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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.
| 701. |
In S.H.M. path length is 4 cm and maximum accelertion is `2pi^(2) cm//s^(2)`. Time period of motion is |
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Answer» Correct Answer - A `A=2cm` `a_(m)=Aomega^(2)` `omega^(2)=sqrt((a_(m))/(A))=sqrt((2pi^(2))/(2))` `therefore omega=pi` `T=(2pi)/(omega)=(2pi)/(T_(1))=2s` |
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| 702. |
Time period of pendulum is 6.28 s and amplitude of oscillation is 3 cm. Maximum acceleration of pendulum isA. `8 cm//s^(2)`B. `0.3 cm//s^(2)`C. `3 cm//s^(2)`D. `58.2 cm//s^(2)` |
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Answer» Correct Answer - C `T=6.28s, A=3cm,a_(m)=? ` `a_(m)=omega^(2)A=(4pi^(2))/(T^(2))A=(3xx4pi^(2))/((2pi)^(2))` `a_(m)=3cm//s^(2)` |
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| 703. |
A block with mass `M` attached to a horizontal spring with force constant `k` is moving with simple harmonic motion having amplitude `A_(1)`. At the instant when the block passes through its equilibrium position a lump of putty with mass `m` is dropped vertically on the block from a very small height and sticks to it. (a) Find the new amplitude and period. (b) Repeat part (a) for the case in which the putty is dropped on the block when it is at one end of its path. |
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Answer» (a) Before the lump of putty is dropped the total mechinical energy fo the blcok and string is `E_(1) = (1)/(2) kA_(1)^(2)`. Since, the block is at the equilibrium position `U = 0`, and the energy is purely kinetic. Let `v_(1)` be the speed of the block at the equilibrium position, we have `E_(1) = (1)/(2) Mv_(1)^(2) = (1)/(2) kA_(1)^(2)` `v_(1) =sqrt((k)/(M)) A_(1)` During the process momentum of the system in horizontal direction is conserved. Let `v_(2)`be the speed of the combined mass, then `(M +m) v_(2) = Mv_(1)` `v_(2) =(M)/(M+m) v_(1)` Now, let `A_(2)` be the amplitude afterwards. Then, `E_(2) = (1)/(2) kA_(2)^(2) = (1)/(2) (M +m) v_(2)^(2)` Substiting the proper vales, we have `A_(2) = A_(1) sqrt((M)/(M+m))` Note: `E_(2) lt E_(1)`, as some energy is lost into heating up the block and putty. Further, `T_(2)=2pi sqrt((M+m)/(k))` (b) When the putty drops on the block, the block is intaneously at rest. All the mechanical enegry is strored in the spring as potential enegry. Again the momentum in horizontal direction is conserved during the process. But now it is zero just before and after putty is dropped. So, in this case, adding the extra mass of the putty has no effect on the mechanical energy, i.e., `E_(2) = E_(1) = (1)/(2) kA_(1)^(2)` and the amplitudes is still `A_(1)`. Thus, `A_(2) = A_(1)` and `T_(2) = 2pi sqrt((M+m)/(k))` |
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| 704. |
The `x-t` graph of a particle undergoing simple harmonic motion is shown in figure. Acceleration of particle at `t = 4//3 s` is A. `(sqrt(3))/(32)pi^(2)cm//s^(2)`B. `(-pi^(2))/(32) cm//s^(2)`C. `(pi^(2))/(32)cm//s^(2)`D. `-(sqrt(3))/(32)pi^(2)cm//s^(2)` |
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Answer» Correct Answer - D `x =sin ((2pi)/(8)t), a = - omega^(2)x` |
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| 705. |
The motion of a particle varies with time according to the relation `y=a (sinomegat + cos omegat)` ,thenA. the motion is oscillatory but not S.H.M.B. the motion is S.M.H. with amplitude AC. the motion is S.M.H. with amplitude `Asqrt(2)`D. the motion is S.M.H. with amplitude 2 A |
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Answer» Correct Answer - C `R=sqrt(A_(1)^(2)+A_(2)^(2))=sqrt(A^(2)+A^(2))=sqrt(2)A` |
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| 706. |
A seimicircular uniform wore of radius `R = pi` can rotate frelly about x-axis. The cnetre of curvature is at origin as shown. The acceleration due to gravity is given by `vec(g)=-x^(2)hatj` units. For small oscillations of the wire its time period is given by `pisqrt(x)`. the value of `x` is ________ |
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Answer» Correct Answer - 6 Let `R d theta `be an element of wire located an angular position of `theta `wrt-ve axis When slightly rotated by angle `beta` torque due to gravity is `vec(tau) = int d vec(tau) =int dm vec(g) xx vec(r)` `=- int [R dtheta(M)/(piR)]x^(2) (R cos theta). sin beta` `rArr tau = int R ((Md theta)/(pi)x^(2)R cos theta) beta`, for small `beta` as `x = R sin theta tau = beta overset((pi)/(2))underset((-pi)/(2))int (MR cos theta)/(pi)R^(2) sin^(2) theta d theta rarr (1)` But `tau= I alpha =(MR^(2))/(2) alpha rarr (2)` where `alpha = (d^(2)beta)/(dt^(2))` `alpha =- omega^(2) beta rArr T = (2pi)/(omega)` |
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| 707. |
A rectangular plate of sides a and b is suspended fropm a ceilling by two parallel strings of length L each figure. The separation between the strings is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute harmonic motion. Find the time period. A. `2pisqrt((L)/(g))`B. `2pi sqrt((L+a)/(g))`C. `2pi sqrt((La)/(g.b))`D. `2pi sqrt((La)/(g.d))` |
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Answer» Correct Answer - A In displaced position `F_("restoring") = mg sin theta as theta` is small : `sin theta rarr theta, cos theta rarr 1` `rArr F_(rest) = mg theta` and `theta = (x)/(L) rArr ma = mg((x)/(L))` `rArr a = ((g)/(L)) x =omega^(2) x, omega = sqrt((g)/(L)), T = 2pi sqrt((L)/(g))` |
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| 708. |
Two masses `m_(1)` and `m_(2)` are suspeded togther by a massless spring of spring constnat `k` (Fig). When the masses are in equilibrium, `m_(1)` is removed. Frequency and amplitude of oscillation of `m_(2)` are A. `omega =sqrt((K)/(m_(1))), A (m_(2)g)/(K)`B. `omega =sqrt((K)/(m_(2))), A (m_(1)g)/(K)`C. `omega =sqrt((K)/(m_(1))), A=(m_(1)g)/(K)`D. `omega = sqrt((K)/(m_(1))), A =(m_(2)g)/(K)` |
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Answer» Correct Answer - B (i) When `m_(1)` is removed only `m_(2)` is left. Therefore, angular frequency: `omega = sqrt((k)/(m_(2)))` (ii) Let `x_(1)` be the extension when only `m_(2)` is left. Then, `kx_(1) = m_(2)g or x_(1) = (m_(2)g)/(k) …(1)` Similarly, let `x_(2)` be extension in equilibrium when both `m_(1)` and `m_(2)` are suspended. Then, `(m_(1)+m_(2)) g = kx_(2), :. x_(2) = ((m_(1)m_(2)g))/(k) ...(2)` From Eqs. (1) and (2) amplitude of oscillation, `A =x_(2) -x_(1) = (m_(1)g)/(k)` |
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| 709. |
For a simple pendulum, a graph is plotted between itskinetic energy `(KE)` and potential energy `(PE)`against its displacement `d`. Which one of the following represents these correctly? (graph are schematic and not drawn to scale)A. B. C. D. |
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Answer» Correct Answer - B A simple pendulum performs simple harmonic motion. For simple harmonic motion. Potential energy goes as `(1)/(2) m omega^(2)x^(2)` The maximum potential energy being `(1)/(2) m omega^(2) A^(2)`. Kinetic energy goes as `(1)/(2)m omega^(1) (A^(2) - x^(2))`. Graph (2) fits these plots. |
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| 710. |
Two bar magnets of identical size have magnetic moments MA and MB . If the magnet A oscillates at twice the frequency of magnet B, then(A) MA = 2MB (B) MA = 8MB(C) MA = 4MB(D) MA = 8MB |
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Answer» Correct option is (C) MA = 4MB |
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| 711. |
If a graph is plotted between velocity (v) and displacement (y) of a particle executing SHM from mean position, then the nature of the graph isA. Straight lineB. ParabolaC. EllipseD. Hyperbola |
| Answer» Correct Answer - 3 | |
| 712. |
The displacement (from intial position) of a particle executing `SHM` with amplitude `A` in half the time period is alwaysA. ZeroB. AC. `A//2`D. any thing from 0 to 2A |
| Answer» Correct Answer - D | |
| 713. |
For a particle executing SHM, the displacement `x` is given by `x = A cos omegat`. Identify the graph which represents the variation of potential energy `(PE)` as a function of time `t` and displacement `x`. (a) `I, III` (b) `II, IV` (c ) `II, III` (d) `I, IV`A. `I.III`B. `II.IV`C. `II,III`D. `I,IV` |
| Answer» Correct Answer - A | |
| 714. |
A particle is executing SHM. Then the graph of acceleration as a function of displacement is …….. (a) straight line (b) circle(c) ellipse (d) hyperbola |
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Answer» (a) straight line In SHM, F ∝ y ⇒ a ∝ y; Thus the graph is a straight line. |
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| 715. |
Velocity -time graph of a particle executing `SHM` is as shown in fig. Select the ocrrect alternatives. (i) at position 1, displacement of particle may be +ve or-ve (ii) at position 2, displacement of particel is -ve (iii) at position 3, acceleration of particel is+v (iv) at position 4, acceleration of particle is +ve A. i, iiB. ii, iiiC. i, ivD. iii,iv |
| Answer» Correct Answer - B | |
| 716. |
If a particle is moving as `vec(r) = ( vec(i) +2 vec(j))cosomega _(0) t ` then,motion of the particleisA. EllipticalB. Along a straight lineC. PeriodicD. Simple harmonic |
| Answer» Correct Answer - 2,3,4 | |
| 717. |
A point moves in the plane `y` according to the law `x = A sin omegat, y = B cos omegat`, where`A,B & omega` are positive constant. The velocity of particle is given byA. `v = sqrt(A^(2) +B^(2)) omega`B. `v = sqrt(A^(2)cos^(2) omegat +B^(2) sin^(2) omegat)omega`C. `v = omega sqrt((A^(2)+B^(2))-(x^(2)+y^(2)))`D. `v = omega sqrt((A+B)^(2)-(x+y)^(2))` |
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Answer» Correct Answer - B `v_(x) = (dx)/(dt),v_(y) = (dy)/(dt) = v = sqrt(v_(x)^(2) +v_(y)^(2))` |
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| 718. |
A particle moves in the `x-y` plane according to the equation `vec(r)=(vec(i)+vec(j)) (A sin omega t +B cos omega t)`. Motion of particle isA. periodicB. `SHM`C. along a straight lineD. all of the above |
| Answer» Correct Answer - D | |
| 719. |
Which of the following expression does not represent simple harmonic motion?A. `x=Acosomega+Bsinomega`B. `x=Acos(omegat+alpha)`C. `x=Bsin(omegat+B)`D. `x=Asinomegatcos^(2)omegat` |
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Answer» Correct Answer - D Simple harmonic motion is represented by a sine function or a cosine function or a linear combination of both, hence, option (a), (b) and (c) represent simple harmonic motion while option (d) is a product of the two functions (sine and cosine) does not represent a simple harmonic motion. |
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| 720. |
Which of the following expressions represent simple harmonic motion?A. `x = a sin(omegat +phi)`B. `x =a cos(omegat +delta)`C. `x =a sin omegat +b cos omegat`D. all of the above |
| Answer» Correct Answer - D | |
| 721. |
Which of the following equation does not represent a simple harmonic motionA. `x=Asinomegat`B. `x=Acosomegat`C. `x=Asinomegat+Bcosomegat`D. `x=Atanomegat` |
| Answer» Correct Answer - D | |
| 722. |
During which part of the simple harmonic motion velocity is positive but the displacement is negative, and vice versa ? |
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Answer» Velocity v is positive (to the right) while displacement x is negative when the particle in SHM is moving from the left extreme towards the mean position. Velocity v is negative (to the left) while displacement x is positive when the particle in SHM is moving from the right extreme towards the mean position. |
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| 723. |
When a mass undergoes simple harmonic motion, there is always a constant ratio between its displacement andA. periodB. massC. accelerationD. velocity |
| Answer» Correct Answer - C | |
| 724. |
The total energy of a particle executing S.H.M is proportional to A) Displacement from equilibrium position B) Frequency of oscillationsC) Velocity in equilibrium positionsD) Square of amplitude of motion |
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Answer» D) Square of amplitude of motion Explanations: Thus, the total energy in the simple harmonic motion of a particle is: • Directly proportional to its mass. • Directly proportional to the square of the frequency of oscillations and. • Directly proportional to the square of the amplitude of oscillation. |
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| 725. |
A particle of mass m executes SHM with amplitude A and frequency n. The average energy in one time period isA. `pi^(2)mn^(2)A^(2)`B. `2pi^(2)mn^(2)A^(2)`C. `pi^(2)m^(2)n^(2)A^(2)`D. `2pimn^(2)A^(2)` |
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Answer» Correct Answer - B `E=(1)/(2) m omega^(2)x^(2)A^(2)=(1)/(2) m(2pin)^(2)A^(2)` `=2pi^(2)MN^(2)a^(2)`. |
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| 726. |
A particle executes simple harmonic motion of amplitude A. (i) At what distance from the mean positio is its kinetic energy equal to its potential energy? (ii) At what points is its speed half the maximum speed?A. 0.81 AB. 0.71 AC. 0.41 AD. 0.91 A |
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Answer» Correct Answer - B Potential energy of a particle executing SHM at a displacement y from the mean position is `PE = (1)/(2)ky^(2)" "...(i)` Final kinetic energy of the particle is `KE = (1)/(2)k(A^(2)-y^(2))" "...(ii)` As PE and KE are equal, so equating them, we get `(1)/(2)ky^(2)=(1)/(2)k(A^(2)-y^(2))` or `A^(2)-y^(2)=y^(2)` `rArr" "2y^(2)=A^(2)` or `y=(A)/(sqrt(2))=0.07 A = 0.71 A` |
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| 727. |
A particle executing S.H.M. starts from midway between mean and extreme position, the phase isA. `(pi)/(3)` radB. `(pi)/(2)` radC. `(pi)/(6)` radD. `pi` rad |
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Answer» Correct Answer - C The phase angle for midway between mean and extreme position is `x=(A)/(2)" is, "delta=(pi)/(6)`. |
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| 728. |
A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilinrium to the end is.A. `pi^(2)mn^(2)A^(2)`B. `2pi^(2)mn^(2)A^(2)`C. `(pi^(2)mn^(2)A^(2))/(2)`D. zero |
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Answer» Correct Answer - A `(ke_("max")+0)/(2)=(1)/(2) ((1)/(2)m omega^(2) A^(2))` `=(1)/(4) m (2pin)^(2)A^(2)` `=pi^(2)mn^(2)A^(2)`. |
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| 729. |
A particle executing S.H.M. has total energy of 20 J. If the potential energy of the particle midway between mean and extreme position is 5 J, then the average potential energy will be |
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Answer» Correct Answer - A `gt P.E.lt =(1)/(2) gt T.E.lt` |
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| 730. |
A particle executing S.H.M. has total energy of 20 J. If the potential energy of the particle midway between mean and extreme position is 5 J, then the average total energy will beA. 10 JB. 20 JC. 15 JD. 7.5 J |
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Answer» Correct Answer - B `gtT.E.lt =T.E.` |
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| 731. |
A particle executing S.H.M. has total energy of 20 J. If the potential energy of the particle midway between mean and extreme position is 5 J, then the average potential energy will beA. 10 JB. 5 JC. 15 JD. 7.5 J |
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Answer» Correct Answer - A `gt K.E.lt =(1)/(2) gt t.e.lt` |
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| 732. |
In S.H.M. the displacement of a particle is half of the amplitude, the share of the potential energy and kinetic energy areA. `50%` and `50%` respectivelyB. `59%` and `41%1` respectivelyC. `25%` and `75%` respectivelyD. `33%` and `66%` respectively |
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Answer» Correct Answer - C `P.E.=(1)/(2) m omega^(2)(A^(2))/(4) at x=(A)/(2)` `=(1)/(2) m omega^(2)A^(2)x25%` `=25% T.E.` `K.E.= 25% T.E.` |
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| 733. |
The ratio of potential energy to kinetic energy of a particle performing S.H.M. at midway between mean and extreme position isA. `3 : 1`B. `1 : 3`C. `6 : 9`D. `4 : 3` |
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Answer» Correct Answer - B `(P.E.)/(K.E.)=((1)/(2)k(A^(2))/(4))/((1)/(2)k(A^(2)-(A^(2))/(4)))=(A^(2))/(4)xx(4)/(3A^(2))=(1)/(3)` |
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| 734. |
A particle of mass 100 g is executing S.H.M. with amplitude of 10 cm. When the particle passes through the mean position at t = 0. Its kinetic energy is 8 mJ. The equation of simple harmonic motion, if initial phase is zero isA. x = 0.1 sin 4tB. x = 0.1 cos 4tC. x = 0.1 sin 2tD. x = 0.1 cos 2t |
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Answer» Correct Answer - A `K.E.=(1//2)momega^(2)A^(2)` `10^(-3)xx8=(1)/(2)xx0.1xx0.01omega^(2)` `therefore omega^(2)=16` `omega=4` `therefore x=0.1sinomegat` |
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| 735. |
The equation of SHM of a particle is \(\frac{d^2y}{dt^2}\) + ky = 0, Where k is a positive constant. The time period of motion is given by ...(a) \(\frac{2π}{\sqrt k}\)(b) \(\frac{2π}{k}\)(c) \(\frac{k}{2π}\)(d) \(\frac{\sqrt k}{2π}\) |
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Answer» Correct answer is (a) \(\frac{2π}{\sqrt k}\) Here k is same as ω2 |
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| 736. |
Define force constant of a spring. |
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Answer» The spring constant of a spring is the change in the force it exerts, divided by the change in deflection of the spring. (K = f/x) |
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| 737. |
How is the frequency of oscillation related with the frequency of change in the K.E. and P.E. of the body in S.H.M.? |
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Answer» P.E. or K.E. completes two vibrations in a time during which S.H.M. completes one vibration or the frequency of R.E. or K.E is double than that of S.H.M. |
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| 738. |
What is the frequency of total energy of a particle in S.H.M. ? |
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Answer» The frequency of total energy of particle is S.H.M. is zero because it remains constant. |
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| 739. |
A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration \(\cfrac{1}{30}\)s after it has crossed the mean position. |
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Answer» Data : A = 4 cm = 4 × 10-2 m, f = 5Hz, t = \(\frac{1}{30}\)s ω = 2πf = 2π (5) = 10π rad/s Therefore, the magnitude of the acceleration, |a| = ω2 x = ω2 A sin ωt = (10π)2 (4 × 102) = 10π2 sin\(\frac{\pi}{3}\) = 10 (9.872)(0.866) = 34.20 m/s2 |
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| 740. |
A particle moves in a circular path with constant speed . Its motion isA. PeriodicB. OscillatoryC. Simple harmonicD. Angular simple harmonic |
| Answer» Correct Answer - 1 | |
| 741. |
If a ball is dropped from height 2 metre on a smooth eleastic floor, then the time period of oscillation isA. `(2)/(g )`B. `(2)/( sqrt(g))`C. `(4)/(g )`D. `(4)/(sqrt(g))` |
| Answer» Correct Answer - 4 | |
| 742. |
A body of mass 1 kg is made to oscillate on a spring of force constant 16 N/m. Calculate (i) the angular frequency (ii) the frequency of oscillation. |
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Answer» Data : m = 1 kg, k = 16 N/m Angular Frequency \(\omega\) = \(\sqrt{\frac km}\) = \(\sqrt{\frac {16}1}\) = 4 rad/s Frequency, f = \(\frac{\omega}{2\pi}\) = \(\frac{4}{2\times3.142}\) = 0.6365 Hz |
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| 743. |
Four mass less spring whose force constant are `2k , 2k, k` and `2k` respectively are attached to a mass `M` kept on a friction less plate (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 `f = (1)/(2pi) sqrt((K_(eff))/(m))` |
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| 744. |
A load of mass 100 gm increases the length of wire by 10 cm. If the system is kept in oscillation, its time period is `(g=10m//s^(2))`A. `0.314 s`B. `3.14 s`C. `0.628 s`D. `6.28 s` |
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Answer» Correct Answer - C `T=2pisqrt((M)/(k))=2pisqrt((0.100)/(10))=2pisqrt(1xx10^(-2))` `=2pixx10^(-1)=0.628s` |
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| 745. |
A particle of mass m is executing SHM about its mean position. The total energy of the particle at given instant isA. `(pi^(2)mA^(2))/(T^(2))`B. `(2pi^(2)mA^(2))/(T^(2))`C. `(4pi^(2)mA^(2))/(T^(2))`D. `(8pi^(2)mA^(2))/(T^(2))` |
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Answer» Correct Answer - B `E=(1)/(2)momega^(2)A^(2)=(m4pi^(2)A^(2))/(T^(2))=(4pi^(2)mA^(2))/(T^(2))` |
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| 746. |
A particle is executing `SHM`. It passes through mean position at the instant `t = 0`. At what instants the speed of it is `50%` of its maximum speed? |
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Answer» `v = v_(max) cos (omega t) rArr (v_(max))/(2) = v_(max) cos ((2pi)/(T)t)` `t = (T)/(2pi) cos^(-2) (+-(1)/(2))` |
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| 747. |
Which one of the following is not a periodic motion?A. Rotation of the earth about its axis.B. A freely suspended bar magnet displaced from its N-S direction and released.C. Motionn of hands of a clock.D. An arrow released from a bow. |
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Answer» Correct Answer - D An arrow released from a bow is not a periodic motion while all others are periodic motion. |
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| 748. |
Fig. 14.23 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion) ? |
| Answer» Figure (b) and (d) represent periodic motion and the time period of each of these is 2 seconds (a), and (c) are non-periodic motions. | |
| 749. |
Which of the following x-t graphs does not represent periodic motion?A. B. C. D. |
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Answer» Correct Answer - B Option (a) represents periodic motion. Option (b) does not represent periodic motion option (c) represents periodic motion. option (d) represents periodic motion. |
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| 750. |
if the frequency of human heart is 1.25 Hz, the number of heart beats in 1 minutes isA. 65B. 75C. 80D. 90 |
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Answer» Correct Answer - B Beat frequency of heart=1.25 Hz `therefore`Number of beats in 1 minute=`1.25xx60=75` |
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