The 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

Question

The 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 <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-vlctktej7l1x.png" /> (A) 4s- 1 (B) (5 / 2 π ) s- 1 (C) (1 / 8 π )s- 1 (D) (1 / 2 π )s- 1

Tardigrade Admin · Accepted Answer

The slope of the curve is (F/x)=-(0.5/5)=-0.1(N/c m)=-10 N/m But F=-mω 2x or F/x=-mω 2 so, -mω 2=-10 or ω 2=(10/m) ∴ ω 2=(10/4 × 10- 1)⇒ ω =(10/2)=5 ∴ f=(ω /2 π )=(5/2 π ) s- 1

Q. The 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

$4 s^{- 1}$

$(5/2 π) s^{- 1}$

$(1/8 π) s^{- 1}$

$(1/2 π) s^{- 1}$

Q. The figure shows the variation of force acting on a particle of mass 400g executing simple harmonic motion. The frequency of oscillation of the particle is

4s−1

(5/2π)s−1

(1/8π)s−1

(1/2π)s−1

The slope of the curve is xF​=−50.5​=−0.1cmN​=−10N/m But F=−mω2x or F/x=−mω2 so, −mω2=−10 or ω2=m10​ ∴ω2=4×10−110​⇒ω=210​=5 ∴f=2πω​=2π5​s−1

Q. The 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

$4 s^{- 1}$

$(5/2 π) s^{- 1}$

$(1/8 π) s^{- 1}$

$(1/2 π) s^{- 1}$