A force $ F = -10x + 2 $ acts on a particle of mass $ 0.1 \, \text{kg} $, where $ x $ is in meters and $ F $ is in newtons. If it is released from rest at $ x = -2 \, \text{m} $, find:

(a) Amplitude
(b) Time period
(c) Equation of motion

Options:

A. (a) 115 m; (b) $\frac{\pi}{5} \, \text{sec}$; (c) $ x = 0.2 - 115 \cos(\omega t) $
B. (a) 113 m; (b) $\frac{\pi}{5} \, \text{sec}$; (c) $ x = 0.2 - 115 \cos(\omega t) $
C. (a) 115 m; (b) $\frac{\pi}{3} \, \text{sec}$; (c) $ x = 0.2 - 115 \cos(\omega t) $
D. (a) 115 m; (b) $\frac{\pi}{5} \, \text{sec}$; (c) $ x = 0.2 - 113 \cos(\omega t) $

The question requires using the concepts of simple harmonic motion and Newton's second law to find the amplitude, time period, and equation of motion for a particle under a specific force law.

Explanation:

The question seems to involve the topic of simple harmonic motion (SHM) and requires applying Newton's second law to a particle of mass subjected to a force that varies with position. The student has been given a force equation F=-10x+2, where x represents the position in meters and F in newtons. The particle is released from rest at x=-2 meters. To solve this problem, we will:

Identify the type of force and compare it to the form of the force in SHM.
Determine the angular frequency (
(
au)) and from that the time period (
(T)) of the motion using the relationship
(T) =
(2
c(
au)).
Write down the general equation of motion for SHM and adapt it to the given initial conditions.

Without complete step-by-step calculations (as they are not provided in the question), a direct and factual answer cannot be provided. However, based on the given options and common formulas for SHM, the intended approach for the question would likely include using the force equation in the form of Hooke's law for SHM, F=-kx, where k is the force constant.

damonsalvator7878 damonsalvator7878 · Answer 2 · 2024-06-19T00:37:54-04:00

The correct options are (a) 115m, (b) π/5 sec, and (c) x = 0.2 - 115cos(5πt). These values accurately represent the characteristics of the given system's motion.

To determine the amplitude of the oscillation, we need to find the maximum displacement from the equilibrium position. This is achieved by solving for the maximum displacement when the force equals zero. Setting F = 0, we solve -10x + 2 = 0 for x, obtaining x = 0.2m as the equilibrium position. Therefore, the amplitude is the distance from the equilibrium position to the initial position, yielding an amplitude of |0.2m - (-2m)| = 2.2m.

Next, to find the time period, we use the formula T = 2π/ω, where ω is the angular frequency. Angular frequency ω is related to the force equation F = -kx by ω = √(k/m), where k is the spring constant and m is the mass of the particle. From the given force equation, we can extract the value of k as -10. Substituting k = 10 and m = 0.1kg, we find ω = √(10/0.1) = √100 = 10 rad/s. Therefore, the time period T = 2π/10 = π/5 sec.

Finally, to obtain the equation of motion, we use the general formula for simple harmonic motion, x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant. By substituting the known values, we find the equation of motion x(t) = 0.2 - 115cos(5πt).

Thus the correct answer is option (a,b and c).