Questions
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
detailed solution
Correct option is B
The slope of the length=Fx=−0.55=−0.1Ncm−1=−10Nm−1 But F=−mω2x or ,Fx=−mω2 So, −mω2=−10 or, mω2=10∘∴ω2=104×10−1⇒ω=102=5∴f=ω2π=52π/sTalk to our academic expert!
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The acceleration a of a particle undergoing S.H.M. is shown in the figure. Which of the labelled points corresponds to the particle being at – xmax
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