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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. |
A solid sphere of radius `R` is floating in a liquid of density `sigma` with half of its volume submerged. If the sphere is slightly pushed and released , it starts axecuting simple harmonic motion. Find the frequency of these oscillations.A. `(1)/(2pi) sqrt((3g)/(2R))`B. `(1)/(2pi) sqrt(2R)/(3g)`C. `(1)/(2pi)sqrt((R )/(g))`D. `(1)/(2pi)sqrt((5g)/(3R))` |
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Answer» Correct Answer - A Half of the volume of sphere is submerged. For equilibrium of sphere, Weight =upthrust `:. Vr_(x)g =(V)/(2) (r_(L)) (g), r_(s) =(rho_(L))/(2) ..(1)` When slightly pushed downwards by `x`, weigth will remain as it is while upthrust will increase. The increased upthrust will become the net restoring force (upwards). `F =-("extra upthrust")` `=- ("extra volume immersed") (r_(L)) (g)` or `ma =- (piR^(2)) xr_(L) (a= "acceleration")` `:. (4)/(3)piR^(3) ((rho)/(2))a =- (piR^(2)) xr_(L)g (a= "acceleraiton")` `:. a =- ((3g)/(2R))x` as the given motion is simple harmonic Frequency of oscillation, `f = (1)/(2pi) sqrt(|(a)/(x)|) =(1)/(2)sqrt((3g)/(2R))` |
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| 502. |
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_(2)^(-1) +t_(2)^(-1)`D. `T^(-2)=t_(1)^(-2)+t_(2)^(-2)` |
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Answer» Correct Answer - B Time period of spring `T = 2pi sqrt(((m)/(k)))` `k` being the force constant of spring. For first spring. `t_(1) = 2pi sqrt(((m)/(k_(1)))) ….(1)` For second spring `t_(2) = 2pi sqrt(((m)/(k_(2)))) …(2)` The effective force constant in their series combination is `k = (k_(1)k_(2))/(k_(1)+k_(2))`. Therefore, time period of combination `T = 2pi sqrt([(m(k_(1)+k_(2)))/(k_(1)k_(2))])` `rArr T^(2) = (4pi^(2)m(k_(1)+k_(2)))/(k_(1)k_(2)) ....(3)` From equations (1) and (2), we obtain `t_(1)^(2) +t_(2)^(2) =4pi^(2) ((m)/(k_(1))+(m)/(k_(2))) :. t_(1)^(2) +t_(2)^(2) =T^(2)` [from eq (3)] |
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| 503. |
An object is attched to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is `15cm//s` and the period is `628` milli seconds. The amplitude of the motion in centimetres isA. `3.0`B. `2.0`C. `1.5`D. `1.0` |
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Answer» Correct Answer - C `v_(m)=15cm//s, T=6.28xx10^(-1)s` `v_(m)=Aomega` `A=(v_(m))/(omega)=(15)/(2pi)xxT` `=(15xx6.28xx10^(-1))/(6.28)=15xx10^(-1)=1.5cm`. |
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| 504. |
The graph shows variation of displacement of a particle performing S.H.M. with time t. Which of the following statements is correct from the graph?(A) The acceleration is maximum at time T. (B) The force is maximum at time 3T/4. (C) The velocity is zero at time T/2. (D) The kinetic energy is equal to total energy at time T/4. |
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Answer» Correct option is: (B) The force is maximum at time 3T/4. |
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| 505. |
Define linear simple harmonic motion. |
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Answer» Definition: Linear simple harmonic motion (SHM) is defined as the linear periodic motion of a body, in which the force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position. OR A particle is said to execute linear SHM if the particle undergoes oscillations about a point of stable equilibrium, subject to a linear restoring force always directed towards that point and whose magnitude is proportional to the magnitude of the displacement of the particle from that point. Examples : The vibrations of the tines (prongs) of a tuning fork, the oscillations of the needle of a sewing machine. |
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| 506. |
From what minimum height `h`must the system be released when spring is unstretches so that after perfectly inelastic collision `(e =0)` with ground, `B` may be lifted off the gound: (Spring constant `=k)` A. `mg //(4k)`B. `4mg//k`C. `mg//(2k)`D. None of the above |
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Answer» Correct Answer - B Just after collision with ground, Applying `COE` `(1)/(2)mv^(2) +mgx +(1)/(2) kx^(2) = (1)/(2)m (2gh) +0 +0` `rArr (1)/(2) mv^(2) gt 0 rArr h gt 4mg//k` |
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| 507. |
The average acceleration over on period of oscialltion for a simple harmonic motion isA. ZeroB. `2 A omega^(2)`C. ` (Aomega^(2))/(2)`D. `A omega^(2)` |
| Answer» Correct Answer - 1 | |
| 508. |
A block of mass 1 kg is placed inside a car of mass 5 kg , as shown . The block can slide smoothly along horizontal direction. If block is displaced slightly and released , then time period of osciallation is A. `2pi sqrt((5)/(2k))`B. ` 2pi sqrt((5)/(12k))`C. `2pi sqrt((12)/(5k))`D. `2pi sqrt((2)/(5k))` |
| Answer» Correct Answer - 2 | |
| 509. |
A particle performs SHM of amplitude 10 cm. Its maximum velocity during oscillations is 100 cm/s. What is its displacement, when the velocity is 60 cm/s? |
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Answer» Data : A = 10 cm, vmax = 100 cm/s, v = 60 cm/s vmax = ωA = 100 cm/s \(\therefore\) ω = \(\frac{v_{max}}A\) = \(\frac{100}{10}\) = 10 rad/s Let x be the displacement when the velocity is v = 60 cm/s. Then, v = ω \(\sqrt{A^2-x^2}\) \(\therefore\) 60 = 10 \(\sqrt{100-x^2}\) \(\therefore\) 6 = \(\sqrt{A^2-x^2}\) \(\therefore\) 36 = 100 - x2 \(\therefore\) x2 = 64 \(\therefore\) x = ± 8 cm |
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| 510. |
Two simple harmonic motions are represented by the equations. `y_(1)=10"sin"(pi)/(4)(12t+1),y_(2)=5(sin3pt+sqrt(3)cos3pt)` the ratio of their amplitudes isA. `1:1`B. `1:2`C. `3:2`D. `2:3` |
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Answer» Correct Answer - A Here, `y_(1)=10"sin"(pi)/(4)(12t+1)` or `y_(1)=10sin(3pit+(pi)/(4))` . . . (i) and `y_(2)=5(sin3pit+sqrt(3)cos3pit)` or `=10(sin3pitxx(1)/(2)+cos3pitxx(sqrt(3))/(2))` `implies y_(2)=10sin(3pit+(pi)/(3))` . . . (ii) Comparing equation (i) and (ii) with standard equation for SHM we get `A_(1)=10 and A_(2)=10 " "because(A_(1))/(A_(2))=(10)/(10)=1/1` |
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| 511. |
Displacement versus time curve for a particle executing SHM is shown in figure. Choose the correct statements. A. Phase of the oscillator is same at t=0 s and t=2 s.B. Phase of the oscillator is same at t=2 s and t=5sC. Phase of the oscillator is same at t=1s and t=7 sD. Phase of the oscillator is same at t=1s and t=5s. |
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Answer» Correct Answer - D The phase of a particle executing SHM is defined as the state of a particle as regards to its position and direction of motion at any instant of time. In the given curve, phase is same when t=1s and t=5s. Also phase is same when t=2s and t=6s. |
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| 512. |
The graph between instantaneous velocity and angular displacement of a particle performing S.H.M. isA. parabolaB. straight lineC. sinusoidalD. circle |
| Answer» Correct Answer - C | |
| 513. |
The graph between instantaneous velocity and displacement of a particle performing S.H.M.with period `2pi sec or omega = 1` isA. parabolaB. straight lineC. ellipseD. circle |
| Answer» Correct Answer - D | |
| 514. |
The graph between instantaneous acceleration and angualr displacement of a particle performing S.H.M. isA. parabolaB. straight lineC. sinusoidalD. circle |
| Answer» Correct Answer - C | |
| 515. |
A graph is plotted between the instantaneous velocity of a particle performing S.H.M. and displacement. Then the period of S.H.M. is A. 2 sB. `pi s`C. `(1)/(pi) s`D. `0.5 s` |
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Answer» Correct Answer - B `v_(m)=8 A=4` `v_(m)=Aomega` `therefore omega=(v_(m))/(A)=(8)/(4)=2` |
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| 516. |
A particle oscillates, according to the equation x = 5 cos `(pit//2)` m. Then the particle moves from equilibrium position to the position of maximum displacement in time ofA. 1 sB. 2 sC. `1//2` sD. 4 s |
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Answer» Correct Answer - A `x=5cos.(pi)/(2)t` Compare it with x = cos `omegat` `therefore omega=(pi)/(2)` `(2pi)/(T)=(pi)/(2)thereforeT=4` `t=(T)/(4)` `=(4)/(4)=1s` |
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| 517. |
Two simple harmonic motions are given by `y_(1) = a sin [((pi)/(2))t + phi]` and `y_(2) = b sin [((2pi)/( 3))t + phi]`. The phase difference between these after `1 s` isA. `pi`B. `pi//2`C. `pi//4`D. `pi//6` |
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Answer» Correct Answer - D `phi_(2)-phi_(1)=((2pi)/(3)t-(pi)/(2)t)=((2pi)/(3)-(pi)/(2))xx1` `=((4pi-3pi)/(6))=(pi)/(6)` |
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| 518. |
The potential energy as a function of displacement for an oscillator `u=8x^(2)` joule. The magnitude of the force on the oscillator of mass 0.5 kg placed at x = 0.2 m isA. 1.6 NB. 3.2 NC. 0.8 ND. 4.8 N |
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Answer» Correct Answer - B `u=8x^(2)` `PE=8x^(2)` `(1)/(2)kx^(2)=8x^(2) therefore k=16` `F=(1)/(2)kx=(1)/(2)xx16xx0.2=1.6N` |
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| 519. |
The displacement of a simple harmonic oscillator is, x = 5 sin `(pit//3) m`. Then its velocity at t = 1 s, isA. `(pi)/(6)m//s`B. `(5pi)/(6)m//s`C. `(6pi)/(6)m//s`D. `(pi)/(2)m//s` |
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Answer» Correct Answer - B `x=5sin.(pit)/(3)` `(dx)/(dt)=5xx(pi)/(3)cos.(pi)/(3)t` `v=(5pi)/(3)xxcos((pi)/(3)xx1)=(5pi)/(3)xx(1)/(2)` `=(5pi)/(6)m//s` |
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| 520. |
The displacement of a simple harmonic oscillator is given by x = 4 cos `(2pit+pi//4)m`. Then velocity of the oscillator at t = 2 s isA. `4pisqrt(2)m//s`B. `(4pi)/(sqrt(2))m//s`C. `(sqrt(2)pi)/(4)m//s`D. `(4sqrt(2))/(pi)m//s` |
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Answer» Correct Answer - A `x=4cos(2pit+(pi)/(4))` `(dx)/(dt)=4xx2pisin(2pit+(pi)/(4))` `v=8pixxsin(2pixx2+(pi)/(4))` `=8pisin.(pi)/(4)` `=8pixx(1)/(sqrt(2))=(4sqrt(2)xxsqrt(2)pi)/(sqrt(2))=4sqrt(2)pi`. |
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| 521. |
A particle of mass 1 kg executing S.H.M. is given by y = 2 cos `(10t+pi//3)` in SI units. Its maximum potential energy isA. 2 JB. 20 JC. 200 JD. 100 J |
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Answer» Correct Answer - C `P.E.=(1)/(2)momega^(2)A^(2)` `=(1)/(2)xx1xx100xx4=200J` |
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| 522. |
In linear SHM, what can you say about the restoring force when the speed of the particle is 1. zero2. maximum ? |
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Answer» The restoring force is 1. maximum 2. zero. |
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| 523. |
The particle performing S.H.M., about mean position it hasA. maximum acceleration and maximum velocityB. minimum acceleration and maximum velocityC. maximum acceleration and minimum velocityD. minimum acceleration and minimum velocity |
| Answer» Correct Answer - B | |
| 524. |
A particle executing SHM passes through the mean position with a velocity of `4 ms^(-1)`. The velocity of the particle at a point where the displacement is half of the amplitude isA. `2 ms^(-1)`B. `2sqrt(3)ms^(-1)`C. `sqrt(3)ms^(-1)`D. `1ms^(-1)` |
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Answer» Correct Answer - B `V=(Aomegasqrt(3))/(2)=(4sqrt(3))/(2)=2sqrt(3)ms^(-1)` |
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| 525. |
The period of SHM of a particle with maximum velocity 50 cm/s and maximum acceleration 10 cm/s2 is (A) 31.42 s (B) 6.284 s (C) 3.142 s (D) 0.3142 s. |
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Answer» Correct option is (C) 3.142 s |
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| 526. |
For a particle in `SHM` the amplitude and maximum velocity are `A` and `V` respectively. Then its maximum acceleration isA. `(V^(2))/(2A)`B. `V^(2)A`C. `(V^(2))/(A)`D. `(V)/(A)` |
| Answer» Correct Answer - C | |
| 527. |
A particle executing SHM of amplitude 5 cm has an acceleration of 27 cm/s2 when it is 3 cm from the mean position. Its maximum velocity is (A) 15 cm/s (B) 30 cm/s (C) 45 cm/s (D) 60 cm/s. |
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Answer» Correct option is (A) 15 cm/s |
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| 528. |
A large horizontal surface moves up and down in SHM with an amplitude of 1 cm . If a mass of 10 kg (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of S.H.M. will beA. `0.5Hz`B. `1.5 Hz`C. `5.0 Hz`D. `10.0Hz` |
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Answer» Correct Answer - C `F_(max) = (1)/(2pi) sqrt(g)/(A)` |
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| 529. |
A block of mass `m` moves with a speed `v` towards the right block which is in equilibrium with a spring attached to rigid wall. If the surface is frictionless and collisions are elastic, the frequency of collisions between the masses will be: A. `(v)/(2L)+(1)/(pi)sqrt((K)/(m))`B. `2[(v)/(2L)+(1)/(2pi)sqrt((K)/(m))]`C. `(2)/([(2L)/(v)+pisqrt((m)/(K))])`D. `(v)/(2l)+(1)/(pi)sqrt((m)/(K))` |
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Answer» Correct Answer - C Time taken to collide on left wall and get back to the mass attached with spring is `t_(1) = (2L)/(v)`. Time to get the spring compressed once and to come back is, `t_(2) = (T)/(2) =(2pi)/(2) sqrt((m)/(K)) =pi sqrt((m)/(K)) :.` Average time between two successive collisions, `t = (t_(1)+t_(2))/(2)` collision freqency `=(1)/(t)` |
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| 530. |
A sphere fo mass m makes SHM in a hemispherical bowl ABC and it moves from A to C and back to A via ABC, so that PB = h. If acceleration due to gravity is g, the speed of the ball when it just crosses the point B is A. `2gh`B. `mgh`C. `sqrt(2gh)`D. `(gh)/(2)` |
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Answer» Correct Answer - C Potential energy at A is converted into K.E. at B `thereforemgh=(1)/(2)mv_(B)^(2),V_(B)^(2)=2gh,v_(B)=sqrt(2gh)` |
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| 531. |
The displacement x is in centimeter of an oscillating particle varies with time t in seconds as x = 2 cos `[0.05pi t+(pi//3)]`. Then the magnitude of the maximum acceleration of the particle will beA. `(pi)/(2)cm//s^(2)`B. `(pi)/(4)cm//s^(2)`C. `(pi^(2))/(200)cm//s^(2)`D. `(pi^(2))/(4)cm//s^(2)` |
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Answer» Correct Answer - C Acceleration max `=(Aomega^(2))` `=2(0.05pi)^(2)` `=2xx25 pi^(2)xx10^(-4)` `=50pi^(2)xx10^(-4)` `=0.5pi^(2)xx10^(-2)` `=(pi^(2))/(200)` |
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| 532. |
In a period of oscillating particle, the number of times kinetic energy and potential energy are equal in magnitude isA. singleB. thriceC. twiceD. four times |
| Answer» Correct Answer - D | |
| 533. |
In a period of kinetic energy, the number of times kinetic energy and potential energy are equal in magnitude isA. single timesB. thriceC. twiceD. four times |
| Answer» Correct Answer - C | |
| 534. |
The potential energy of a particle performing S.H.M. in its rest position is 15 J. If the average kinetic energy is 5 J, then the total energy of the particle performing S.H.M. will beA. 5 JB. 10 JC. 20 JD. 15 J |
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Answer» Correct Answer - B `lt K.E. gt =(T.E.)/(2)` `therefore T.E.=2ltK.E. gt =2xx5=10J` |
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| 535. |
Kinetic energy of a particle performing S.H.M.A. leads the potential energy by a phase of `pi`B. leads the potential energy by a phase of `pi//2`C. lags the potential energy by a phase of `pi//2`D. lags the potential energy by a phase of `pi` |
| Answer» Correct Answer - B | |
| 536. |
The graph between potential energy and displacement of a particle performing S.H.M. isA. parabolaB. straight lineC. ellipseD. circle |
| Answer» Correct Answer - A | |
| 537. |
The graph between kinetic energy and displacement of a particle performing S.H.M. isA. parabolaB. straight lineC. ellipseD. circle |
| Answer» Correct Answer - A | |
| 538. |
The graph between total energy and displacement of a particle performing S.H.M. isA. parabolaB. straight lineC. ellipseD. circle |
| Answer» Correct Answer - B | |
| 539. |
Potential energy of the particle performing S.H.M. isA. harmonic motion and oscillatoryB. periodic motion but not oscillatoryC. oscillatory motion but not periodicD. periodic and oscillatory motion |
| Answer» Correct Answer - B | |
| 540. |
A body of mass 0.50kg is executing SHM Its period is 0.1 s and amplitude 10cm. When the body is at a distance of 5cm from the mean postion, find i its accelration ii force acting upon it and ii its potential energy. |
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Answer» Correct Answer - (i) `197.2ms^(-2)` (ii). `98.6N` (iii). 2.465J |
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| 541. |
If the two particles performing S.H.M. with different initial phase angle and amplitude, then the initial phase angle of resultant motion depends onA. initial phase angle onlyB. initial phase angle and amplitude of individualC. amplitude of individual onlyD. neither amplitude nor initial phase angle |
| Answer» Correct Answer - B | |
| 542. |
The total energy of a partical executing simple harmonic motion of period `2pi` seconds is 10,240erg. The displacement of the particle at `pi//4` second is `8sqrt(2)cm`. Calculate the amplitutde of motion and mass of the particle |
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Answer» Correct Answer - 0.16m; 0.08 kg |
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| 543. |
A spring of force constant k is cut into four equal parts. The force constant of each part will beA. kB. 4 kC. `k//4`D. 16 k |
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Answer» Correct Answer - B Let k be the force constant for each piece. `(1)/(k_(s))=(1)/(k)+(1)/(k)+(1)/(k)+(1)/(k)` `therefore (1)/(k_(s))=(4)/(k) therefore k_(s)=4k` |
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| 544. |
If the two particles performing S.H.M. with same amplitude and initial phase angle, then the initial phase angle of resultant motion depends onA. initial phase angle onlyB. initial phase angle and amplitude of individualC. amplitudeof individual onlyD. neither amplitude nor initial phase angle |
| Answer» Correct Answer - A | |
| 545. |
For a body of mass m attached to the spring the spring factor is given byA. `m//omega^(2)`B. `momega^(2)`C. `m^(2)omega`D. `m^(2)omega^(2)` |
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Answer» Correct Answer - B `K=momega^(2)` |
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| 546. |
1 kg weight is suspended to a weightless spring and it has time period T. If now 4 kg weight is suspended from the same spring the time period will beA. TB. `T//2`C. `2T`D. 4T |
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Answer» Correct Answer - C `T_(1)=2pisqrt((m)/(k))=2pisqrt((1)/(k))` `T_(2)=2pisqrt((4)/(k))=2pixx2sqrt((1)/(k))` `(T_(2))/(T_(1))=(2)/(1) therefore T_(2)=2T_(1)=2T` |
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| 547. |
Define angular frequency. Give its S.I. unit. |
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Answer» It is the angle covered per unit time or it is the quantity obtained by multiplying frequency by a factor of 2π. ω = 2πn, S.I. unit is rad s–1. |
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| 548. |
The differential equation of angular S.H.M. is in the order ofA. 2B. 0C. 3D. 1 |
| Answer» Correct Answer - A | |
| 549. |
The ratio of maximum acceleration to the maximum velocity of a particle performing S.H.M. is equal toA. amplitudeB. angular velocityC. square of amplitudeD. square of angular velocity |
| Answer» Correct Answer - B | |
| 550. |
A particle executes S.H.M. of amplitude 25 cm and time period 3 s. What is the minimum time required for the particle to move between two points 12.5 cm on either side of the mean position ?A. 0.25 sB. 0.5 sC. 0.75 sD. 1 s |
| Answer» Correct Answer - B | |