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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.
| 1. |
A body is performing simple harmonic motion. Then itsA. average total energy per cycle is equal to its maximum kinetic energy.B. average kinetic energy per cycle is equal to half of its maximum kinetic energy.C. average total energy per cycle is equal to half of its malximum kinetic energy.D. None of these. |
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Answer» Correct Answer - B When a body performs `S.H.M.` total energy of the body remain constant and kinetic energy of the body is maximum at mean position its maximum value is equal to total oscillation energy of the body .Hence average total oscilation energy per cycle is equal its the maximum `K.E.` of the body hence (a) is connect Obviously (c) is wrong If amplitude of oscillations is equal to a and angular frequency is to then maximum velocity of the body will be equal to a as and it varius according to sine over a completed cycle will be equal to `(1)/(sqrt(2)) a omega` Avarage `K.E. = (1)/(2) mv_(max)^(2) = (1)/(4) ma^(2) a omega` which is half of maximum `K.E.` |
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| 2. |
A verticle mass-spring system executed simple harmonic ascillation with a period `2s` quantity of this system which exhibits simple harmonic motion with a period of `1sec` areA. velocityB. potential energyC. phese difference between acceleration and displacementD. difference between kinetic energy and potential energy. |
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Answer» Correct Answer - B The time period of potential energy and kinetic energy is half of time period of `S.H.M.` |
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| 3. |
Four types of oscillatory system a simple pendlum a physic pendlum a torsional pendlum and a spring mass system each of same time period are taken to the mass if line period will have it unchanged?A. only spring - mass system.B. spring - mass system and torsional pendulum.C. spring - mass system and physical pendulum.D. None of these |
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Answer» Correct Answer - B Both the spring - mass and torsional pendulum have no dependence gravitational acceleraton for their time periods. |
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| 4. |
A body is executing `SHM` under action of the a force of whose maximum is `50 N`. magnitude of force acting on the particle at the time when its energy is half kineic energy and half potential is (Assume potential energy to be zero at mean position).A. `12.5 sqrt(2) N`B. `12.5 N`C. `25 N`D. `25 sqrt(2) N` |
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Answer» Correct Answer - D `F_(max) = m omega^(2) A = 50 N` Given `(1)/(2) kx^(2) = (1)/(2) [(1)/(2) KA^(2)] rArr x = A//sqrt(2)` `F = m omega^(2) x = (m omega^(2)A)/(sqrt(2)) = (50)/(sqrt(2)) = 25 sqrt(2)N` |
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| 5. |
A `4 kg` particle is moving along the x- axis under the action of the force `F = - ((pi^(2))/(16)) x N` At `t = 2 sec ` the particle passes through the origin and `t = 10sec`, the speed is `4sqrt(2)m//s` The amplitude of the motion isA. `(32sqrt(2))/(pi)m`B. `(16)/(pi)m`C. `(4)/(pi)m`D. `(16sqrt(2))/(pi)m` |
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Answer» Correct Answer - A `a = -((pi^(2))/(64)) x rArr omega rArr sqrt((pi^(2))/(64)) = (pi)/(8)` `rArr T = (2pi)/(omega) = 16 sec` There is a time difference of `712` between `t = 2sec` to `t = 10sec`, Hence particle is again passing through the mean position of `SHM` where itsw speed is maximum `i.e. V_(max) = A omega = 4sqrt(2)` `rArr A = (4sqrt(2))/(pi//8) = (32sqrt(2))/(pi) m` |
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| 6. |
A particle of mass `10gm` is placed in a potential field given by `V = (50x^(2) + 100)J//kg`. The frequency of oscilltion in `cycle//sec` isA. `(10)/(pi)`B. `(5)/(pi)`C. `(100)/(pi)`D. `(50)/(pi)` |
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Answer» Correct Answer - B Potential energy `U = mV` ` rArr U = (50x^(2) + 100)10^(-2)` `F = (dU)/(dx) = (- 100x)10^(-2)` `rArr m omega^(2)x = -(100 xx 10^(-2))x` ` 10 xx 10^(-3) omega^(2) x = 100 xx 10^(-2)x` `rArr omega^(2) = 100 :. omega = 10` `rArr f = (omega)/(2pi) = (10)/(2pi) = (5)/(pi)` |
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| 7. |
The potential energy of a particle of mass `1kg` in motion along the x- axis is given by: `U = 4(1 - cos 2x)`, where `x` in metres. The period of small oscillation (in sec) isA. `2pi`B. `pi`C. pi//2`D. `pi//4` |
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Answer» Correct Answer - C `F = -(dU)/(dx) = - 8 sin 2s` For small oscilation `S` in `2x = 2s` i.e. `a = - 16x` Since `a = 1x` the oscilation are `SH` in rature `T = 2pi sqrt(|(x)/(a)|) = 2pi sqrt((1)/(16)) = (pi)/(2) sec` |
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| 8. |
A mass (M) is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes `(5T)/3`. Then the ratio of `m/M` is .A. `(5)/(3)`B. `(3)/(5)`C. `(25)/(9)`D. `(16)/(9)` |
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Answer» Correct Answer - D `T prop sqrt(m) rArr (T_(2))/(T_(1)) = sqrt((m_(2))/(m_(1))) rArr (5)/(3) = sqrt((M +m)/(M))` `rArr (25)/(6) = (M+m)/(M) rArr (m)/(M) = (16)/(9)` |
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| 9. |
A mass `M` is suspended from a spring of negiliglible mass the spring is pulled a little and then released so that the mass executes simple harmonic oscillation with a time period `T` If the mass is increases by `m` the time period because `((5)/(4)T)`,The ratio of `(m)/(M)` isA. `9//16`B. `25//16`C. `4//5`D. `5//4` |
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Answer» Correct Answer - A `T = 2pi sqrt((m)/(k)) rArr m prop T^(2) rArr (m_(2))/(m_(1)) = (T_(2)^(2))/(T_(1)^(2))` ` rArr (M +m)/(M) = = (((5)/(4)T)/(T))^(2) rArr (m)/(M) = (9)/(16)` |
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| 10. |
Two simple pendulum of length `1m` and `16m` respectively are both given small displacement in the same direction of the same instant. They will be phase after one shorter pendulum has complated a oscillations. The value of `n` isA. `1//3`B. `2//3`C. `1`D. `4//3` |
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Answer» Correct Answer - D `T_(1) = 2 pi sqrt((1)/(g)) , T_(2) = 2piu sqrt((16)/(g)) = 4T_(1)` at any time `t` phase of pendulums are `phi_(1) = omega_(1)t = (2pi)/(T_(1)) t, phi_(2) = omega_(2)t = (2pi)/(T_(2)) t` First pendulum is faster both will be in same phase again when faster pendulum completated one oscillation more than slower pendulum `rArr (2pi)/(T_(1)) t - (2pi)/(4T_(1)) t = 2 pi rArr t = (4T_(1))/(3)` No of oscillation completed by shorter pendulum in time `t` `n = (t)/(T_(1)) = (4)/(3)` |
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| 11. |
STATEMENT-1 : The height of a liquid column in a `U`-tube is `0.3 m`. If the liquid in one of the limbs is depressed, and then released, the time period of liquid column will be `1.1 sec`. STATEMENT-2 : this follows from the relation. `T = 2pisqrt((h)/(g))`A. If both assertion and reason are true and the reasopn is correct explanation of the assertionB. If both assertion and reason are true and but not the correct explanation of assertionC. If the assertion is true but reason is falseD. If both the assertion and reason are false |
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Answer» Correct Answer - A Time period of a liquid column `T = 2pi sqrt((h)/(g)) = 2 xx (22)/(7) xx sqrt((0.3)/(9.8)) = 1.1 s` |
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| 12. |
A particle of mass `m` is released from rest and follow a particle part as shown Accuming that the displacement of the mass from the origin is small which graph correctlly depicts the position of the particle as a function of time? A. B. C. D. |
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Answer» Correct Answer - D The given velocity position graph depends the motion of the particle is `SHM` is `SHM` at `t = 0 , v = 0` and `x= v_(max)` so option d is correct |
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| 13. |
A particle is in a line `SHM` acceleration and the corresponding velocity if this particle are `x` and `y` then the graph relation to these in force value isA. B. C. D. |
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Answer» Correct Answer - C The equation giving relation between acceleration and speed as function of distance from mean position is `a = - omega^(2)x`…(1) `v = omega sqrt(a^(2) - x^(2))`…(2) From equation (1) and (2) we get `a^(2) = omega^(2) (A^(2) - (v^(2))/(omega^(2)))` |
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| 14. |
Assertion : A hole were drilled through the center of each and a ball is dropped into the hole at one will not get other out of other end of the hole Reason : It will come out of the end normallyA. If both assertion and reason are true and the reasopn is correct explanation of the assertionB. If both assertion and reason are true and but not the correct explanation of assertionC. If assertion is true but the reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - C The ball wil not go out other end the hole becomes it will execute `SHM` One reasing the other end the hole its velocity becomes and acceleration of ball will be maximum and will be direction toward the center of earth |
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| 15. |
The length of a simple pendulum executing simple harmonic motion is increased by `21%`. The percentage increase in the time period of the pendulum of increased lingth is.A. `10%`B. `21%`C. `30%`D. `50%` |
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Answer» Correct Answer - A If initial length `l_(1) = 100` then `t_(2) = 121` By using `T = 2pi sqrt((1)/(g)) rrArr (T_(1))/(T_(2)) = sqrt((I_(1))/(I_(2)))` Hence `(T_(1))/(T_(2)) = sqrt((100)/(121)) rArr T_(2) = 1.1T_(1)` `%` increase `= (T_(2) - T_(1))/(T_(1)) xx 100= 10%` Alternative Time period of simple pendulum `T = 2pi sqrt((1)/(g))` `:. (Delta T)/(T) = (1)/(2) (DeltaI)/(I)` since `(DeltaI)/(I) = 21%` `:.Delta T)/(T) = (1)/(2) xx 21% = 10%` |
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| 16. |
A partcle is performing simple harmonic motion along x- axis with amplitude `4cm` and time period `1.2sec` The minimum time period taken by the again is given byA. `0.6sec`B. `0.4 sec`C. `0.3 sec`D. `0.2 sec` |
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Answer» Correct Answer - B Time taken by period to move from `x = 0` (mean pisition) to `x = 4` (extermeposition) `= (T)/(4) = (1.2)/(4) = 0.3 sec` Let `t` be the time taken by the particle to move from `x = 0 to x= 2 cm` `y = a sin omega t rArr 2 = 4 sin"(2pi)/(T)t rArr (1)/(2) = sin"(2pi)/(1.2)t` `rArr (pi)/(6) = (2pi)/(1.2) t rArr = 0.1 sec` Hence time to move from `x = 2 to x= 4` will be equal to `0.3 - 0.1 = 0.2 sec` Hence total time to move from`x = 2 to x= 4`and back again ` 2 xx 0.2 = 0.4sec` |
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| 17. |
The resultant of two rechangular simple harmonic motion of the same frequency and unequal amplitude but differing in phase by `pi//2` isA. Simple harmonicB. CircularC. EllipticalD. Parabolic |
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Answer» Correct Answer - C If first equation is `y_(1) = a_(1) sin omega t` `rArr sin omega t = (y_(1))/(a_(1))` then second equation will be `y_(2) = a_(2) sin (sin omega t cos (pi)/(2))` `= a_(2)[sin omega t cos "(pi)/(2)+ cos omega t sin"(pi)/(2)]= a_(2) cos omega t ` `rArr cos omega t = (y_(2))/(a_(2))` By squiring and adding (i) and (ii) `sin ^(2) omega t + cos^(2) omega t = (y_(1)^(2))/(a_(1)^(2)) + (y_(2)^(2))/(a_(2)^(2))` `rArr (y_(1)^(2))/(a_(1)^(2)) + (y_(2)^(2))/(a_(2)^(2)) = 1`. This is the equation of ellipse |
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| 18. |
If a simple harmonic motion is erpresented by `(d^(2)x)/(dt^(2))+ax=0`, its time period is.A. `(2pi)/(k)`B. `2pi k`C. `(2pi)/(sqrt(k)`D. `2pi sqrt(k)` |
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Answer» Correct Answer - C On comparing with standered equation `(d^(2)y)/(dt^(2) + omega^(2) y = 0` we get `omega^(2) = k rArr omega = (2pi)/(T) = sqrt(k) rArr = (2pi)/(sqrt(k)` |
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| 19. |
A particle at the end of a spring executes S.H,M with a period `t_(2)` If the period of oscillation with two spring in .A. `T = t_(1) +t_(2)`B. `T^(2) = t_(1)^(2) +t_(2)^(2)`C. `T^(-1) = t_(1)^(-1) +t_(2)^(-1)`D. `T^(-2) = t_(1)^(-2) +t_(2)^(-2)` |
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Answer» Correct Answer - B `t_(1) = 2pi sqrt((m)/(k_(1))) and t_(2) 2 pi sqrt((m)/(k_(2)))` in series effective spring constant `k = (k_(1)k_(2))/(k_(1) +k_(2))` `:. Time period T = 2pi sqrt((m(k_(1) + k_(2)))/(k_(1)k_(2)))`…..(i) Now `t_(1)^(2) + t_(2)^(2)= 4pi^(2)m((1)/(k_(1)) + (1)/k_(2)) = (4pi^(2)m(k_(1) + k_(2)))/(k_(1)k_(2))` `t_(1)^(2) + t_(2)^(2)= T^(2)`. [Using equation (ii)] |
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| 20. |
Assertion : Two unequal of same instrial are loaded with same load The longer one will have longer value of time period Reason: The concept will follow if we made a expect to measureA. If both assertion and reason are true and the reasopn is correct explanation of the assertionB. If both assertion and reason are true and but not the correct explanation of assertionC. If assertion is true but the reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - C When a longer spring loded with weight `mg` the extersion `1` will be more Therefore acceleration to the expression `T = 2pi sqrt((1)/(g))`, the longer spring will have a large value of the time period |
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| 21. |
A particle is `SHM` is discribed by the displacement function `x(t) = a cos (Delta omega + theta)` If the initial `(t = 0)` position of the particle `1cm` and its initial velocity is `picm//s` The angular frequency of the particle is `pi rad//s`, then its amplitude isA. `1cm`B. `sqrt(2)cm`C. `2 cm`D. `2.5 cm` |
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Answer» Correct Answer - B `x = a cos(omega t +theta)`…(i) And `v = (dx)/(dt) = - aomegasin(omega t + theta)`…(ii) Given at `t = 0 , x = 1 cm and v = pi` and `omega = pi` Putting these value in equation (i) and (ii) , are will get `sin theta = (-1)/(a)` and `cos theta = (1)/(A)` `rArr sin^(2) theta + cos^(2) theta = (-(1)/(a))^(2) +((1)/(a))^(2) rArr a = sqrt(2) cm` |
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| 22. |
The velocity `v` of a particle of mass in moving along a straigh line change withh time `t` as `(d^(2)v)/(dt^(2)) = - Kv` where `K` is a particle constant which of the folloewing statement is correct?A. The particle does not perform `SHM`B. The particle perform `SHM`with time period `2pi sqrt((m)/(k))`C. The particle perform `SHM`with time frequency`sqrt(k)/(2pi)`D. The particle perform `SHM`with time period `(2pi)/(K)` |
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Answer» Correct Answer - C `(d^(2)v)/(dt) = - kv^(2)` this equation has standard solution `v = v_(0) sin(sqrt(k)t + theta)`where `= omega = sqrt(k)` Hence the particle executes `SHM` with angular frequency `omega = sqrt(k) ` or frequency `f = sqrt(k)/(2pi)`. |
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| 23. |
Two pearticle are oscillation along two close parallel straght lines side by side with the same frequency and amplitude They pass each other moving in opposite directions when their displacement is half of the on a stright line perpendicular to the part of the two particle The phase difference isA. zeroB. `(2pi)/(5)`C. `pi`D. `(pi)/(6)` |
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Answer» Correct Answer - B Equation of `SHM` is given by `x = A sin (omega t + delta)` `(omega t = delta)` is called phase. When `x = (A)/(2)` then `sin (omega t + delta) = (1)/(2) implies omega t + delta = (pi)/(6)` or `phi_(1) = (pi)/(6)` For second particle, `phi_(2) = pi - (pi)/(6) = (5 pi)/(6)` `:. phi = phi_(2) = phi_(1)` `= (4pi)/(6) = (2 pi)/(3)` |
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| 24. |
A particle free to move along the (x - axis) hsd potential energy given by `U(x)= k[1 - exp(-x^2)] for -o o le x le + o o`, where (k) is a positive constant of appropriate dimensions. Then.A. for small total displacement from `x = 0`, the motion is simple harmonicB. if its total machanical energy is `k//2`, it has its maximum kinetic energy at the origin.C. for any final norezero value of `s` there is a force directed away from the origin.D. at points away from the origin the particle is in unstable equilibrium. |
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Answer» Correct Answer - A Since `F = (dU)/(dv)= 2kx exp(-x^(2))` `F = 0` (at equilibrium as x = 0) `U` is minimum at `x = 0` and `U` and `U_(min) = 0` U is minimum at `x rarr +- oo and U_(min) = k` The particle would oscillation about `x = 0` for small displacement from the origin and it is in stable equilibrium at the origin |
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| 25. |
A mass `m` is suspended separately by two different spring of spring constant `K_(1)` and `k_(2)` given the time period `t_(1)` and`t_(2)` respectively if the same mass `m` is shown in the figure then time period `t` is given by the relation A. `t = t_(1) + t_(2)`B. `t =( t_(1)t_(2))/(t_(1) + t_(2))`C. `t^(2) = t_(1)^(2) + t_(2)^(2)`D. `t^(-2) = t_(1)^(-2) + t_(2)^(-2)` |
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Answer» Correct Answer - D `t_(1) = 2pi sqrt((m)/(k_(1)))` and `t_(2) = 2pi sqrt((m)/(k_(2)))` Equivalant constant for shown combination is , `(K_(1) +K_(2)` so time period `t` is given by `t= 2pi sqrt((m)/(K_(1) + K_(2)))` By solving these equations we get `T^(2) = t_(1)^(2) + t_(2)^(2)` |
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| 26. |
Out of the following functions representing motion of a particle which represents SHM I. `y = sin omega t - cos omega t` II. `y = sin^(3)omega t` III. `y = 5 cos ((3 pi)/(4)-3 omega t)` IV. `y = 1 + omega t + omega^(2)t^(2)`A. only(iv) does not represent `SHM`B. (i) and (iii)C. (i) and (ii)D. only (i) |
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Answer» Correct Answer - B For a simple harmonic motion `(d^(2)y)/(dt^(2)) prop - y` Hence equation `y = sin omega t` and `y = 5 cos ((3pi)/(4) - 3 omegat)` setisfy this contation and equation `y = 1 + omega t + omega^(2) t^(2)` is not peridlum `y = sin^(2)omega^(2)` periode but not `SHM` option c is correct |
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| 27. |
A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. proportional to `(1)/(sqrt(a))`B. proportional to `a`C. proportional to `sqrt(a)`D. proportional to `a^(1//2)` |
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Answer» Correct Answer - A `U = k(x)^(2) rArr T = - (dU)/(dx) = - 3k(x)^(2)`….(i) Also the `SHM x = a sin omega t amA(A^(2)x)/(dt^(2)) + omega^(2)x = 0` `rArr "acceleration" (d^(2)x)/(dt^(2)) = - omega^(2)x rArr F = ma` `= m(d^(2)x)/(dt^(2)) = - m omega^(2)x` From equation (i) and (ii) we get `omega = sqrt((3kx)/(m))` `rArr T = (2pi)/(omega) = 2pi sqrt((m)/(3kx)) = 2pi sqrt((m)/(3k(a sin omegat)) rArr T prop (1)/(sqrt(a))` |
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| 28. |
The displacement `y` of a particle executing periodic motion is given by `y = 4 cos^(2) ((1)/(2)t) sin(1000t)` This expression may be considereed to be a result of the superposition ofA. TwoB. ThreeC. FourD. Five |
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Answer» Correct Answer - B `y = 4 cos^(1)((t)/(2)) sin1000t` `rArr y = 2(1 + cos t)sin 1000t` `y = 2sin 1000t + 2cos t sin 1000t` `y = 2sin 1000t + 2sin 999t sin 1000t` it is a sum of time `S.H.M` |
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| 29. |
Which of the following is greater in `SHM` (assuming potential energy` = 0` at mean position )?A. Average kinetic energy with respect to spaceB. Average potential energy with respect to spaceC. Average kinetic energy with respect to timeD. Average potential energy with respect to time |
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Answer» Correct Answer - A Avarage kinetic energy with respect to space `= (1)/(3) kA^(2)` Average potential energy with respect to space `= (1)/(6) kA^(2)` Avarage kinetic energy with respect to time `= (1)/(4) kA^(2)` Average potential energy with respect to time `= (1)/(4) kA^(2)`. |
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| 30. |
A simple pendulum hangs the celling of a car if the car acceleration with a uniform acceleration , the frequency of the simple pendulum willA. increaseB. decreaseC. become infiniteD. remain constrant |
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Answer» Correct Answer - A In this case frequency of oscillation is given by `n = (1)/(2pi) sqrt(sqrt(g^(2) +a^(2))/(l))` where a is the acceleration of car if a increase then `n` also increase |
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| 31. |
Two simple harmonic are represented by the equation `y_(1)=0.1 sin (100pi+(pi)/3) and y_(2)=0.1 cos pit`. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is.A. `(-pi)/(3)`B. `(pi)/(6)`C. `(-pi)/(6)`D. `(pi)/(3)` |
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Answer» Correct Answer - C `v_(1) = (dy_(1))/(dt) = - 0.1 xx 100pi cos(100 pi t + (pi)/(3))` `v_(2) = (dy_(2))/(dt) = - 0.1 pi sin pi t = 0.1 pi cos(pi t + (pi)/(2))` phase difference of velocity of first particle with respect to the velocity of `2^(nd)` particle at `t = 0` is `Delta phi = phi_(1) - phi_(2)= (pi)/(3) - (pi)/(2)= -(pi)/(6)` |
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| 32. |
Two pendulum of different angle are in pgase at mean position at a cartain The minimum and after which they will be again in phase is `5T//4` where in the period of shaorter pendulum find the ratio length of the two pendulumA. `1:16`B. `1:4`C. `1:2`D. `1:25` |
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Answer» Correct Answer - D Let `T_(1) = T and T_(2) = KT,(2pi)/(T) t - (2pi)/(KT) t = 2pi` `t((K - 1)/(KT)) = 1, ((K - 1)/(KT)) = 1 which given K = 5` If length of first pendulum `= t, T_(1) = 2pi sqrt(1//g)` `T_(2) = 5 xx sqrt(1//g) = 2pi sqrt(25//g)` So ratio of length `= 1: 25` |
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| 33. |
In the figure all spring are indentical having spring constant `k` and mass `m` each Th eblock also mass `m` The frequency of oscilation of the block is A. `(1)/(2pi)sqrt((3k)/(m))`B. `(1)/(2pi)sqrt((3k)/(2m))`C. `2pi sqrt((3m)/(3k))`D. None of these |
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Answer» Correct Answer - B If spring has the mass thebn `f = (1)/(2pi) sqrt((k)/(m +(M_("spring"))/(3))` spring equivalent `= 3k` (As all are in perallel) `f = (1)/(2pi) sqrt((2k)/(m +((3m))/(3))` rArr `f = (1)/(2pi)sqrt((3k)/(2m))` |
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| 34. |
A force of `6.4N` streches a vertical spring by `0.1m` The mass that must be supended from the spring so that it oscilation with a time period of `pi//4` secondA. `(pi)/(4)kg`B. `(4)/(pi)kg`C. `1 kg`D. `10kg` |
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Answer» Correct Answer - C spring constant `K = (6.4)/(0.1) = 64 N//m`. Now `T = 2pi sqrt((m)/(k))` or `(pi)/(4) = 2pi sqrt((m)/(64)) :. m = 1kg` |
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| 35. |
Which one of the following statement is true for the speed `v` and the acceleration a of a particle executing simple harmonic motion?A. When `v` is maximum `a` is maximumB. value of `a` is zero wherever may be the value of `v`C. When `v` is zero `a` is zeroD. When `v` is maximum `a` is zero |
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Answer» Correct Answer - A In simple harmonic motion the displacement equation is `y = A sin omega t` Where `A` is amplitude of the motion Velocity `v = (dv)/(dt) = A cos omega t` `v = A omega sqrt(1 - sin^(2) omega t)` `v = omega sqrt(A^(2) - y^(2))` ….(i) Acceleration `a = (dv)/(dt) = (d)/(dt) (A omega t cos omega t)` `a = -A omega ^(2) sin omega t` `a = - omega^(2) t`....(ii) When `y = 0 , y = A omega = v_(max)` `a= 0 = a_(min)` when `y = A, y = 0 = v_(min)` `a = omega^(2) A = a_(max)` Hence it is clean that when `x` is maximum then `a` is minimum (i.e. zero) at vics versa. |
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| 36. |
The equation of motion of a particle executing simple harmonic motion is `a+16pi^(2)x = 0` In this equation, a is the linear acceleration in `m//s^(2)` of the particle at a displacement x in meter. The time period in simple harmonic motion isA. `(3)/(4)sec`B. `(1)/(2)sec`C. `1sec`D. `2 sec` |
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Answer» Correct Answer - B `a = -16 pi^(2) s` standared equation of `SHM` is `A = - omega^(2) s` Hence camparing two equatios, we get `omega = 4 pi` `T = (2pi)/(omega) = (2 pi)/(4 pi) = (1)/(2) sec` |
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| 37. |
A charged particle as deflected by two mutually perpendicular oscillation electeic field such that the displecement of the particle that in each one at then given by `x = A sin (omega t) and y = A sin (omega t + (pi)/(6))` respectively. The frequency following by the changed particle isA. a circle with equation `x^(2) + y^(2) = A^(2)`B. a straight line with equation ` y=sqrt(3)a`C. an ellipse with equation `x^(2) + y^(2) - xy= (3)/(4)A^(2)`D. an ellipse with equation `x^(2) + y^(2) - sqrt(3)xy = (1)/(4)A^(2)` |
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Answer» Correct Answer - D When two `SHM` (which one perpendicular to end other ) are given by `x = omega t and y = A_(2) sin (omega t + delta)` and when they are combirted the trangectory of the particle is given by `(x^(2))/(A_(1)^(2)) + (y^(2))/(A_(2)^(2)) - (2xy)/(A_(1)A_(2)) cos delta = sin^(2) delta` (This is equation of ellipse) in the given equation , phase different is `delta = (pi)/(6)` and `A_(1) = A_(2) = A` `:.(x^(2))/(A_(1)^(2)) + (y^(2))/(A_(2)^(2)) - (2xy)/(A_(1)A_(2)).sqrt(3)/(2) = (1)/(4)` `x^(2) + y^(2) - sqrt(3)xy = (1)/(4) A^(2)` and this in an elipse |
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| 38. |
The equation of the resulting oscillation obtained by the summation at two mutally perpendicular oscillation with the same frequency `f_(1) = f_(2) = 5 Hz` and same initial phase `delta_(1) = delta_(2) = 60^(@) ` is (Given their amplitude are `A_(1) = 0.1m and A_(2) = 0.05m`A. `0.15sin(10pi t +(pi)/(6))`B. `0.05sin(10pi t +(2pi)/(3))`C. `0.112 sin(10pi t +(pi)/(3))`D. `0.313 sin(10pi t +(pi)/(2))` |
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Answer» Correct Answer - C When `x = A_(1) sin (omega t + phi_(1))` and `y = A_(2) sin (omegat + phi_(2))` then `(x^(2))/(A_(1)^(2)) + (y^(2))/(A_(2)^(2)) - (2xy)/(A_(1) A_(2)) cos delta = sin ^(2) delta` Here phase i.e. `delta = 0` `:.(x^(2))/(A_(1)^(2)) + (y^(2))/(A_(2)^(2)) - (2xy)/(A_(1) A_(2)) = 0` `rArr ((x)/(A_(1)) - (y)/(A_(2)))^(2) = 0 rArr y = (A_(2))/(A_(1)) x` REsulting amplitude is `A = sqrt(A_(1)^(2) + A_(2)^(2)) = sqrt((0.1)^(2) + (0.05)^(2)) = 0.112m` Thus the equation of resulting `SHM` ` r = 0.112 sin (10pi t + (pi)/(3))` |
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| 39. |
A simple pendulum hung from the calling of a trak moving at constant speed has a period `T` it the train start acceleration or decelerating then what be the effect on time period of pendulum ?A. Decreases only when train acceleratesB. Decreases only when train decaleratesC. Decreases in both casesD. Increases in both cases |
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Answer» Correct Answer - C `T = 2pi sqrt((1)/((vec g + vec a))) vec a` is in horizontal direction in both cases Hence `|vec g + vec a|gt g` so `T` decrease in both cases. |
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| 40. |
At `t = 0` aparticle of mass `m` start moving from rest due to a force `vecF = F_(0) sin (omega t)hati` :A. Particle perform `SHM` about its initial position of restB. Particle perform `SHM` with initial position as extreme position with angular frequency `omega`C. At any instant, distance moved by the particle equals its displacement, from the initial positionD. initial velocity of particle increases with time but after time `t = 2 pi// omega` it becomes constant |
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Answer» Correct Answer - C If a particle `S.H.M.` If comes to rest at exterm position but at external position , acceleration of the particle is maximum position .Therefore resultantforce on the particle will also the maximum pasition But force on the given by `F = F_(s)` on as and at `t = 0` the particle was at when it is at rest Therefore this particle cannot `S.H.M.` Therefore (a) and (b) are wrong In the particle case the force is zero at the initial moment and always act along `x- axis` velocity the force start position `x-` dirfection` Therefore particles start to acceleration along `x-` dirfectionand continutes position up to the instant given by `omega t = pi` after particle moment form because negative hence position start to retard but the impule of the force from ` t = 0` to `t = pi//omega t` is numerically to its impluse from `t = (pi)/(omega)` to `t = (2pi)/(omega)` hence the particle will comes to rest particle contrinuiously increase and and then it decrease particle move because negative hence the particle contitues to move along positive direction of a `x-`axis hebnce its displacement at any instant will be equal to the distance moved by the particle from the initial position Hence c is correct After time `t = (2pi)/(omega)` force because position again it mean the particle start to acceleration again along position direction of a `x-` axis .Therefore velocity never becomes correct |
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| 41. |
A particle oscillation is given by `(f_(0)) = kpl^(2)` with for constant `k` and an amplitude A The maximum velocity during the oscillation a preperitiaonal to :A. `A`B. `A^(2)`C. `A^(3//2)`D. none of these |
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Answer» Correct Answer - C Let at any time displacement is `y` and velocity is `s` since total energy can be the maximum `PE` we get `(1)/(2) mv^(2) + ky^(3) = kA^(3) rArr = sqrt((2k(A^(2)- y^(3)))/(m))` Velocity maximum is therefore `(at y = 0)` `v_(max) = sqrt((2kA^(2))/(m)) rArr v_(max) prop A^(1//2)` |
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| 42. |
A linear harmonic oscillator of force constant `2 xx 106Nm^(-1)` and amplitude `0.01m` has a total mechanical energy `160 J`. Among the followinhg statement , which are correct? i Maximum `PE`is `100J` ii Maximum `KE`is `100J` iii Maximum `PE` is `160J` iv Maximum `PE` is zeroA. Both (i) and (iv)B. Both (ii) and (iii)C. Both (i) and (ii)D. Both (ii) and (iv) |
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Answer» Correct Answer - B Total mechanical energy is `E_(T) = 160 J` `:. U_(max) = 160 J` At extreme position `KE` is zero , Work done by spring force extreme position to mean position is `W = (1)/(2) kA^(2)` `:. K_(max) = W = (1)/(2) (2 xx 10^(6) )(0.01)^(2) = 100 J` `:. U _(min) = 160-100 = 60J` |
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| 43. |
A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. independent of aB. proportional to `sqrt(a)`C. proportional to `a^(3//2)`D. proportional to `(1)/(sqrt(a))` |
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Answer» Correct Answer - D Potential energy `U = k|x|^(2)` Hence force `SHM F = (dU)/(dx) = -3k k|x|^(2)`….(i) Also for `SHM x= a sin omega t` and `d^(2)x)/(dt^(2)) + omega^(2) x= 0` `rArr` Acceleration `omega = ((d^(2)x)/(dt^(2)) = - omega^(2)x`…(ii) From (i) and (ii) we get `omega = sqrt((3ks)/(m))` `rArr T = (2pi)/(omega) = 2pi sqrt((m)/(3ks)) = 2pi sqrt((m)/(3k(a sin omega t)))` `rArr T prop (1)/(sqrt(a))` |
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| 44. |
A particle of mass `m` is executing oscillations about origin on the x axis amplitude A its potential energy is given as`U(x) = beta x^(4)` where `beta` is constant x cooridirate of the particle where the potential energy is one third of the kinetic energy isA. `+- (A)/(2)`B. `+- (A)/(sqrt(2)`C. `+- (A)/(3)`D. `+- (A)/(sqrt(3)` |
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Answer» Correct Answer - B `U = beta x^(4)` (given) `:. U_(max) = beta .A^(4)` `k(x) = U_(max) - U(x) = betaA^(4) - betax^(4) = beta (A^(4) - x^(4))` `U(x) = (1)/(3)K(k)` (Given) `:.betax^(4) = (1)/(3) beta(A^(4) - x^(4))` `rArr 3 betax^(4) = beta A^(4) - betax^(4)` ` 4beta x^(4) = beta A^(4)rArr x^(4) =(A^(4))/(4)` `rArr x = +- (A)/(sqrt(2))` |
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| 45. |
A particle executing `SHM` while moving from executy it found at distance `x_(1) x_(2)` and `x_(2)` from comes at the and of three successive second The period of oscilation is where `theta = cos^(-1)((x_1 + x_2)/(2x_(2)))`A. `2 pi//theta`B. `pi//theta`C. `theta`D. `pi//2theta` |
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Answer» Correct Answer - A Displacement time equation of the particle will be `X = A cos omega t` Given that `x_(1) = A cos omega` `and x_(2) = A cos 2omega` `Now (x_(1) + x_(3))/(2x_(2)) = (A (cos theta + cos 3 omega))/(2A cos 2omega)` `= (2A cos 2 omega cos omega)/(2A cos 2 omega)` `= cos omega` `omega = cos^(-1)((x_(1) +x_(2))/(2x_(2))) = (2pi)/(T)` `or T = (2pi)/(theta) where theta = cos^(-1)((x_(1) +x_(2))/(2x_(2)))` |
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| 46. |
A plane oscillation oscillation with time period `T` suddenly another aplte put on the first plate, then time operiodA. will decreaseB. will increaseC. will be sameD. None of these |
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Answer» Correct Answer - C Time period is independent of mass of pendulum |
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| 47. |
Assertion : The gerph between velocity and displacement for a harmonic oscillation is a parabola Reason : Velocity does not change uniformly with displacement in simple harmonic motionA. If both assertion and reason are true and the reasopn is correct explanation of the assertionB. If both assertion and reason are true and but not the correct explanation of assertionC. If assertion is true but the reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - C While executing `SHM` from relation `v = omega sqrt((A^(2) - y^(2))` `v^(2) = omega (A^(2) - y^(2))` `v^(2) = omega^(2) A^(2) -omega^(2) y^(2)` Dividing both side the `omega^(2) A^(2)` `(v^(2))/(omega^(2)A^(2)) + (y^(2))/(A^(2)) = 1`...(1) it is understood that Eq `(1)` repressent the equation of an elipe so the graph between `v` (velocity) and `y` (displacement) is not a perabole Angle from Eq (1) displacement `(y)` increase the velocity decrease becomes zero at maximum displace and again increase with in `SHM` the velocity Thus with displacement in `SHM` the velocity does not change uniformly |
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| 48. |
Assertion: The simple harmonic motion is to and the fro and periodic Reason : The motion of the earth is periodicA. If both assertion and reason are true and the reasopn is correct explanation of the assertionB. If both assertion and reason are true and but not the correct explanation of assertionC. If assertion is true but the reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - B From the defibnition of `SHM` a particle execute `SHM` when it goes to end fro about a mean position under the resulting force hence assertion is true The earth complate one revolution aroound the sun after a fixed interval of `1` year. Therefore it is a period motion Reason is false is also true. |
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| 49. |
A particle doing simple harmoonic motion amplitude `= 4 cm` time peiod `= 12sec` The ratio between time taken by it in going from its mean position to `2cm ` and from `2cm` to exterme position isA. `1`B. `1//3`C. `1//4`D. `1//2` |
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Answer» Correct Answer - B `omega = (2pi)/(T) = (2pi)/(12) = (pi)/(6)(rod)/(sec)` (for `y = 2 cm) 2 = 4(sin"(pi)/(6)t_(1))` By solving ` t_(1) = 1` sec(For `y = 4 cm`) `t_(2) = 3 sec` So time taken by particle from `2cm` to extreme position is `t_(2) - t_(1) = 2 sec` Hence required ratio will be `(1)/(2)` |
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| 50. |
The potential energy of a harmonic oscillation when is half way to its and end point is (where `E` it’s the total energy)A. `(1)/(8)E`B. `(1)/(4)E`C. `(1)/(2)E`D. `(2)/(3)E` |
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Answer» Correct Answer - B `(U)/(g) = ((1)/(2)m omega^(2)y^(2))/((1)/(2) m omega^(2)a^(2)) = (y^(2))/(a^(2)) = (((a)/(2))^(2))/(a^(2)) = (1)/(4) rArr U = (E )/(4)` |
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