The acceleration of a particle performing SHM is 12 cm sec-2 at a distance of 3 cm from the mean position. Its time period is:

1. 2.0s

2. 3.14s

3. 0.5s

4. 1.0s

Subtopic:  Simple Harmonic Motion |
 89%
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The acceleration d2x/dt2 of a particle varies with displacement x as d2xdt2=-kx

where k is a constant of the motion. The time period T of the motion is equal to :

1. 2πk

2. 2πk

3. 2π/k

4. 2π/k

Subtopic:  Simple Harmonic Motion |
 87%
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A coin is placed on a horizontal platform, which undergoes horizontal SHM about a mean position O. The coin placed on the platform does not slip, the coefficient of friction between the coin and the platform is μ. The amplitude of oscillation is gradually increased. The coil will begin to slip on the platform for the first time:

1. at the mean position

2. at the extreme position of oscillations

3. for an amplitude of μg/ω2

4. for an amplitude of g/μω2

Subtopic:  Simple Harmonic Motion |
 72%
From NCERT
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A particle moves according to the law, \(x = r \mathrm{cos}\left(\frac{\pi t}{2}\right )\). The distance covered by it in the time interval between \(t=0\) to \(t = 3~\text{s}\) is:
1. \(r\)
2. \(2r\)
3. \(3r\)
4. \(4r\)

Subtopic:  Simple Harmonic Motion |
From NCERT
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A particle is executing SHM of period 24 sec and of amplitude 41 cm with O as equilibrium position. The minimum time in seconds taken by the particle to go from P to Q, where OP=-9cm  and OQ=40cm is:

1. 5

2. 6

3. 7

4. 9

Subtopic:  Simple Harmonic Motion |
 61%
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The figure shows the circular motion of a particle which is at the topmost point on the y-axis at t=0. The radius of the circle is B and the sense of revolution is clockwise.  The time period is indicated in the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is:
                   

(1) x(t) = Bsin 2πt30

(2) x(t) = Bcos πt15

(3) x(t) = Bsin πt15+π2

(4) x(t) = Bcosπt15+π2

Subtopic:  Phasor Diagram |
 52%
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1.00×10-20 kg particle is vibrating with simple harmonic motion with a period of 1.00×10-5 s and a maximum speed of 1.00×103 m/s. The maximum displacement of the particle from the mean position is:

(1) 1.59 mm

(2) 1.00 cm

(3) 10 m

(4) None of these

Subtopic:  Simple Harmonic Motion |
 73%
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Which one of the following equations does not represent SHM, x = displacement, and t = time? Parameters a, b and c are the constants of motion.

(1) x = a sin bt

(2) x = a cos bt + c

(3) x = a sin bt + c cos bt

(4) x = a sec bt + c cosec bt

Subtopic:  Types of Motion |
 74%
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The bob of a simple pendulum of length L is released at time t = 0 from a position of small angular displacement. Its linear displacement at time t is given by :

(1) X=a sin2πLg×t

(2) X=a cos2πgL×t

(3) X=a singL×t

(4) X=a cosgL×t

Subtopic:  Simple Harmonic Motion |
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A particle in SHM is described by the displacement function x(t)=A cos(ωt+ϕ), ω=2π/T. If the initial (t = 0) position of the particle is 1 cm, its initial velocity is π cm s-1 and its angular frequency is πs-1, then the amplitude of its motion is:

(1) π cm

(2) 2 cm

(3) 2 cm

(4) 1 cm

Subtopic:  Simple Harmonic Motion |
 59%
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