The velocity of a particle (v) moving with simple harmonic motion, at any instant is given by (where, r = Amplitude of motion and y = Displacement of the particle from mean position.)

The velocity of a particle moving with simple harmonic motion, at any instant is given by (where $$\omega $$ = Displacement of the particle from mean position)

A

$$\omega \sqrt {{{\text{x}}^2} - {{\text{r}}^2}} $$

B

$$\omega \sqrt {{{\text{r}}^2} - {{\text{x}}^2}} $$

C

$${\omega ^2}\sqrt {{{\text{x}}^2} - {{\text{r}}^2}} $$

D

$${\omega ^2}\sqrt {{{\text{r}}^2} - {{\text{x}}^2}} $$

A particle in simple harmonic motion is described by the displacement function x(t)=Acos⁡(ωt+θ). If the initial (t=0) position of the particle is 1cm and its initial velocity isπcm/s, what is its amplitude? The angular frequency is the particle is πrad/s.

A

1 cm

B

√2 cm

C

2 cm

D

2.5 cm

Statement: In simple harmonic motion, the velocity is maximum, when the acceleration is minimum. Reason: Displacement and velocity in simple harmonic motion is differ in phase by π/2.

A

Both statement and reason are true and the reason is the correct explanation of the statement

B

Both statement and reason are true but the reason is not the correct explanation of the statement

C

Statement is true, but the reason is false

D

Statement and reason are false

The acceleration of the particle moving with simple harmonic motion is inversely proportional to the displacement of the particle from the mean position.

A

True

B

False

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

A

1/ 2π√3

B

2π√3

C

2π/√3

D

√3/2π

A particle executes simple harmonic motion with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is?

A

T/12

B

T/8

C

T/4

D

T/2

একটি কণা A বিস্তারে সরল ছন্দিত স্পন্দনে স্পন্দিত হচ্ছে। সাম্যাবস্থা থেকে কত দূরত্বে কণাটির বিভব শক্তি তার মোট শক্তির এক-চতুর্থাংশ হয়? (A particle is executing Simple Harmonic Motion with an amplitude A. At what displacement from the equilibrium position, is the potential energy of the particle one-fourth of its total energy?)

A

A/4

B

A/2

C

A/√2

D

A√2

The velocity of a particle (v) moving with simple harmonic motion, at any instant is given by

A

ω √r2 − x2

B

ω √x2 − r2

C

ω2 √r2 − x2

D

ω2√x2 − r2

A particle is executing simple harmonic motion of amplitude 10cm. Its time period of oscillation is π seconds. The velocity of the particle when it is 2 cm from extreme position is ___________

A

10 cm/s

B

12 cm/s

C

16√16 cm/s

D

16 cm/s

A particle executes simple harmonic oscillation. Its amplitude is a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is___________

A

T/8

B

T/12

C

T/2

D

T/4

The velocity of a particle (v) moving with simple harmonic motion, at any instant is given by (where, r = Amplitude of motion and y = Displacement of the particle from mean position.)

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