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Acceleration of a particle executing S.H.M. at its mean position.

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प्रश्न

Acceleration of a particle executing S.H.M. at its mean position.

विकल्प

  • Is infinity

  • Varies

  • Is maximum

  • Is zero

MCQ
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उत्तर

The acceleration of a particle executing S.H.M. at its mean position is zero. 

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अध्याय 5: Oscillations - MCQ’s

संबंधित प्रश्न

Choose the correct option:

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 particle is performing S.H.M. of amplitude 5 cm and period of 2s. Find the speed of the particle at a point where its acceleration is half of its maximum value.


For a particle performing SHM when displacement is x, the potential energy and restoring force acting on it is denoted by E and F, respectively. The relation between x, E and F is ____________.


The displacement of a particle from its mean position (in metre) is given by, y = 0.2 sin(10 πt + 1.5π) cos(10 πt + 1.5π).

The motion of particle is ____________.


In U.C.M., when time interval δt → 0, the angle between change in velocity (δv) and linear velocity (v) will be ______.


A body performing a simple harmonic motion has potential energy 'P1' at displacement 'x1' Its potential energy is 'P2' at displacement 'x2'. The potential energy 'P' at displacement (x1 + x2) is ________.


The relation between time and displacement for two particles is given by Y1 = 0.06 sin 27`pi` (0.04t + `phi_1`), y2 = 0.03sin 27`pi`(0.04t +  `phi_2`). The ratio of the intensity of the waves produced by the vibrations of the two particles will be ______.


If 'α' and 'β' are the maximum velocity and maximum acceleration respectively, of a particle performing linear simple harmonic motion, then the path length of the particle is _______.


The distance covered by a particle undergoing SHM in one time period is (amplitude = A) ____________.


Which of the following represents the acceleration versus displacement graph of SHM?


A particle executing S.H.M. has amplitude 0.01 m and frequency 60 Hz. The maximum acceleration of the particle is ____________.


A body of mass 5 g is in S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm/s. Its velocity will be 50 cm/s at a distance of, ____________.


The length of the second's pendulum is decreased by 0.3 cm when it is shifted from place A to place B. If the acceleration due to gravity at place A is 981 cm/s2, the acceleration due to gravity at place B is ______ (Take π2 = 10)


A body is executing S.H.M. Its potential energy is E1 and E2 at displacements x and y respectively. The potential energy at displacement (x + y) is ______.


The displacement of the particle performing S.H.M. is given by x = 4 sin πt, where x is in cm and t is in second. The time taken by the particle in second to move from the equilibrium position to the position of half the maximum displacement, is ______.

`[sin30^circ=cos60^circ=0.5, cos30^circ=sin60^circ=sqrt3/2]`


A particle performs linear SHM at a particular instant, velocity of the particle is 'u' and acceleration is a while at another instant velocity is 'v' and acceleration is 'β (0 < α < β). The distance between the two position is ______.


A particle of mass 5 kg moves in a circle of radius 20 cm. Its linear speed at a time t is given by v = 4t, t is in the second and v is in ms-1. Find the net force acting on the particle at t = 0.5 s.


The displacement of a particle of mass 3 g executing simple harmonic motion is given by Y = 3 sin (0.2 t) in SI units. The kinetic energy of the particle at a point which is at a distance equal to `1/3`​ of its amplitude from its mean position is ______.


Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by x = `5sin((pit)/3)`m.


State the expressions for the displacement, velocity and acceleration draw performing linear SHM, starting from the positive extreme position. Hence, their graphs with respect to time.


Which one of the following is not a characteristics of SHM?


A particle executing SHM has velocities v1 and v2 when it is at distance x1 and x2 from the centre of the path. Show that the time period is given by `T=2pisqrt((x_2^2-x_1^2)/(v_1^2-v_2^2))`


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