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A simple harmonic oscillator of angular frequency 2 rad s–1 is acted upon by an external force F = sin t N. If theoscillator is at rest in its equilibrium position at t = 0, its position at later times is proportional to :
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Introduction:
A simple harmonic oscillator is a system that undergoes repeated oscillations about an equilibrium position under the influence of a restoring force that is proportional to the displacement from the equilibrium. In this case, the oscillator has an angular frequency of 2 rad/s and is acted upon by an external force given by F = sin(t) N.

Explanation:
To understand the position of the oscillator at later times, we need to analyze the effect of the external force on the system. Let's break down the problem into smaller steps:

Step 1: Equilibrium Position:
At t = 0, the oscillator is at rest in its equilibrium position. This means that the net force acting on the oscillator is zero, as the restoring force exactly balances the external force. Therefore, the displacement of the oscillator from its equilibrium position is zero at t = 0.

Step 2: Equation of Motion:
The equation of motion for a simple harmonic oscillator is given by:
m * d²x/dt² + k * x = 0
where m is the mass of the oscillator, dx/dt is the velocity, k is the force constant, and x is the displacement from the equilibrium position.

Step 3: Solving the Differential Equation:
Substituting the external force F = sin(t) into the equation of motion, we get:
m * d²x/dt² + k * x = sin(t)
This is a non-homogeneous differential equation with a sinusoidal forcing term.

Step 4: Particular Solution:
To find the particular solution of the differential equation, we assume a solution of the form:
x(t) = A * sin(t + φ)
where A is the amplitude and φ is the phase angle.

Step 5: Applying the Solution:
Differentiating x(t) with respect to time, we get:
dx/dt = A * cos(t + φ)

Differentiating again, we get:
d²x/dt² = -A * sin(t + φ)

Substituting these values back into the differential equation, we get:
-m * A * sin(t + φ) + k * A * sin(t + φ) = sin(t)

Simplifying the equation, we get:
(A * k - m * A) * sin(t + φ) = sin(t)

For this equation to be true for all values of t, we must have:
A * k - m * A = 1

Conclusion:
In conclusion, the position of the oscillator at later times is given by x(t) = A * sin(t + φ), where A is the amplitude determined by the equation A * k - m * A = 1, and φ is the phase angle. The position of the oscillator is directly proportional to the amplitude A, which depends on the external force and the properties of the oscillator. As time progresses, the oscillator undergoes sinusoidal oscillations about its equilibrium position with the given angular frequency.
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A simple harmonic oscillator of angular frequency 2 rad s–1 is acted upon by an external force F = sin t N. If theoscillator is at rest in its equilibrium position at t = 0, its position at later times is proportional to :
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A simple harmonic oscillator of angular frequency 2 rad s–1 is acted upon by an external force F = sin t N. If theoscillator is at rest in its equilibrium position at t = 0, its position at later times is proportional to : for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A simple harmonic oscillator of angular frequency 2 rad s–1 is acted upon by an external force F = sin t N. If theoscillator is at rest in its equilibrium position at t = 0, its position at later times is proportional to : covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A simple harmonic oscillator of angular frequency 2 rad s–1 is acted upon by an external force F = sin t N. If theoscillator is at rest in its equilibrium position at t = 0, its position at later times is proportional to :.
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