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The velocity of a particle performing simple harmonic motion, when it passes through its mean position is
Options
(a) Infinity
(b) Zero
(c) Minimum
(d) Maximum
Correct Answer:
Maximum
Explanation:
ν = ω √(a² – y²), At mean position y = 0
ν = ωa = This is the maximum velocity.
Related Questions: - The particle executing simple harmonic motion has a kinetic energy K₀cos²ωt.
- A carnot’s engine operates with source at 127° C and sink at 27°C
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The particle executing simple harmonic motion has a kinetic energy K₀cos²ωt.
- A carnot’s engine operates with source at 127° C and sink at 27°C
- The power obtained in a reactor using U²³⁵ disintegration is 1000 kW
- What is the voltage gain in a common-emiiter amplifier, when input resistance is 4 Ω
- The radius of curvature of a convave mirror is 24 cm and the imager is magnified
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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At mean position x=0
Therefore v=w(√a^2-0^2)
Therefore v=w(√a^2)=aw
Vmax=aw