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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.UnattemptedWe 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 timeperiod is
CorrectWe 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 timeperiod 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 timeperiod of simple harmonic motion (T) = 2π / ω = 2π / √αUnattemptedWe 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 timeperiod 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} ms^{2}, then angular frequency of the particle is
CorrectWe have amplitude (a) = 0.01 m and maximum acceleration (α_{max}) = 144 π^{2} ms^{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 rads^{1}.IncorrectWe have amplitude (a) = 0.01 m and maximum acceleration (α_{max}) = 144 π^{2} ms^{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 rads^{1}.UnattemptedWe have amplitude (a) = 0.01 m and maximum acceleration (α_{max}) = 144 π^{2} ms^{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 rads^{1}.  Question 4 of 6
4. Question
1 pointsA particle executes simple harmonic motion with an angular frequency and maximum acceleration of 3.5 rads^{1} and 7.5 ms^{2} respectively. Amplitude of the oscillation is
CorrectWe have Angular frequency (ω) = 3.5 rads^{1} and maximum acceleration (α_{max}) = 7.5 ms^{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 rads^{1} and maximum acceleration (α_{max}) = 7.5 ms^{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 mUnattemptedWe have Angular frequency (ω) = 3.5 rads^{1} and maximum acceleration (α_{max}) = 7.5 ms^{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 rads1. The velocity of the body at 20 mm displacement, when amplitude of motion is 60 mm, is
CorrectWe have angular frequency (ω) = 2 rads1; 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 mms^{1}.IncorrectWe have angular frequency (ω) = 2 rads1; 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 mms^{1}.UnattemptedWe have angular frequency (ω) = 2 rads1; 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 mms^{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 rads^{1}. The maximum acceleration of the particle is
CorrectWe 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} ms^{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} ms^{2}.UnattemptedWe 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} ms^{2}.