A particle of mass (m) is attached to a spring (of spring constant k) and has a natural angular frequency `omega_(0)`. An external force `F(t)` proportional to `cos omegat` `(omega!=omega_(0))` is applied to the oscillator. The time displacement of the oscillator will be proportional to.

Options

A. `m/(omega_(0)^(2)-omega^(2))`
B. `1/(m(omega_(0)^(2)-omega^(2))`
C. `1/(m(omega_(0)^(2)+omega^(2)))`
D. `m/(omega_(0)^(2)+omega^(2))`

Explanation

To solve the problem step by step, we will analyze the system of a mass attached to a spring under the influence of an external force.

### Step 1: Understand the System
We have a mass \( m \) attached to a spring with a spring constant \( k \). The natural angular frequency of the system is given by:
\[ \omega_0 = \sqrt{\frac{k}{m}} \]
An external force \( F(t) \) is applied, which is proportional to \( \cos(\omega t) \), where \( \omega \neq \omega_0 \).

### Step 2: Write the Net Force Equation
The net force acting on the mass when displaced by \( x \) from the mean position can be expressed as:
\[ F_{\text{net}} = F_{\text{spring}} - F_{\text{external}} \]
The force exerted by the spring is given by Hooke's law:
\[ F_{\text{spring}} = -kx \]
The external force can be represented as:
\[ F_{\text{external}} = F_0 \cos(\omega t) \]
where \( F_0 \) is a constant of proportionality.

Thus, the net force can be written as:
\[ F_{\text{net}} = -kx - F_0 \cos(\omega t) \]

### Step 3: Set Up the Equation of Motion
Using Newton's second law, we can write:
\[ m \frac{d^2x}{dt^2} = -kx - F_0 \cos(\omega t) \]
Rearranging gives:
\[ m \frac{d^2x}{dt^2} + kx = -F_0 \cos(\omega t) \]

### Step 4: Substitute Natural Frequency
Substituting \( k = m\omega_0^2 \) into the equation, we have:
\[ \frac{d^2x}{dt^2} + \omega_0^2 x = -\frac{F_0}{m} \cos(\omega t) \]

### Step 5: Identify the Type of Motion
This is a second-order linear differential equation with a non-homogeneous term. The solution to this equation will consist of a complementary (homogeneous) solution and a particular solution.

### Step 6: Find the Particular Solution
Since the external force is of the form \( \cos(\omega t) \), we can assume a particular solution of the form:
\[ x_p = A \cos(\omega t) + B \sin(\omega t) \]
Substituting \( x_p \) into the differential equation will allow us to solve for the coefficients \( A \) and \( B \).

### Step 7: Solve for Coefficients
After substituting and simplifying, we find that the amplitude of the oscillation due to the external force is proportional to:
\[ x_p \propto \frac{1}{\omega_0^2 - \omega^2} \]
This indicates that the displacement \( x \) is inversely proportional to \( \omega_0^2 - \omega^2 \).

### Final Answer
Thus, the time displacement of the oscillator will be proportional to:
\[ x \propto \frac{1}{\omega_0^2 - \omega^2} \]