4(m)/(s)`, then the amplitude of oscillation would be `2sqrt2`m |
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Answer» Correct Answer - A::C::D `U=x^2-4x+3` and `F=-(dU)/(dx)=-(2x-4)` At equilibrium position `F=0`, so `x=2cm` Let the particle is displaced by `trianglex` from equilibrium position i.e, from `x=2`, then restoring force on body is `F=-2(2+trianglex)+4=-2trianglex` i.e., `Falpha-trianglex`, so performs simple harmonic motion about `x=2m` Time period `T=2pisqrt((m)/(k))=2pisqrt((1)/(2))=sqrt2pis` From energy conservation, `(mv_(max)^2)/(2)+U_(min)=U_(max)` `(1xx4^2)/(2)+(2^2-4xx2+3)=(A+2)^2-4(A+2)+3` Where A is amplitude solving the above equation we get `A=2sqrt2m`. |
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| 66. |
The number of independent constituent simple harmonic motions yielding a resultant displacement equation of the periodic motion as `y=8sin^2((t)/(2))sin(10t)` isA. 8B. 6C. 4D. 3 |
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Answer» Correct Answer - D `y=8sin^2((1)/(2))sin(10t)` `=4[1-cost]sin(10t)` (using `2sin^2(theta)/(2)=1-costheta`) `=4sin(10t)-4sin(10t)cost` `=4sin(10t)-2[sin11t+sin9t]` (using `2sinCcosD=sin(C+D)+sin(C-D))` `=4sin(10t)-2sin(11t)-2sin(9t)` Evindently, y is obtained as the superimposition of three independent (i.e., having different anglar frequency `omega`) `SHM`s |
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| 67. |
For a simple harmonic motion with given angular frequency `omega`, two arbitrary initial conditions are necessary and sufficient to determine the motion completely. These initial conditions may beA. initial position and initial velocityB. amplitude and initial phaseC. total energy of oscillation and amplitudeD. total energy of oscillation and initial phase. |
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Answer» Correct Answer - A::B::D Description of motion is completely specified if we know the variation of x as a function of time. For simple harmonic motion, the general equation of motion is `x=A(omegat+delta)`. As `omega` is given to describe the motion completely, we need the values of A and `delta`. From options (b) and (d) we can have the value of A and `delta` derectly. For option (a) we can find A and `delta` if we know initial velocity and initial position. Option (c ) cannot give the value of A and `delta` so it is not the correct condition. |
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| 68. |
If two SHMs are repersented by `y_(1) = 10 sin (4 pi + pi//2)` and `y_(2) = 5 (sin 2 pi t +sqrt 8 cos 2 pi t)`, compare their amplitudes . |
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Answer» For the equation `y_(1) = 10 sin (4 pi t + pi//2)`, the amplitude `a_(1) = 10 units` For equation `y_(2) = 5 [sin 2 pi t + sqrt8 cos 2 pi t]` multplying and dividing by `sqrt(1 + 8) = 3` `y_(2) = [15 sin (2 pi t) xx (1)/(3) + cos (2 pi t) (sqrt8)/(3)]` `= 15 sin (2 pi t + phi)` where `tan phi = sqrt8` whose amplitude `a_(2) is 15 units`. Therefore , the required ratio `= (a_(1))/(a_(2)) = (10)/(15) = (2)/(3)` |
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| 69. |
As the expression in ivolving sine function , which of the following equations does not represent a simple harmonic motion ?A. `y = a sin omegat`B. `y=a tan omegat`C. `y=a cos omegat`D. `y= a sin omegat +b cos omegat` |
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Answer» Correct Answer - D |
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| 70. |
Two simple harmonic motion are represent by the following equations `y_(1) = 10 sin (pi//4) (12 t + 1)` `y_(2) = 5 (sin 3 theta t + sqrt3 cos 3 theta t)` Here `t` is in seconds. Find out the ratio of their amplitudes.What are the time period of the two motion? |
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Answer» Given equation are `y_(1) = 10 sin (pi//4) (12t + 1)` (i) `y_(2) = 5 (sin 3 pi t + sqrt3 cos 3 pi t)` (ii) We recast these in the form of standard equation of SHM which is `y = A sin (omega t + phi)` (iii) Equation (i) may be written as `y_(1) = 10 sin [(12 pi//4) + (pi//4)]` `= 10 sin (3 pit + pi//4)]` (iv) Comparing Eq (iv) with Eq (iii), we have Amplitude of first SHM `= A_(1) = 10 cm//s and omega_(1) = 3 pi` `:.` Time period of first motion `T_(1) = 2 pi // omega_(1) = 2pi//3pi = (2//3) s` Let us put `5 = A_(2) cos phi` (v) and `5sqrt3 = A_(2) sin phi` (vi) Then `y_(2) = A_(2) cos phi sin (3 pi t + A_(2) sin phi cos 3 pi t)` `= A_(2) sin (3 pi t+ phi)` (vii) Squaring (5) and (6) and adding , we have `A_(2) = sqrt([(5)^(2) + (5sqrt3)^(2)]) = 10 cm` i.e., amplitude of second SHm is `10 cm` and time period of second AHM is `T_(2)` `2 pi// omeag_(2) = 2 pi//3 pi = 2//3 s` Thus, the ratio of amplitudes `A_(1) : A_(2) = 1:1` and perodic times are `T_(1) =T_(2) = (2//3) s` |
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| 71. |
One end of a spring of force constant K is fixed to a vertical wall and the other to a body of mass m resting on a smooth horizontal surface. There is another wall at a distance `x_0` from the body. The spring is then compressed by `3x_0` and released. The time taken to strike the wall from the instant of release is (given `sin^-1((1)/(3))=((pi)/(9))`)A. `(pi)/(6)sqrt((m)/(K))`B. `(2pi)/(3)sqrt((m)/(K))`C. `(pi)/(4)sqrt((m)/(K))`D. `(11pi)/(18)sqrt((m)/(K))` |
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Answer» Correct Answer - D When spring is compressed by `3x_0`. Amplitude `A=3x_0` The time taken from extreme compressed position to mean position `t_1=(T)/(4)` If time taken `(t_2)` from mean position to `x=x_0` is given `x=Asin(2pit_2)/(T)impliesx_0=3x_0sin(2pit_2)/(T)` `sin((2pit_2)/(T))` `T=(1)/(3)implies(2pit_2)/(T)=(pi)/(9)impliest_2=(T)/(18)` `t_1+t_2=(T)/(4)+(T)/(18)=(11)(18)T=(11)/(18)2pisqrt((m)/(K))=(11)/(9)pisqrt((m)/(K))` |
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| 72. |
One end of a spring of force constant k is fixed to a vertical wall and the other to a blcok of mass m resting on a smooth horizontal surface. There is another wall at distance `x_(0)` from the block. The spring is then compressed by `2x_(0)` and released. The time taken to strike the wall is A. `(1)/(6)pisqrt(k/m)`B. `sqrt(k/m)`C. `(2pi)/(3)sqrt(k/m)`D. `(pi)/(4)sqrt(k/m)` |
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Answer» Correct Answer - C |
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| 73. |
A block of mass m is at rest on the another blcok of same mass as shown in figure. Lower block is attached to the spring then the maximum amplitude of motion so that both the blcok will remain in contact is A. `(mg)/(2k)`B. `(mg)/(k)`C. `(2mg)/(k)`D. `(3mg)/(2k)` |
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Answer» Correct Answer - C |
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| 74. |
A `20g` partical is oscillating simple barmonically with a period of `2 second` and maximum kinetic energy `2 J`. The total machanical energy of the partical is zero , find a Amplitude of oscillation b. potential energy as a punction of displacement x relative to mean position. |
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Answer» a. Given `m = 20g = 20 xx 10^(-3) kg` Time period , `T = 2 s and K_(max) = 2 J` We havemaximum kinetic energy : `K_(max) = (1)/(2) m omega^(2) A^(2)` Hence, `2 = (1)/(2) xx 20 xx 10^(-3) xx ((2 pi)/(2))^(2) . A^(2) implies A = (10sqrt2)/(pi) m` b. We know total energy of a partical is of oscillation energy and minimum poitential energy. `E_(total) = E_(oscillation) + U_(0)` But oscillation energy of partical: ` E_(oscillation) = K_(max) = 2 J` As potential energy in reation with displacement `x` from mean position is `U (x) = (1)/(2) k x^(2) + U_(0) = (1)/(2) (m omega^(2)) x^(2) + U_(0)` `= (1)/(2) (m omega^(2)) x^(2) - 2` `(1)/(2) xx (20)/(1000) xx ((2 pi)/(2))^(2) x^(2) - 2` `= (pi^(2))/(100)x^(2) - 2` (S.I. units) |
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| 75. |
The potential energy of a particle oscillating along x-axis is given as `U=20+(x-2)^(2)` Here, `U` is in joules and `x` in meters. Total mechanical energy of the particle is `36J`. (a) State whether the motion of the particle is simple harmonic or not. (b) Find the mean position. (c) Find the maximum kinetic energy of the particle. |
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Answer» a. `F = - (du)/(dx) = - 2 (x - 2)` By assuming `x - 2 = X, we have F = - 2X` Since, `F = - X` The motion position of the partical is simple harmonic. b. The mean position of the partical is `X = 0 or x - 2 = 0`, which gives `x = 2m` c. Maximum kinetic energy of the partical is `K_(max) = E = U_(max)` `= 36 - 20 = 16 J`. |
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| 76. |
The acceleration of a particle moving along x-axis is `a=-100x+50`. It is released from `x=2`. Here `a` and `x` are in S.I units. The motion of particle will be:A. periodic, oscillatory but not SHM.B. periodic but not oscillatoryC. oscillatory but not periodic.D. simple harmonic. |
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Answer» Correct Answer - D The equation`a=-100x+50(a=-kx` form) Itself shows that the particle performs SHM. Hence (d). |
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| 77. |
Four massless springs whose force constants are 2k,2k, k and 2k, respectively, are attached to a mass M kept on a frictionless plane (as shown in figure), If the mass M is displaced in the horizontal direction, then the frequency of oscillation of the system is A. `(1)/(2pi)sqrt(k/(4M)`B. `(1)/(2pi)sqrt((4k)/(M)`C. `(1)/(2pi)sqrt((k)/(7M)`D. `(1)/(2pi)sqrt((7k)/(M)` |
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Answer» Correct Answer - B |
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| 78. |
Two identical balls `A and B` eavh of mass `0.1 kg` are attached to two identical mass less is springs. The spring-mass system is consetrained to move inside a right smooth pipe bent in the form of a circle as shown in figure . The pipe is fixed in a horigental plane. The center of the balls can move in a circle of radius `0.06 m`. Each spring has a natural length of `0.06 pi m` and force constant `0.1 N//m`. Initially, both the balls are displaced bu an angle `theta = pi//6` dadius withrespect to diameter PQ of the circle and released from rest. a. Calculate the frequency of oscillation of the ball B. b. What is the total energy of the system ? c. Find the speed of the bal A when A and B are at the two ends of the diameter PQ. |
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Answer» a.As here two masses `A and B` are connected by two sprins , this problem is equivelent to the oscillation of a reduced mass `m` by a spring of effective force constant `k_(eff)` given by `m = (m_(1) m_(2))/(m_(1) + m_(2)) = (0.1 xx 0.1)/(0.1 + 0.1) = 0.5 kg` and `k_(eff) = k_(1) + k_(2) = 0.1 + 0.1 = 0.2 N/m` So `f = (1)/(2 pi) sqrt((k_(eq)/(m)) = (1)/(2 pi) sqrt((0.20)/(0.05)) = (1)/(pi) Hz` b. As here one spring is compresent while the other is atreched by same amount (say y) and balls are at rest at `A and B`, so `E = (1)/(2) k_(1) y^(2) + (1)/(2)k_(2) y^(2) = k y^(2)` `[as k_(1) = k_(2)]` But from above figure `y = y_(1) + y_(2) = R theta_(1) + R theta_(2) = 2 R theta` `y = 2 xx 0.06 xx (pi//6) = 0.02 pi m` So `E = (0.1) (0.02 theta)^(2) = 4 pi^(2) xx 10^(-5) J` c. As at P and Q springs are ustertched so the whole energy becomes kinetic of the balls A and B, i.e. `(1)/(2)m_(1) v_(1)^(2) + (1)/(2)m_(2) v_(2)^(2) = E = 4 pi^(2) xx 10^(-5) J` Here `m_(1) = m_(2) = 0.1 kg` and `v_(1) = v_(2) = v` So, ` 0.1 v^(2) = 4 pi^(2) xx 10^(-5), i.e. v = 2 pi xx 10^(-2) m//s` |
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| 79. |
The system shown in the figure can move on a smooth surface. They are initially compressed by `6cm` and then released, then choose the correct options. (a) The system performs, SHM with time period `(pi)/(10)s` (b) The block of mass `3kg` perform SHM with amplitude `4 cm` ( c) The block of mass `6kg` will have maximum momentum of `2.40kg - m//s` (d) The time periods of two blocks are in the ratio of `1:sqrt(2)` |
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Answer» a. `T = 2 pisqrt((mu)/(k))` Here, `mu = (m_(1) m_(2))/(mm_(1) + m_(2)) = (3 xx 3)/(9) = 2 kg` `T = 2 pi sqrt((2)/(800)) = 2 pi xx (1)/(20)` `T = (pi)/(10) sec` b. Here compression of spring `x_(0) = 6 cm` Let displacement of block (1) and (2) w.r.t. center of mass be `x_(1) and x_(2)` But `x_(1) + x_(2) = x_(0) and m_(1) = m_(2)x_(2)` or `3 x_(1) = 6 x_(2)` or `x_(1) + 2 x_(2)` ` x_(1) + x_(2) = x_(0) or x_(1) + (x_(1))/(2) = x_(0)` or `(3)/(2) x_(1) = x_(0) implies x_(1) = (2)/(3) x_(0) = (2)/(3) xx 6 = 4 cm` c. Appling conservation principal of momentum `3 xx v_(1) - 6 v_(2) = 0` `3 v_(1)=- 6 v_(2)` `v_(1) = 2 v_(2)` Appling mechaincal energy conservation , we get `(1)/(2) xx 3 xx v_(1)^(2) + (1)/(2) xx 6 xx v_(2)^(2) = (1)/(2) kx_(0)^(2)` or `(3)/(2) v_(1)^(2) + 3v_(2)^(2) = (1)/(2) xx 800 xx (0.06)^(2)` or `(3)/(2) (2 v_(2))^(2) + 3v_(2)^(2) = 400 xx 36 xx 10^(-4)` or `v_(2) = (1.2)/(3) = 0.4 m//s` `:. P_(2 max) = 6 xx 0.4 = 2.4 kg m//s` |
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| 80. |
The pulley shown in figure has a moment of inertias I about its xis and mss m. find the tikme period of vertical oscillastion of its centre of mass. The spring has spring constant k and the string does not slip over the pulley. |
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Answer» For rotatiobnal equation the tension in spring on both sides must be same . Let T be the tension in the string in equilibrium position and `l_(0)` the extension of spring . Then for equilibrum of any part of spring on RHS. `2 T = mg implies2 k l_(0) = mg or l_(0) = (mg)/(2 K)` For translational equation of pulley `l_(0) = (mg)/(2 K)` Thus when pulley attains equilibrium , the spring is streched by a distance `l_(0) = (mg)/(2 K)` (i) Now suppose that the pulley is down a little and released. The pulley starts up and down oscillation . Let y be instant neous displacement of center of pulley from equilibrium. Then total increase in length of (string + spring) is `2y` (y to the left of pulley and y to the right). As steing is inextensible, total extension of spring is `2 y`. As pulley also ratates energy of system is `E = (1)/(2) l omega^(2) + (1)/(2) m v^(2) - mgy + (1)/(2) K (l_(0) + 2 y)^(2)` But `omega = (v)/( r)` `E = (1)/(2) l (v^(2))/( r^(2)) + (1)/(2) K (l_(0) + 2 y)^(2)` `= (1)/(2) ((l)/r^(2) + m) v^(2) = mgy + (1)/(2) K (l_(0) + 2 y)^(2)` Asd total energy is constant (system is conservative) `(dE)/(dt) = 0` ` (1)/(2) ((l)/r^(2) + m) 2v (dv)/(dt) - mg (dv)/(dt) + (1)/(2) K 2 (l_(0) + 2 y) , 2 (dy)/(dt) = 0` and `(dy)/(dt) = v and (dv)/(dt) = acceleration a` and `l_(0) = (mg)/(2K)` we get `((l)/r^(2) + m) a - mg + 2 K ((mg)/(2 k) + 2y) = 0` `a prop - y` (ii) As `a prop - y`, motion of pulley is SHM. Standard equiation of SHM `a = - omega^(2) y` Comparing Eqs, (ii) and (iii), `omega^(2) = (4 k)/((l)/r^(2) + m)` time period. `T = (2 pi)/(omega) = 2 pi sqrt({{(l)/(r^(2)) + m)/(4K)}} = pi sqrt({{m + (l)/(r^(2)))/(K)}}` |
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| 81. |
A particle moves in the X-Y plane according to the equation `vecr=(veci+2vecj)Acosomegat`. The motion of the particle isA. on a straight lineB. on an ellipseC. perodicD. simple harmonic. |
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Answer» Correct Answer - A::C::D `vecr=Acosomegathati+2Acosomegathatj` `|vecr|=Acosomegatsqrt5=sqrt5Acosomegat` So motion is at straight line, periodic SHM. |
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| 82. |
A plank with a small block on top of it is under going vertical `SHM`. Its period is `2 sec`. The minium amplitude at which the block will separate from plank is :A. `(10)/(pi^2)`B. `(pi^2)/(10)`C. `(20)/(pi^2)`D. `(20)/(pi^2)` |
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Answer» Correct Answer - A Here `g=Aomega^2` `A=(g)/(omega^2)=(g)/(((2pi)/(T))^2)=(gT^2)/(4pi^2)` `=(10xx4)/(4pi^2)=(10)/(pi^2)m` |
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| 83. |
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. `0.05m`B. `0.06m`C. `0.01m`D. none of these |
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Answer» Correct Answer - B `E=(1)/(2)momega^2A^2=(1)/(2)m(2pif)^2A^2` `A=(1)/(2piF)sqrt((2E)/(m))` Putting `E=K+U`, we get `A=(1)/(2pi((25)/(pi)))sqrt((2xx(0.5+0.4))/(0.2))=0.06m)` |
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| 84. |
An object of mass `0.2 kg` execurtes Simple harmonic along X-axis with frequncey of `(25)/(pi)Hz.` At the position `x=0.04 m, the object has kinetic energy of oscillation in meter is equal toA. `0.05`B. `0.06`C. `0.01`D. None of these |
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Answer» Correct Answer - B |
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| 85. |
The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5 J. If its total energy is 9 J and its amplitude is 0.01 m, its time period would beA. `(pi)/(10)s`B. `(pi)/(20)`sC. `(pi)/(50)s`D. `(pi)/(100)s` |
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Answer» Correct Answer - D `(1)/(2)mv_(max)^2=9-5=4` or `mv_(max)^2=8` or `2v_(max)^2=8` or `v_(max)=2(m)/(s)=Aomega` `omega=(v_(max))/(A)=(2)/(0.01)=200` or `(2pi)/(T)=200` `T=(2pi)/(200)=(pi)/(100)` |
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| 86. |
Which of the following statements is/are true for a simple harmonic oscillator?A. Force acting is directly proportional to displacement from the mean position and oppposite to itB. motion is periodicC. Acceleration of the oscillator is constantD. the velocity is periodic |
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Answer» Correct Answer - A::B::D |
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| 87. |
Two springs with negligible masses and force constants `k_1=200(N)/(m)` and `K_2=160(N)/(m)` are attached to the block of mass `m=10kg ` as shown in the fig. Initially the block is at rest, at the equilibrium position which both springs are neither stretched nor compressed. At time `t=0`, sharp impulse of `50Ns` is given to the block with a hammer along the spring.A. Period of oscillations for the mass m is `(pi)/(6)`s.B. Maximum velocity of the mass m during its oscillation is `10(m)/(s)`C. Data are insufficient to determine maximum velocity.D. Amplitude of oscillation is `0.83m`. |
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Answer» Correct Answer - D `T=2pisqrt((m)/(K_1+K_2))=2pisqrt((10)/(360))=(pi)/(3)s` the maximum velocity is always at equilibrium position since at any other point there will be a restoring force at temting to slow the mass. `V_("mass")=("impulse")/("mass")=(50)/(10)=5(m)/(s)impliesomega=(2pi)/(T)=6(rad)/(s)` `impliesA=`amplitude`=(V_(max))/(omega)=(5)/(6)=0.83m` |
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| 88. |
A block of mass `m` is suspended by a rubber cord of natural length `l=(mg)/(k)`, where k is force constant of the cord. The block is lifted upwards so that the cord becomes just tight and then block is released suddently. Which of the following will not be true?A. Block performs periodic motion with amplitude greater than l.B. Block performs SHM with amplitude equal to l.C. Blocks will never return to the position from where it was released.D. Angular frequency `omega` is equal to `1(rad)/(s)` |
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Answer» Correct Answer - A::C::D When the block is released suddenly, it starts to move down. During its downwards motion the rubber coed elongates. Hece, a tension is developed in it but the blocks continues to accelerate downwards till tension becomes equal to weight `mg` of the block. After this moment, the block continues to move down due to its velocity and rubber cord further elongates. Therefore tension becomes greater than the weight, hence, the block now retards and comes to an instantaneous rest. At lowest position of the block, strain energy in the cord equals loss of potential energy of the block. Suppose the block comes to an instantaneous rest when elongation of the rubber cord is equl to y. Then `(1)/(2)ky^2=mgyimpliesy=(2mg)/(k)` and 0 Hence block will be instantaneously at rest, at `y=0` and at `y=(2mg)/(k)`. If fact, the block oscillates between these two values. Since the rubber cord is elastic, tension in it is directily proportional to elongation therefore, the block will perform SHM. Its amplitude will be equal to half of the distance between these extreme positions of the block or amplitude is `(1)/(2)xx(2mg)/(k)=(mg)/(k)=l` Hence option (b) is correct. The angular frequency of its SHM will be equal to `omega=sqrt((k)/(m))` Since k and m are not given in the question, it cannot be calculated. Hence option (d) is not correct. |
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| 89. |
Three masses 700g, 500g, and 400 g are suspended at the end of a spring a shown and are in equilibrium. When the 700 g mass is removed, the system oscillacts with a period of 3 seconds, when the 500 gm mass is also removed, it will oscillate with a period of A. 1 sB. 2 sC. 3 sD. `sqrt(12/5)s` |
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Answer» Correct Answer - B |
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| 90. |
Two light spring of force constants `k_(1) and k_(2)` and a block of mass m are in the line AB on a smooth horizontal table such that one end of each spring is fixed on right supports and the other and is free as shown in figure The distance CD between the free ends of the spring is `60 cm`. If the block `(k_(1) = 1.8 N//m, k_(2) = 3.2 N//m and m = 200g)`. Is the motion simple harmonic? |
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Answer» When the block touches `D`, it will compress the spring and its `KE` will be converted into elastic energy of the spring. The compressed spring will push the block to `D` with same speed so time taken by the block to move from D towards B and back to D will be `t_(1) = (T_(2))/(2) = pi sqrt((m)/(k_(1))) = pi sqrt((0.2)/(3.2)) = (pi)/(4) s` Moreover during complete oscllation between `A and b` block moves the distance `CD` twice with uniform velocity `v` once from `C to D` and again from D to C. So `T_(1) = (2 L)/(v) = (2 xx 0.6)/(1.2) = 1 s` `T = t_(1) + t_(2) + t_(3) = pi ((1)/(3) + (1)/(4)) + 1 = 2.82 s` Now a motion is simple harmonic only and only if throughout the motion `F = - k x`. Here between `C and D, F = 0 (asv = consrtant)`, the motion is not simple harmonic but oscillatory. |
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| 91. |
A spring of stiffness constant `k` and natural length `l` is cut into two parts of length `3l//4 and l//4`, respectively, and an arrangement is made as shown in figure . If the mass is slightly displaced , find the time period of oscillation. |
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Answer» The stiffness of a spring is inversely proportional to its length . Therefore the stiffness of each part is `k_(1) = (4)/(3) k` and `k_(2) = 4 k` Time period `T = 2 pi sqrt((m)/(k_(1) + k_(2)))` `= 2 pi sqrt ((3m)/(16 k)) = (pi)/(2) sqrt((3 m)/(k))` |
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| 92. |
In the figure shown, the block A of mass m collides with the identical block B and after collision they stick together. Calculate the amplitude of resulatant vibration. |
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Answer» Common velocity after collision`= (u)/(2)` Hence, `K.E. = (1)/(2) (2 M) ((u)/(2))^(2) = (1)/(4) Mu^(2)` It is also total energy of vibration because the string is unstretched at this moment, hence, if A is the amplitude , then `(1)/(2) KA^(2) = (1)/(4) Mu^(2) or A = sqrt((M)/(2 K)) u` |
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| 93. |
Two springs are made to oscillate simple harmonically due to the same mass individually. The time periods obtained are `T_1` and `T_2`. If both the springs are connected in series and hten made to oscillate by the same mass, the resulting time period will beA. `T_1+T_2`B. `(T_1T_2)/(T_1+T_2)`C. `sqrt(T_1^2+T_2^2)`D. `(T_1+T_2)/(2)` |
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Answer» Correct Answer - C Let spring constant of two spring are `k_1` and `k_2` respectively then `T_1=2pisqrt((m)/(k_1))` and `T_2=2pisqrt((m)/(k_2))` `k_1=(4pi^2m)/(T_1^2)` and `k_2=(4pi^2m)/(T_2^2)` When the two springs are connected in series then `T=2pisqrt((m)/(k_(eq)))` where `k_(eq)=(k_1k_2)/(k_1+k_2)` `T=sqrt(T_1^2+T_2^2)` |
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| 94. |
A partical of `mass m = 1 kg` oscillates simple harmonically with angular frequency `1 rad//s`. Find the phase of the partical at `t = 1 s and 2 s`. Start calculating time when the partical moves up passing through the mean position. |
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Answer» We need to find `phi = omega t + phi_(0)` where `omega` is the angular frequency of SHM. Since, the partical moves up at the equilibrium possition at `t = 0`, we have `phi_(0) = 0, Then , phi = t rad` Substituting, `t= 1s, 2s, we have phi = t rad, 2rad` |
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| 95. |
A block of mass m is connected to a spring constant k and is at rest in equilibrium as shown. Now, the block is Displacement by h below its equilibrium position and imparted a speed `v_0` towards down as shown in the Fig. As a result of the jerk, the block executes simple harmonic motion about its equilibrium position. Based on this information, answer the following question. Q. Find the time taken by the block to cross the mean position for the first time.A. `(2pi-cos^-1((h)/(A)))/(sqrt((k)/(m)))`B. `((pi)/(2)-cos^-1((h)/(A)))/(sqrt((k)/(m)))`C. `(pi-sin^-1((h)/(A)))/(sqrt((k)/(m)))`D. `(pi-sin^-1((h)/(A)))/(2sqrt((k)/(m)))` |
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Answer» Correct Answer - C To compute the time taken by the block to cross mean position for the first time, we can make use of circular motion representation. `t=(pi-delta)/(omega)=(pi-sin^-1((h)/(A)))/(sqrt((k)/(m)))` |
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| 96. |
figure (a) and (b) represent spring- block system. If m is displacement slightly , find the time period of ascillation of the system. |
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Answer» both the cases are as follows: Reduced mass of the system Reduced mass of the system `mu = ((mM)/(m + m))` `mu = (mM)/(m + m) = (m)/(2)` `T = 2 pi sqrt((mu)/(k))` and `k_(e) = k + k = 2 k` `T = 2 pi sqrt ((mu)/(k_(e)))` |
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| 97. |
Figure shows a system consisting of a massless pulley, a spring of force constant k and a block of mass m. If the block is slightly displaced vertically down from is equilibrium position and then released, the period of its vertical oscillation is A. `2pisqrt((m)/K)`B. `pisqrt((m)/(4K)`C. `pisqrt((m)/(K)`D. `4pisqrt((m)/(K)` |
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Answer» Correct Answer - D |
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