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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:
In a simple harmonic oscillator, the position of the oscillator at later times is determined by the external force acting on it. In this case, the external force is given by F = sin t N, where t is the time in seconds. We need to determine the position of the oscillator at later times, considering that it is initially at rest in its equilibrium position at t = 0.

Explanation:
To find the position of the oscillator at later times, we can use the equation of motion for a simple harmonic oscillator:

m(d^2x/dt^2) + kx = F

where m is the mass of the oscillator, x is the position of the oscillator, k is the spring constant, and F is the external force.

Step 1: Determine the spring constant:
Since the angular frequency of the oscillator is given as 2 rad s–1, we can use the formula:

k = mω^2

where ω is the angular frequency. Plugging in the values, we get:

k = m(2^2) = 4m

Step 2: Solve the differential equation:
Substituting the given external force F = sin t N into the equation of motion, we have:

m(d^2x/dt^2) + 4mx = sin t

This is a second-order linear homogeneous ordinary differential equation with constant coefficients. The general solution to this equation is of the form:

x(t) = A*cos(2t) + B*sin(2t) + xs

where A and B are constants determined by the initial conditions and xs is the particular solution to the non-homogeneous equation.

Step 3: Determine the particular solution:
To find the particular solution xs, we can assume it has the same form as the external force F, but with unknown coefficients:

xs = Asin(t) + Bcos(t)

Differentiating xs twice with respect to t, we get:

d^2xs/dt^2 = -Asin(t) - Bcos(t)

Substituting this into the differential equation, we have:

-m(Asin(t) + Bcos(t)) + 4m(Asin(t) + Bcos(t)) = sin(t)

Simplifying the equation, we get:

(3m - 4mB)sin(t) + (4mA - m)cos(t) = sin(t)

For the equation to hold true for all values of t, the coefficients of sin(t) and cos(t) on both sides must be equal. Therefore, we have:

3m - 4mB = 1
4mA - m = 0

Solving these two equations simultaneously, we find:

B = 3/4
A = 1/16

Step 4: Substitute the values into the general solution:
Substituting the values of A and B into the general solution, we get:

x(t) = (1/16)*sin(t) + (3/4)*cos(t) + A*cos(2t) + B*sin(2t)

Simplifying further, we have:

x(t) = (1/16)*sin(t) + (3/4)*cos(t) + (3/4)*sin(2t) + (1
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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 Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 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 Class 11 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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