Questions
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 at a distance of 5 m from the mean position. The time period of oscillation is
detailed solution
Correct option is D
In case of SHM a→ = -ω2y→Given a = 20 m/s2 at y = 5 mHence |a→| = ω2|y→|20 = ω2×5or ω2 = 4 ⇒ ω = 2 rad/secBut ω = 2πT2= 2πT or T = π secTalk to our academic expert!
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The identical springs of constant ‘K’ are connected in series and parallel as shown in figure A mass M is suspended from them. The ratio of their frequencies of vertical oscillations will be
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