Figure shows the variation of force acting on a particle of mass 400 g executing simple harmonic motion. The frequency of oscillation of the particle is
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a
4s−1
b
(5/2π)s−1
c
(1/8π)s−1
d
(1/2π)s−1
answer is B.
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Detailed Solution
The slope of the length is Fx=−0.55=−0.1N/cm=−10N/mBut F=−mω2x or F/x=−mω2so, −mω2=−10 or mω2=10or, ω2=10/m∴ω2=104×10−1⇒ω=102=5∴f=ω2π=52π/s