Question

A particle having mass \(1 \mathrm{~kg}\) is executing S.H.M. with an amplitude of \(0.01 \mathrm{~m}\) and a frequency of \(60 \mathrm{hz}\). The maximum force acting on this particle is \(\ldots \ldots . . \mathrm{N}\) (A) \(144 \pi^{2}\) (B) \(288 \pi^{2}\) (C) \(188 \pi^{2}\) (D) None of these. (A) \(x=a \sin 2 p \sqrt{(\ell / g) t}\) (B) \(x=a \cos 2 p \sqrt{(g / \ell) t}\) (C) \(\mathrm{x}=\mathrm{a} \sin \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\) (D) \(\mathrm{x}=\mathrm{a} \cos \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\)

Short answer

The maximum force acting on the particle is \(14400\pi^2\) N, so the correct answer is (D) None of these.

Step-by-step solution

Calculate the angular frequency (ω)

To find the angular frequency, we'll use the formula: ω = 2πf Here, f is the frequency of the motion, which is given as 60 Hz. Plugging in the values, we have: ω = 2π(60) ω = 120π

Calculate the maximum acceleration (amax)

The maximum acceleration of the particle in S.H.M. can be found using the formula: amax = ω²A Here, A is the amplitude of motion, given as 0.01 m. Plugging in the values, we have: amax = (120π)²(0.01) amax = 14400π²

Calculate the maximum force (Fmax)

Using Newton's second law of motion (F = ma), we can find the maximum force acting on the particle: Fmax = m*amax Here, m is the mass of the particle, given as 1 kg. Plugging in the values, we have: Fmax = (1)(14400π²) Fmax = 14400π² N From the given options, the closest value is (A) 144π² N. However, the correct answer is actually 14400π² N, none of the options provided are exactly accurate. Thus, the answer is (D) None of these.