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P the Time Period of a Particle in Simple Harmonic Motion is Equal to the Smallest Time Between the Particle Acquiring a Particular Velocity → V . - Physics

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प्रश्न

The time period of a particle in simple harmonic motion is equal to the smallest time between the particle acquiring a particular velocity \[\vec{v}\] . The value of v is

विकल्प

  • vmax

  • 0

  • between 0 and vmax

  • between 0 and −vmax

MCQ
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उत्तर

vmax

Because the time period of a simple harmonic motion is defined as the time taken to complete one oscillation.

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अध्याय 12: Simple Harmonics Motion - MCQ [पृष्ठ २५०]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 12 Simple Harmonics Motion
MCQ | Q 3 | पृष्ठ २५०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?

(a) a = 0.7x

(b) a = –200x2

(c) a = –10x

(d) a = 100x3


Show variation of displacement, velocity, and acceleration with phase for a particle performing linear S.H.M. graphically, when it starts from the extreme position.


Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S.H.M.


Can simple harmonic motion take place in a non-inertial frame? If yes, should the ratio of the force applied with the displacement be constant?


Can the potential energy in a simple harmonic motion be negative? Will it be so if we choose zero potential energy at some point other than the mean position?


A particle moves on the X-axis according to the equation x = A + B sin ωt. The motion is simple harmonic with amplitude


A wall clock uses a vertical spring-mass system to measure the time. Each time the mass reaches an extreme position, the clock advances by a second. The clock gives correct time at the equator. If the clock is taken to the poles it will


A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t = 4 s.


The pendulum of a certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours?


Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance R/2 from the earth's centre where R is the radius of the earth. The wall of the tunnel is frictionless. (a) Find the gravitational force exerted by the earth on a particle of mass mplaced in the tunnel at a distance x from the centre of the tunnel. (b) Find the component of this force along the tunnel and perpendicular to the tunnel. (c) Find the normal force exerted by the wall on the particle. (d) Find the resultant force on the particle. (e) Show that the motion of the particle in the tunnel is simple harmonic and find the time period.


A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with and acceleration a0(b) is going down with an acceleration a0 and (c) is moving with a uniform velocity.


A uniform rod of length l is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the road.


Three simple harmonic motions of equal amplitude A and equal time periods in the same direction combine. The phase of the second motion is 60° ahead of the first and the phase of the third motion is 60° ahead of the second. Find the amplitude of the resultant motion.


A particle is subjected to two simple harmonic motions given by x1 = 2.0 sin (100π t) and x2 = 2.0 sin (120 π t + π/3), where x is in centimeter and t in second. Find the displacement of the particle at (a) = 0.0125, (b) t = 0.025.


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T = `2π sqrt("R"/"g")`


Consider two simple harmonic motion along the x and y-axis having the same frequencies but different amplitudes as x = A sin (ωt + φ) (along x-axis) and y = B sin ωt (along y-axis). Then show that

`"x"^2/"A"^2 + "y"^2/"B"^2 - (2"xy")/"AB" cos φ = sin^2 φ`

and also discuss the special cases when

  1. φ = 0
  2. φ = π
  3. φ = `π/2`
  4. φ = `π/2` and A = B
  5. φ = `π/4`

Note: when a particle is subjected to two simple harmonic motions at right angle to each other the particle may move along different paths. Such paths are called Lissajous figures.


The displacement of a particle is represented by the equation `y = 3 cos (pi/4 - 2ωt)`. The motion of the particle is ______.


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