find the distance as a function of time if initially an end of a bunch of spring is join with a block of mass m. total mass of spring is 2m. initial velocity of the block is v.system is on a horizontal plane.

To find the distance as a function of time for a system where one end of a spring is attached to a block of mass \( m \) and the total mass of the spring is \( 2m \), we need to consider the dynamics of the system. This involves understanding how the spring behaves when it is stretched or compressed and how that affects the motion of the block.

Understanding the System Dynamics

In this scenario, we have a block of mass \( m \) that is initially moving with a velocity \( v \) on a horizontal plane. The spring, which has a total mass of \( 2m \), is attached to the block. When the spring is at its natural length, the block moves freely. However, as the block moves, it will stretch the spring, creating a restoring force that acts on the block.

Applying Newton's Second Law

To analyze the motion, we can apply Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:

• Net Force (F): The force exerted by the spring when it is stretched can be described by Hooke's law, which states that the force exerted by a spring is proportional to its displacement from the equilibrium position:
• F = -kx: Here, \( k \) is the spring constant, and \( x \) is the displacement of the spring from its equilibrium position.

For the block of mass \( m \), the equation of motion can be written as:

m \frac{d^2x}{dt^2} = -kx

Setting Up the Differential Equation

Rearranging the equation gives us:

\(\frac{d^2x}{dt^2} + \frac{k}{m}x = 0\)

This is a second-order linear differential equation that describes simple harmonic motion. The general solution to this type of equation can be expressed as:

x(t) = A \cos(\omega t + \phi)

where:

• A: Amplitude of the motion, which depends on the initial conditions.
• \(\omega = \sqrt{\frac{k}{m}}\): Angular frequency of the oscillation.
• \(\phi:\ Phase constant, determined by initial conditions.

Determining Initial Conditions

To find the specific solution for our problem, we need to apply the initial conditions:

• At \( t = 0 \), the initial position \( x(0) = 0 \) (assuming the spring is at its natural length initially).
• At \( t = 0 \), the initial velocity \( \frac{dx}{dt}(0) = v \).

Using these conditions, we can find \( A \) and \( \phi \). From the initial position, we have:

0 = A \cos(\phi)

This implies that \( \cos(\phi) = 0 \), leading to \( \phi = \frac{\pi}{2} \) (or \( \frac{3\pi}{2} \)). Thus, we can express the position function as:

x(t) = A \sin(\omega t)

Next, we differentiate to find the velocity:

\(\frac{dx}{dt} = A \omega \cos(\omega t)\)

Applying the initial velocity condition:

v = A \omega \cos(0) = A \omega

From this, we can solve for \( A \):

A = \frac{v}{\omega} = \frac{v}{\sqrt{\frac{k}{m}}} = v \sqrt{\frac{m}{k}}

Final Expression for Distance

Substituting \( A \) back into the position function gives us:

x(t) = \frac{v}{\sqrt{\frac{k}{m}}} \sin\left(\sqrt{\frac{k}{m}} t\right)

This equation describes the distance of the block from the equilibrium position of the spring as a function of time. The motion is periodic, characteristic of simple harmonic motion, with the block oscillating back and forth as the spring stretches and compresses.

In summary, the distance as a function of time for the block attached to the spring can be expressed as:

x(t) = \frac{v}{\sqrt{\frac{k}{m}}} \sin\left(\sqrt{\frac{k}{m}} t\right)

This formula captures the essence of the system's dynamics, illustrating how the initial velocity and the properties of the spring influence the motion of the block over time.