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.

Key concepts

Angular Frequency

In simple harmonic motion (SHM), angular frequency (\(\omega\)) plays a crucial role in describing how quickly the oscillations are happening. It is directly tied to the frequency of the oscillation, which is how many cycles or vibrations occur per unit time. The relationship between the ordinary frequency (\(f\)) and the angular frequency is given by:\[ \omega = 2\pi f \]

• Here, \(\pi\) is a constant approximately equal to 3.14159.
• The factor of \(2\pi\) arises because a full cycle of a circular motion (or one complete oscillation) corresponds to a traversal of \(2\pi\) radians.

For the given exercise, since the frequency \(f\) is 60 Hz, we calculate angular frequency as \(\omega = 120\pi\) rad/s. This means the particle completes 120\pi radians per second, which tells us how fast the particle repeats its oscillatory path.

Maximum Acceleration

In SHM, the maximum acceleration (\(a_{max}\)) occurs when the particle reaches its maximum displacement (amplitude). This is because acceleration is proportional to displacement in SHM but in the opposite direction, according to Hooke's law.The formula to find the maximum acceleration is given by:\[ a_{max} = \omega^2 A \]

• Here, \(A\) is the amplitude of the motion.
• \(\omega^2\) accentuates the quadratic influence of the angular frequency on acceleration.

Given that \(\omega = 120\pi\) rad/s and \(A = 0.01\) m, the maximum acceleration calculated was: \[a_{max} = (120\pi)^2 \times 0.01 = 14400\pi^2 \text{ m/s}^2\]This high acceleration reflects the vigorous nature of rapid oscillations.

Newton's Second Law

Newton's second law of motion is a fundamental principle that connects the concepts of force, mass, and acceleration through the equation:\[ F = ma \]

• Here, \(F\) represents force, \(m\) is the mass of the object, and \(a\) is its acceleration.
• It states that the force exerted by or on an object is directly proportional to the acceleration inversely proportional to its mass.

In the context of this exercise, the law helps us understand how the mass affects the force experienced by the particle during SHM. Given a mass of 1 kg and the calculated maximum acceleration, the second law allows us to find the force that acts on the particle at its extreme positions.

Force Calculation

Calculating the maximum force (\(F_{max}\)) in SHM involves using the determined values of maximum acceleration and mass within the framework provided by Newton's second law: \[ F_{max} = m \times a_{max}\]Given that the mass (\(m\)) of the particle is 1 kg and the maximum acceleration (\(a_{max}\)) is \(14400\pi^2\) m/s², the force acting on the particle reaches: \[ F_{max} = 1 \times 14400\pi^2 = 14400\pi^2 \text{ N}\]

• This result underscores how force is a product of the combined effects of mass and acceleration.
• The particle experiences this maximal force at the turnaround points of its motion where displacement and acceleration are greatest.

Therefore, while the options presented in the original problem list a force of \(144\pi^2\) N, the correctly calculated maximum force based on provided parameters is actually \(14400\pi^2\) N.