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**Name: **Oscillations Objective Question Answer Practice Test

**Subject: **Physics

**Topic: **Harmonic Motion

**Questions: **6 Objective type numerical

**Time Allowed: **12 Minutes

**Important for: **students of Class 11th, 12th, B. Sc., M. Sc., Aspirants of IIT JEE, AIPMT etc.

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- Question 1 of 6
##### 1. Question

1 pointsWhich one of the following statements is true for the speed (v) and the acceleration (α) of a particle executing simple harmonic motion?

CorrectWe have speed of the particle = v and acceleration of the particle = α

We know that velocity of a particle at some displacement (v) = ω√a2 – y2.

Therefore at mean position y = 0, and velocity is maximum.

We also know that acceleration of the particle at that point (α) = -ω^{2}y.

Therefore at mean position y = 0, and acceleration is also zero.

Thus option ‘ When v is maximum, α is zero’ is true.IncorrectWe have speed of the particle = v and acceleration of the particle = α

We know that velocity of a particle at some displacement (v) = ω√a2 – y2.

Therefore at mean position y = 0, and velocity is maximum.

We also know that acceleration of the particle at that point (α) = -ω^{2}y.

Therefore at mean position y = 0, and acceleration is also zero.

Thus option ‘ When v is maximum, α is zero’ is true. - Question 2 of 6
##### 2. Question

1 pointsIf a simple harmonic motion is represented by d

^{2}x / dt^{2}+ αx = 0, its time-period isCorrectWe have the equation of simple harmonic motion = d

^{2}x / dt^{2}+ αx = 0

We know that the standard equation of simple harmonic motion is = d^{2}x / dt^{2}+ ω^{2}x = 0

Comparing the given equation with the standard equation, we get ω^{2}= α or ω= √α

We also know that time-period of simple harmonic motion (T) = 2π / ω = 2π / √αIncorrectWe have the equation of simple harmonic motion = d

^{2}x / dt^{2}+ αx = 0

We know that the standard equation of simple harmonic motion is = d^{2}x / dt^{2}+ ω^{2}x = 0

Comparing the given equation with the standard equation, we get ω^{2}= α or ω= √α

We also know that time-period of simple harmonic motion (T) = 2π / ω = 2π / √α - Question 3 of 6
##### 3. Question

1 pointsA particle executing simple harmonic motion has an amplitude 0.01 m. If maximum acceleration of the particle is 144 π

^{2}m-s^{-2}, then angular frequency of the particle isCorrectWe have amplitude (a) = 0.01 m and maximum acceleration (α

_{max}) = 144 π^{2}m-s^{-2}.

We know that maximum acceleratioiun of the particle (α_{max}) = aω^{2}or 144 π^{2}= 0.01 ω^{2}

or ω^{2 }= 144π^{2}/ 0.01 = 14400 π^{2}or ω = 120 π rad-s^{-1}.IncorrectWe have amplitude (a) = 0.01 m and maximum acceleration (α

_{max}) = 144 π^{2}m-s^{-2}.

We know that maximum acceleratioiun of the particle (α_{max}) = aω^{2}or 144 π^{2}= 0.01 ω^{2}

or ω^{2 }= 144π^{2}/ 0.01 = 14400 π^{2}or ω = 120 π rad-s^{-1}. - Question 4 of 6
##### 4. Question

1 pointsA particle executes simple harmonic motion with an angular frequency and maximum acceleration of 3.5 rad-s

^{-1}and 7.5 m-s^{-2}respectively. Amplitude of the oscillation isCorrectWe have Angular frequency (ω) = 3.5 rad-s

^{-1}and maximum acceleration (α_{max}) = 7.5 m-s^{-2}.

We know that maximum acceleration (α_{max}) = aω^{2}or 7.5 = a × (3.5)^{2}= 12.25 a or

a = 7.5 / 12.25 = 0.61 mIncorrectWe have Angular frequency (ω) = 3.5 rad-s

^{-1}and maximum acceleration (α_{max}) = 7.5 m-s^{-2}.

We know that maximum acceleration (α_{max}) = aω^{2}or 7.5 = a × (3.5)^{2}= 12.25 a or

a = 7.5 / 12.25 = 0.61 m - Question 5 of 6
##### 5. Question

1 pointsA body is executing simple harmonic motion with an angular frequency of 2 rad-s-1. The velocity of the body at 20 mm displacement, when amplitude of motion is 60 mm, is

CorrectWe have angular frequency (ω) = 2 rad-s-1; Displacement (y) = 20 mm and amplitude (a) = 60 mm.

We know that velocity of the body at the displacement (v) = ω√a^{2}– y^{2}

= 2√(60)^{2}– (20)^{2}= 2√3200 = 113 mm-s^{-1}.IncorrectWe have angular frequency (ω) = 2 rad-s-1; Displacement (y) = 20 mm and amplitude (a) = 60 mm.

We know that velocity of the body at the displacement (v) = ω√a^{2}– y^{2}

= 2√(60)^{2}– (20)^{2}= 2√3200 = 113 mm-s^{-1}. - Question 6 of 6
##### 6. Question

1 pointsA particle is executing simple harmonic motion with an amplitude of 2 cm and angular frequency 100π rad-s

^{-1}. The maximum acceleration of the particle isCorrectWe have amplitude (a) = 2cm = 0.02 m and angular frequency (ω) = 100 π rad – s

^{-1}.

We know that maximum acceleration of the particle (α_{max}) = aω^{2}= 0.02 × (100 π)^{2}= 200 π^{2}m-s^{-2}.IncorrectWe have amplitude (a) = 2cm = 0.02 m and angular frequency (ω) = 100 π rad – s

^{-1}.

We know that maximum acceleration of the particle (α_{max}) = aω^{2}= 0.02 × (100 π)^{2}= 200 π^{2}m-s^{-2}.