Harmonic Motion Multiple Questions and Answers.
Exercise Questions ::
Oscillations
1. A spring of force constant k is cut into two pieces, such that one piece is double the length of the other. Then the long piece will have a force constant of?
A. (2/3) k B. (3/2) k C. 3k D. 6k
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Answer: Option B
Explanation:
Force constant, k = F/x The length of the long piece is 2x/3 So, its force constant is k‘ =F/(2x/3)=3F/2x=3/2×k.
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2. In a simple harmonic oscillator, at the mean position?
A. Kinetic energy in minimum, potential energy is maximum B. Both kinetic and potential energies are maximum C. Kinetic energy is maximum, potential energy is minimum D. Both kinetic and potential energies are minimum
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Answer: Option C
Explanation:
At mean position, the kinetic energy is maximum and potential energy is minimum.
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3. What is the maximum acceleration of the particle executing the simple harmonic motion, y = 2sin[(πt/2)+φ], where y is in cm?
A. π/2 cm/s2 B. π2 /2 cm/s2 C. π/4 cm/s2 D. π2 /4 cm/s2
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Answer: Option B
Explanation:
y=2sinâ¡[(πt/2)+] y=Asin(ωt+φ) A=2cm,ω=(π/2 rad)/s amax =ω2 A=π2 /2 cm/s2 .
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4. When the maximum kinetic energy of a simple pendulum is K, then what its displacement (in terms of amplitude a) when its kinetic energy is K/2?
A. a/√2 B. a/2 C. a/√3 D. a/3
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Answer: Option A
Explanation:
Here Kmax =1/2×ka2 =K Let 1/2×ky2 =K/2=1/2×1/2×ka2 y=a/√2.
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5. The mass and the radius of a planet are twice that of earth. Then, period of oscillation of a second pendulum on that planet will be?
A. 1/√2s B. 2√2s C. 2s D. 1/2s
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Answer: Option B
Explanation:
On earth, g=GM/R2 On planet, g‘ =G2M/(4R)2 = g/2 T‘ /T=√(g/g‘ ) = √((g/g)/2) = √2 T‘ =√2×T=2√2 s.
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