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Q.

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

$4 s^{- 1}$

b

$(5 / 2 π) s^{- 1}$

c

$(1 / 8 π) s^{- 1}$

d

$(1 / 2 π) s^{- 1}$

answer is B.

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Detailed Solution

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$The slope of the length is \frac{F}{x} = - \frac{0.5}{5} = - 0.1 N / cm = - 10 N / m$

$\begin{array}{l} But F = - m ω^{2} x or F / x = - m ω^{2} \\ so, - m ω^{2} = - 10 or m ω^{2} = 10 \\ or, ω^{2} = 10 / m \\ ∴ ω^{2} = \frac{10}{4 \times 10^{- 1}} \Rightarrow ω = \frac{10}{2} = 5 \\ ∴ f = \frac{ω}{2 π} = \frac{5}{2 π} / s \end{array}$

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