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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.
| 501. |
Calculate the angular frequency of the system shown in fingure. Friction is absent everywhere and the threads, spring and pulleys are massless. Given that `m_(A) = m_(B) = m`. |
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Answer» Correct Answer - A::B::D Let `x_(0)` be the extension in the spring in equilibrium. Then equilibrium of `A` and `B` give, `T = kx_(0) + mg sin theta` …(i) and `2T = mg` …(ii) Here, `T` is the tension in the string. Now, suppose `A` is further displaced by `(x)/(2)` and speed of `B` at this instant will be `(v)/(2)`. Total energy of the system in this position will be, `E = (1)/(2) k(x + x_(0))^(2) + (1)/(2)m_(A)v^(2) + 1/2mB((v)/(2))^(2) + m_(A)gh_(A) - m(B)gh_(B)` or `E = (1)/(2)k(x + x_(0))^(2) + 1/2mv^(2) + 1/8mv^(2) + mgxsin theta -mg x/2` or `E = (1)/(2)k(x + x_(0))^(2) + 5/8mv^(2) + mgxsin theta - mg x/2` Since, `E` is contant, `(dE)/(dt) = 0` or `0 = k(x + x_(0))(dx)/(dt) + 5/4mv ((dv)/(dt)) + mg (sin theta)((dx)/(dt)) - (mg)/(2)((dx)/(dt))` Substituting, `(dx)/(dt) = v` rArr `(dv)/(dt) = a` and `kx_(0) + mg sin theta = (mg)/(2)` [From Eqs. (i)and(ii)] We get, `(5)/(4)m a = - kx` Since, `a prop - x` Motion is simple harmonic, time period of which is, `T = 2pi sqrt|(x)/(a)|` `= 2pi sqrt((5m)/(4k))` `:. omega = (2pi)/(T) = sqrt((4k)/(5m))` |
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| 502. |
The function sin^(2) (omegat) represents.A. a simple harmonic motion with a period `2pi//omega`B. a simple harmonic motion with a period `pi//omega`C. a periodic, but not simple harmonic, motion with a period `2pi//omega`D. a periodic, but not simple harmonic, motion with a period `pi//omega` |
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Answer» Correct Answer - B |
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| 503. |
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 |
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| 504. |
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 |
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| 505. |
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. (a) `(pi)/3`B. (b) (-pi)/6`C. (c ) `(pi)/6`D. (d) (-pi)/3` |
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Answer» Correct Answer - B (b) `v_(1)=(dy_(1))/(dt)=0.1xx100pi cos(100pi+(pi)/3)` `v_(2)=(dy_(2)/(dt)=-0.1pi sint=0.1 cos(pit+(pi)/2)` :. Phase diff=phi_(1)=(pi)/3=(pi)/2=(2pi-3pi)/6=-(pi)6`. |
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| 506. |
A block is on a piston, which executes simple harmonic motion in the vertical plane with a period of 1s. At what amplitude will the block and the piston separate? If the piston has an amplitude of 0.05m, what is the maximum frequency at which the black and piston will remain in contact continously? [hint: the block and piston will separate when acceleration of piston = acceleration due to gravity.] |
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Answer» Correct Answer - 0.25m,2.23 |
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| 507. |
The balance wheel of a watch vibrates with an angualr amplitude `pi` radians and a period of 0.5s. Find a) the maximum angular speed of the wheel, b) the angular speed of the wheel when its displacement is `pi/2` radians and c) the angular acceleration of the wheel when its displacement is `pi/4` radians. [hint: use formulae of linear SHM for the angular SHM of the wheel] |
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Answer» Correct Answer - 39.5rad/s, 34.2rad/s,`124rad//s^(2)`. |
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| 508. |
A cubical block of mass M vibrates horizontally with amplitude of 4.0 cm and a frequency of 2.0 Hz. A small block of mass is placed on the bigger block. In order that the smaller block does not side on the bigger block, the minimum value of the coefficient of static friction between the two blocks is (0.16x). Find the valule of x(Take `pi^(2)`=10 and g=10m/`s^(2)`)A. 0.36B. 0.4C. 0.64D. 0.72 |
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Answer» Correct Answer - C (c) `mug gt ` maximum acceleration in SHM ` gt omega^(2)A` `therefore mu gt (omega^(2)A)/(g) implies mu gt ((2pif)^(2)xx A)/(g)` `implies mu gt ((2xx3.14xx2)^(2)xx 4xx 10^(-2))/(10)` `implies mu gt 0.631 ~~ 0.64` |
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| 509. |
The potential energy of a harmonic oscillator of mass 2kg in its equilibrium position is 5 joules. Its total energy is 9 joules and its amplitude is 1cm. Its time period will beA. 6.28 sB. `3.14 xx 10^(-2)`sC. 3.14 sD. 0.314 s |
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Answer» Correct Answer - B |
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| 510. |
Two springs with negligible massess and force constant of `k_(1)= 200 Nm^(-1)` and `k_(2)=160Nm^(-1)` are attached to the block of mass m = 10kg as shown in the figure. Initially the block is at rest at the equilibrium position the block is at rest at the equilibrium position ir. Which both springs are neither stretched nor compressed. At time t = 0, sharp impulse of 50 N-s is given to the block in horizontal direction. A. period of oscialltions for the mass m is `(pi/6)` sB. maximum velocity of the mass m during its oscillation is `10ms^(-1)`.C. maximum velocity is 6m/sD. amplitude of oscillations is `5/6`m. |
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Answer» Correct Answer - D |
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| 511. |
Two springs with negligible massess and force constant of `k_(1)= 200 Nm^(-1)` and `k_(2)=160Nm^(-1)` are attached to the block of mass m = 10kg as shown in the figure. Initially the block is at rest at the equilibrium position the block is at rest at the equilibrium position nor compressed. At time t=0, sharp impulse of 50 N-s is given to the block. A. Period of oscillations for the mass m is `pi/3` sB. maximum velocity of the mass m during its oscillations is `5ms^(-1)`C. Data are insufficient to determine maximum velocityD. Amplitude of oscillation is 0.42 m |
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Answer» Correct Answer - A::B |
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| 512. |
The velocity of a particle in simple harmonic motion at displacement y from mean position isA. `omegasqrt(a^(2)+y^(2)`B. `omegasqrt(a^(2)-y^(2)`C. `omegay`D. `omega^(2)sqrt(a^(2)-y^(2)` |
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Answer» Correct Answer - B |
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| 513. |
The time period of a particle in simple harmonic motion is T. Assume potential energy at mean position to be zero. After a time of `(T)/(6)` it passes its mean position its,A. velocity will be one half its maximum velocityB. displacement will be one half of its amplitudeC. Acceleration will be nearly `86^(@)` of its maximum accelerationD. kinetic energy = potential energy |
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Answer» Correct Answer - A::C |
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| 514. |
At `x = (A)/(4)`, what fraction of the mechanical energy is potential ? What fraction is kinetic ? Assume potential energy to be zero at mean position. |
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Answer» Correct Answer - A `E = (1)/(2)kA^(2)` `U = (1)/(2)kx^(2) = (1)/(2)k((A)/(4))^(2)` `= (1)/(16)((1)/(2)kA^(2))` `= (1)/(16)E` Hence, `(1)/(16)` fraction is potential energy and `(15)/(16)` fraction is kinetic energy. |
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| 515. |
`a - x` equation of a particle in SHM is `a + 4x = 0` Here, `a` is in `cm//s^(2)` and `x` in cm.Find time period in second. |
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Answer» `a = - 4x` Comparig with a `= - omega^(2)x`, we get `omega = 2 rad//s` Now, `T = (2pi)/(omega) = (pi)sec` |
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| 516. |
The amplitude of a particle executing SHM about `O` is `10 cm`. ThenA. when the `KE` is `0.64` times of its maximum`KE`, its displacement is `6cm` from `O`B. its speed is half the maximum speed when its displacement is half the maximum displacementC. Both (a) and (b) are correctD. Both (a) and (b) are wrong |
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Answer» Correct Answer - A (a) Kinetic energy is `0.64` times, it means speed is `0.8` times `0.8 omega A = omega sqrt(A^(2)) - (x^(2))` `:. x = 0.6 A = 6cm` (b) `(omegaA) /(2) = omega sqrt(A^(2) - x^(2))` `:. x = sqrt(3)/(2) A` |
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| 517. |
In the previous question, the angular frequency of the simple harmonic motion is `omega`. The coefficient of friction between the coin and the platform is `mu`. The amplitude of oscillation is gradually increased. The coin will begin to slip on the platform for the first time (i) at the extreme positions of oscillations (ii) at the mean position (iii) for an amplitude of `(mug)/(omega^2)` (iv) for an amplitude of `(g)/(muomega^2)`A. (i),(iii)B. (ii), (iii)C. (i), (iv)D. (ii),(iv) |
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Answer» Correct Answer - A `momega^2Alemumg` `Ale(mug)/(omega^2)impliesA_(max)=(mug)/(omega^2)` |
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| 518. |
In the arrangement shown in figure, pulleys are light and spring are ideal. `K_(1)`, `k_(2)`, `k_(3)`and `k_(4)` are force constant of the spring. Calculate period of small vertical oscillations of block of mass `m`. |
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Answer» Correct Answer - A::B::C::D When the mass `m` is displaced from its mean position by a distance `x`, let `F` be the restoring (extra tension) force produced in the string. By this extra tension further elongation in the springs are `(2F)/(k_(1))` , `(2F)/(k_(2))`, `(2F)/(k_(3))` and `(2F)/(k_(4))` respectively. Then, `x = 2((2F)/(k_(2))) + 2((2F)/(k_(2))) + 2((2F)/(k_(3))) + 2((2F)/(k_(4)))` or `F((4)/(k_(1)) + (4)/(k_(2)) + (4)/(k_(3)) + (4)/(k_(4))) = - x` Here negative sign shows the restoring nature of force. `a = - (x)/(m((4)/(k_(1)) + (4)/(k_(2)) + (4)/(k_(3)) + (4)/(k_(4)))` `T = 2pi sqrt |(x)/(a)|` `= 4pi sqrt (m((1)/(k_(1)) + (1)/(k_(2)) + (1)/(k_(3)) + (1)/(k_(4))))` |
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| 519. |
In a simple pendulum the period of oscillation `(T)` is related to the length of the pendulum `(L)` asA. `(l)/(T)` =constantB. `(l^(2))/(T)` =constantC. `(l)/(T^(2))` =constantD. `(l^(2))/(T^(2))` =constant |
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Answer» Correct Answer - C (c) `T=2pisqrt((l)/(g))implies (l)/(T^(2))=(g)/(4pi^(2))`=constant |
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| 520. |
When a particle is mass `m` moves on the `x-` axis in a potential of the from `V(x) = kx^(2)`, it performs simple harmonic motion. The corresponding thime periond is proportional to `sqrt((m)/(k))`, as can be seen easily asing dimensional analysis. However, the motion of a pariticle can be periodic even when its potential enem increases on both sides `x = 0` in a way different from `kx^(2)` and its total energy is such that the particel does not escape to infinity. consider a particle of mass `m` moving onthe `x-`axis . Its potential energy is `V(x) = omega (alpha gt 0`) for `|x|` near the origin and becomes a constant equal to `V_(0)` for `|x| ge X_(0)` (see figure) If the total energy of the particle is `E`, it will perform is periodic motion why if :A. `E lt 0`B. `E gt 0`C. `V_(0) gt E gt 0`D. `E gt V_(0)` |
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Answer» Correct Answer - C When `0 lt E lt V_(0)` there will be acting a restoring force to perform oscillation because in this case paticle will be in the region `|x|lex_(0)`. |
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| 521. |
When a particle is mass `m` moves on the `x-` axis in a potential of the from `V(x) = kx^(2)`, it performs simple harmonic motion. The corresponding thime periond is proportional to `sqrt((m)/(k))`, as can be seen easily asing dimensional analysis. However, the motion of a pariticle can be periodic even when its potential enem increases on both sides `x = 0` in a way different from `kx^(2)` and its total energy is such that the particel does not escape to infinity. consider a particle of mass `m` moving onthe `x-`axis . Its potential energy is `V(x) = omega (alpha gt 0`) for `|x|` near the origin and becomes a constant equal to `V_(0)` for `|x| ge X_(0)` (see figure) If the total energy of the particle is `E`, it will perform is periodic motion why if :A. propotional to `V_(0)`B. propotional to `V_(0)/(mX_(0))`C. propotional to `sqrt((V_(0))/(mX_(0)))`D. zero |
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Answer» Correct Answer - D `F = -(dU)/(dx)` as for `|x| gt x_(0) V = V_(0) =` constant `rArr (dU)/(dx) = 0 rArr F = 0`. |
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| 522. |
When a particle is mass `m` moves on the `x-` axis in a potential of the from `V(x) = kx^(2)`, it performs simple harmonic motion. The corresponding thime periond is proportional to `sqrt((m)/(k))`, as can be seen easily asing dimensional analysis. However, the motion of a pariticle can be periodic even when its potential enem increases on both sides `x = 0` in a way different from `kx^(2)` and its total energy is such that the particel does not escape to infinity. consider a particle of mass `m` moving onthe `x-`axis . Its potential energy is `V(x) = omega (alpha gt 0`) for `|x|` near the origin and becomes a constant equal to `V_(0)` for `|x| ge X_(0)` (see figure) If the total energy of the particle is `E`, it will perform is periodic motion why if :A. `Asqrt((m)/(alpha))`B. `(1)/(A)sqrt((m)/(alpha))`C. `Asqrt((alpha)/(m))`D. `(1)/(A)sqrt((alpha)/(m))` |
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Answer» Correct Answer - B `V = alphaX^(4)` `T.E. = (1)/(2) momega^(2)A^(2) = alphaA^(4)` (not stricltly applicable just for dimension matching it is used) `omega^(2) = (2alphaA^(2))/(m) rArr T prop (1)/(A)sqrt((m)/(alpha))` |
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| 523. |
Under the action of a force `F=-kx^(3)` , the motion of a particle is (k=a positive constant)A. simple harmonic motionB. uniformly acceleration motionC. not periodicD. periodic but not simple harmonic |
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Answer» Correct Answer - D (d) At x=0 , F=0 . For `x gt 0` , force is in negagtive direction and for `x lt 0` , force is in positive x-direction . Therefore, motion of the particle is periodic about mean position x=0 |
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| 524. |
A particle of mass 2 kg executing SHM has amplitude 10 cm and time period is 1 s.Find (i) the angular frequency (ii) the maximum speed (ii) the maximum acceleration (iv) the maximum restoring force (v) the speed when the displacement from the mean position is 8 cm (vi) the speed after `(1)/(12)` s the particle was at the extreme position (vii) the time taken by the particle to go directly from its mean position to half the amplitude (viii) the time taken by the particle to go directly from its exterme position to half the amplitude. |
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Answer» Given, `m=2kg` , amplitude A=10cm , T=1 s `(i) omega=(2pi)/(T)=2pis^(-1)=6.28 s^(-1)` (ii) `v_("max")=Aomega=(10 cm)(2pis^(-1))=0.628 ms^(-1)` `(iii)a_("max")=Aomega^(2)=(10 cm)(2 pis^(-1))^(2)=4 ms^(-2) " "("take" , pi^(2)=10)` (vi) `F_("max")=ma_("max")=mAomega^(2)=(2kg)(4 ms^(-2))=8N` (v) `v=omegasqrt(A^(2)-x^(2))=(6.28 s^(-1))sqrt((10 cm)^(2)-(8 cm)^(2))` `=(6.28 s^(-1))(6 cm)` =37.68 cm/s (vi) Suppose `x=A"cos"omegat`, then the particle will be at the extreme position at time t=0, `v=-Aomega"sin"omegat` `therefore " At "t=(1)/(12) s, v=-(10 cm)(6.28 s^(-1))"sin"(2pis^(-1).(1)/(12)s)` `=(6.28 cms^(-1))"sin"(pi)/(6)=-31.4 cms^(-1)` Negative sign indicates that velocity is directed towards the mean position if the particle starts to the move from the extreme right. (vii) When time is taken from the mean position, we take `x=A"sin"omegat` Suppose, the particle reaches `x=+(A)/(2)` , at time t, then `(A)/(2)=A"sin"omegatimplies"sin"omegat=(1)/(2)implies omegat=(pi)/(6) s` or , `t=(pi)/(6 omega)=(pi)/(6xx2pi//T)=(T)/(12)=(1 s)/(12)` (viii) When time is taken from the extreme position , we take `x=A"cos"omegat` At t=0, x=A, i.e., the particle is at the extreme right Suppose at time t, the particle reaches `x=(A)/(2)`, then `(A)/(2)=A"cos"omegat` `implies"cos"omegat=(1)/(2)impliesomegat=(pi)/(3)` or `t=(pi)/(3omega)=(pi)/(3((2pi)/(T)))=(T)/(6)=(1)/(6)s` |
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| 525. |
The displacement of a body executing `SHM` is given by `x=A sin (2pi+pi//3)`. The firs time from `t=0` when the velocity is maximum isA. `0.33 sec`B. `0.16sec`C. `0.25 sec`D. `0.5 sec` |
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Answer» Correct Answer - A |
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| 526. |
The equation of a particle executing `SHM` is `(d^(2)x)/(dt^(2))=-omega^(2)x`. Where `omega=(2pi)/("time period")`. The velocity of particle is maximum when it passes through mean position and its accleration is maximum at extremeposition. The displacement of particle is given by `x=A sin(omegat+theta)` where `theta`-initial phase of motion. `A`-Amplitude of motion and T-Time period The accleration is half of its maximum value at an amplitude ofA. `(A)/(sqrt2)`B. `(sqrt3A)/(2)`C. `(A)/(sqrt3)`D. `(A)/(2)` |
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Answer» Correct Answer - D |
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| 527. |
A uniform disc of radius r is to be suspended through a small hole made in te disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the centre for it to have minimum time period? |
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Answer» Correct Answer - B Let mass of uniform disc =m Then minimum time period `T=2pi sqrt(1/(mg.sqrt2r))` `=2pi sqrt((2mr^2)/(sqrt2 mgr))` `=2pi sqrt(2r^2)/g)` clearlyl for minimum time period distance of hole from the centre `=r/sqrt2`. |
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| 528. |
Displacement between maximum potential energy position energy potential andmaximum kinetic energy position for a particle executing `S.H.M` isA. ` -a `B. `+a`C. `+-a`D. `+-a/4` |
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Answer» Correct Answer - C |
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| 529. |
The maximum acceleratioin of a particle in SHM is made two times keeping the maximum speed to be constant. It is possible whenA. amplitude of oscillation is doubled while frequency remains constantB. amplitude is doubled while frequency is halvedC. frequency is doubled while amplitude is halvedD. frequency of oscillatin is doubled while amplitude remains constant. |
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Answer» Correct Answer - C |
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| 530. |
The maximum acceleration of a particle un SHM is made two times keeping the maximum speed to be constant. It is possible when .A. amplitude of oscillation is double while frequency remains constantB. amplitude is double while frequency is halvedC. frequency is doubled while amplitude is halvedD. frequency of oscillation is doubled while amplitude remians constant |
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Answer» Correct Answer - C (c) `omegaA`= constant and `omega^(2)A` is made two times. This is possible when `omega` is double and A is halved. |
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| 531. |
A particle performing `SHM` is found at its eqilbrium at `t=1 sec`.and it is found to have a speed of `0.25 m//s` at `t=2sec`. If the period of oscillation is `6sec`. Calculate amplitude of oscillation-A. `(2)/(2pi)m`B. `(3)/(2pi)m`C. `(6)/(pi)m`D. `(3)/(8pi)` |
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Answer» Correct Answer - A |
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| 532. |
The maximum acceleration of a particle in `SHM` is motion two times keeping the maximum speed in the constant it is position whenA. Amplitude of oscillation is doubled while frequency remains constantB. amplitude is doubled while frequency is halvedC. frequency is doubled while amplitude is halvedD. frequency is doubled while amplitude remains constant |
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Answer» Correct Answer - C |
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| 533. |
A particle moves along the x-axis according to the equation `x=4+3sin(2pit)` Hence, x in cm and t in seconds. Select the correct alternativesA. The motion of the particle is simple harmonic with mean position at x=0B. the motion of the particle is simple harmonic with mean position at x=4cmC. The motion of the particle is simple harmonic with mean position at x=`-4cm`D. Amplitude of osciallation is 3cm |
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Answer» Correct Answer - B::D |
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| 534. |
Equation of two S.H.M. `x_(1)=5 sin (2pi t+pi//4),x_(2)=5sqrt(2)(sin 2pit+cos 2pit)` ratio of amplitude & phase difference will beA. `2:1,0`B. `1:2,0`C. `1:2,pi//2`D. `2:1,pi//2` |
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Answer» Correct Answer - B `x_(1)=5 sin (2pit+pi//4)` `x_(2)=5.sqrt(2)(sin 2pi t+cos 2pi t)` `x_(2)=10 sin (2pit+pi//4)` So, ratio amplitude 1:2 and phase difference is zero |
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| 535. |
A particle is executing S.H.M. from mean position at 5 cm distance, acceleration is `20 cm//sec^(2)` then value of angular velocity will beA. 2 rad/secB. 4 rad/secC. 10 rad/secD. 15 rad/sec |
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Answer» Correct Answer - A `a =omega^(2) x ` `omega=2 rad//sec` |
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| 536. |
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency `omega`. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first timeA. At the highest position of the platformB. At the mean position of the platformC. Form an amplitude of `g//omega^(2)`D. For an amplitude of `g^(2)//omega^(2)` |
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Answer» Correct Answer - C |
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| 537. |
As shown in figure in a simple harmonic motion oscillator having identical four springs has time period A. `T=2pisqrt((m)/(4k))`B. `T=2pisqrt((m)/(2k))`C. `T=2pisqrt(m/k)`D. `T=2pisqrt((2m)/(k))` |
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Answer» Correct Answer - C The upper two springs are in parallel, `(k_1)_(eq)=k+k=2k` The lower two springs are in parallel, `(k_2)_(eq)=k+k=2k` `(k_1)_(eq)` and `(k_2)_(eq)` are in series `k_(eq)=(2k*2k)/(2k+2k)=k` `T=2pisqrt((m)/(k_(eq)))=2pisqrt(m/k)` |
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| 538. |
The total energy of a particle, executing simple harmonic motion is. where x is the displacement from the mean position, hence total energy is independent of x.A. `propx`B. `propx^(2)`C. independent of `x`D. `prop x^(1//2)` |
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Answer» Correct Answer - C The total energy of a harmonic oscillator is a constant and it is expressed as `E = (11)/(2) momega^(2)("Amplitude")^(2)` E is independen to x instantaneous displacement. |
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| 539. |
The equation of SHM of a particle is `(d^2y)/(dt^2)+ky=0`, where k is a positive constant. The time period of motion isA. `(2pi)/(sqrt3)`B. `(2pi)/(k)`C. `k/2`D. `(sqrtk)/(2pi)` |
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Answer» Correct Answer - A `(d^2y)/(dt^2)=-ky=-omega^2y` `omega=sqrtk` `T=(2pi)/(sqrtk)` |
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| 540. |
A particle at the end of a spring executes simple harmonic motion with a period `t_(1)` while the corresponding period for another spring is `t_(2)` if the oscillation with the two springs in series is T thenA. `T =t_(2) + t_(2)`B. `T^(2) = t_(1)^(2) + t_(2)^(2)`C. `T^(2) = t_(1)^(-1) + t_(2)^(-1)`D. `T^(2) = t_(1)^(-2) + t_(2)^(-2)` |
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Answer» Correct Answer - B |
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| 541. |
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. (a) `T^(-1)=t_(1)^(1)+t_(2)^(-1)`B. (b) `T^(2)=t_(1)^(2)+t_(2)^(2)`C. (c )`T=t_(1)+t_(2)`D. (d) T^(-2)=t_(1)^(-2)+t_(2_^(2)` |
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Answer» Correct Answer - B (b) For first spring, t_(1)=2pisqrtm/k_(1)`, For second spring, `t_(2)=2pisqrtmk_(2)`. when springs are in series then, `(k_(1)k_(2))/(k_(1)k_(2)` :. `T=2pisqrt((m(k_(1)+k_(2))))/((k_(1)k_(2)`. :. `T=2pisqrtm/k_(2)+m/k_(1) =2pisqrt((t_(2)^(2))/(2pi^(2)+t_(1)^(2)/(2pi)^(2))` `rArr T^(2)=t(1)^(2)+t_(1)^(2)`. where x is the displacement from the mean position. |
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| 542. |
A particle connected at the end of a spring executes SHM with a period `T_(1)`. While the corresponding period for another spring is `T_(2)`. When the two springs are connected in series and particle is connected at the one T. Find the relation between `T, T_(1)` and `T_(2)`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)` |
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Answer» Correct Answer - B |
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| 543. |
A particle of mass m is free to move along x axis under the influence of a conservative force. The potential energy of the particle is given by `U=-ax^(x) e^(-bx)` [a and b are positive constants] Find the frequency of small oscillations of the particle about its equilibrium position |
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Answer» Correct Answer - `f=(1)/(2pi) sqrt((ae^(-n)n^(n-1))/(mb^(n-2)))` |
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| 544. |
Potential energy `(U)` of a body of unit mass moving in a one-dimension conservative force field is given by, `U=(X^(2)-4X+3)`. All units are in `S.I.` |
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Answer» Correct Answer - (i) `X_(0)=2m`; (ii) ``T=sqrt(2) pi sec.`; (iii) `2sqrt(3)` |
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| 545. |
the two block shown here rest on a frictionless surface. If they are pulled apart by a small distance and released at `t=0`, the time when `1kg` block comes to rest can be- A. `(2pi)/(3)sec`.B. `pi sec`.C. `(pi)/(3) sec`.D. `(pi)/(9)sec`. |
| Answer» Correct Answer - A::B::C | |
| 546. |
What is the maximum acceleration of the particle doing the SHM `gamma=2sin[(pit)/2phi]` where gamma is in cm?A. `(pi)/(2)cm//s^(2)`B. `(pi^(2))/(2)cm//s^(2)`C. `(pi)/(4)cm//s^(2)`D. `(pi)/(4)cm//s^(2)` |
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Answer» Correct Answer - B `y=2 sin ((pit)/(2)+phi)` Comparing the equation with the standard equation `y=A sin (omegat+phi)` So `A=2 cm, omega=(pi)/(2)` Acceleration of particle is `a=omega^(2)x" "["numerically"]` At `x=+A, a=a_("max") `therefore" "a_("max")=omega^(2)A=((pi)/(2))^(2)xx2` `=2xx(pi^(2))/(4)=(pi^(2))/(2)cm//s^(2)` |
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| 547. |
What is the maximum acceleration of the particle doing the SHM `gamma=2sin[(pit)/2phi]` where gamma is in cm?A. `pi /2 cm //s^(2)`B. `pi^(2) /2 cm //s^(2)`C. `pi /4 cm //s^(2)`D. `pi /4 cm //s^(2)` |
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Answer» Correct Answer - B |
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| 548. |
A particle executes linear simple harmonic motion with an amplitude of 2 cm . When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds isA. `1/(2pisqrt3)`B. `2pisqrt3`C. `(2pi)/(sqrt3)`D. `(sqrt3)/(2pi)` |
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Answer» Correct Answer - C |
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| 549. |
A `2kg` block hangs without vibrating at the bottom end of a spring with a force constant of `400 N//m`. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of `5 m//s^(2)` when the acceleration suddenly ceases at time `t = 0` and the car moves upward with constant speed `(g = 10 m//s^(2))` What is the angular frequencyof the block after the acceleration ceases ?A. `10sqrt(2)rad//s`B. `20 rad//s`C. `20 sqrt(2) rad//s`D. `32rad//s` |
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Answer» Correct Answer - A `omega = sqrt((k)/(m)) = sqrt((400)/(2)) = 10 sqrt(2) rad//s` |
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| 550. |
A `2kg` block hangs without vibrating at the bottom end of a spring with a force constant of `400 N//m`. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of `5 m//s^(2)` when the acceleration suddenly ceases at time `t = 0` and the car moves upward with constant speed `(g = 10 m//s^(2)` The amplitude of the oscillation isA. `7.5cm`B. `5 cm`C. `2.5 cm`D. `1 cm` |
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Answer» Correct Answer - C `kx_(0) = ma` ` x_(0) = (ma)/(K) = (2 xx 5)/(400)m` ` = 0.025m` ` = 2.5cm` `:. A = x_(0) = 2.5 cm` |
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