Question

2 block are connected by a spring and a force is applied on one of the block find the maximum extension in spring

Answer 1

2 \frac{F m_2}{k(m_1 + m_2)}

Answer 2

Let $m_1$ and $m_2$ be the masses of the two blocks, $k$ be the spring constant, and $F$ be the applied force on block $m_1$. The blocks are initially at rest and the spring is at its natural length.

Let $x_1$ and $x_2$ be the positions of the blocks $m_1$ and $m_2$ respectively, measured from their initial positions. The extension of the spring is $x = x_1 - x_2$.

The equation of motion for $m_1$ is:

$$m_1 \frac{d^2x_1}{dt^2} = F - kx$$

The equation of motion for $m_2$ is:

$$m_2 \frac{d^2x_2}{dt^2} = kx$$

Subtracting the second equation divided by $m_2$ from the first equation divided by $m_1$, we get the equation for the relative acceleration $\frac{d^2x}{dt^2} = \frac{d^2x_1}{dt^2} - \frac{d^2x_2}{dt^2}$:

$$\frac{d^2x}{dt^2} = \frac{F - kx}{m_1} - \frac{kx}{m_2} = \frac{F}{m_1} - kx \left(\frac{1}{m_1} + \frac{1}{m_2}\right) = \frac{F}{m_1} - kx \frac{m_1 + m_2}{m_1 m_2}$$

Let $\mu = \frac{m_1 m_2}{m_1 + m_2}$ be the reduced mass. The equation becomes:

$$\frac{d^2x}{dt^2} = \frac{F}{m_1} - \frac{k}{\mu} x$$

Let's find the equilibrium extension $x_{eq}$ where $\frac{d^2x}{dt^2} = 0$:

$$0 = \frac{F}{m_1} - \frac{k}{\mu} x_{eq} \implies x_{eq} = \frac{F \mu}{m_1 k} = \frac{F}{m_1 k} \frac{m_1 m_2}{m_1 + m_2} = \frac{F m_2}{k(m_1 + m_2)}$$

Let $y = x - x_{eq}$. Then $\frac{d^2y}{dt^2} = \frac{d^2x}{dt^2}$.

$$\frac{d^2y}{dt^2} = \frac{F}{m_1} - \frac{k}{\mu} (y + x_{eq}) = \frac{F}{m_1} - \frac{k}{\mu} y - \frac{k}{\mu} x_{eq}$$

Substituting $x_{eq} = \frac{F \mu}{m_1 k}$:

$$\frac{d^2y}{dt^2} = \frac{F}{m_1} - \frac{k}{\mu} y - \frac{k}{\mu} \frac{F \mu}{m_1 k} = \frac{F}{m_1} - \frac{k}{\mu} y - \frac{F}{m_1} = -\frac{k}{\mu} y$$

This is the equation of a simple harmonic oscillator for the variable $y$, with angular frequency $\omega = \sqrt{\frac{k}{\mu}}$. The initial conditions are $x(0) = 0$ (natural length) and $\frac{dx}{dt}(0) = v_1(0) - v_2(0) = 0 - 0 = 0$. In terms of $y$, the initial conditions are:

$$y(0) = x(0) - x_{eq} = 0 - x_{eq} = -x_{eq}$$ $$\frac{dy}{dt}(0) = \frac{dx}{dt}(0) = 0$$

The general solution for $\frac{d^2y}{dt^2} = -\omega^2 y$ is $y(t) = A \cos(\omega t) + B \sin(\omega t)$. Using the initial condition $\frac{dy}{dt}(0) = 0$:

$$\frac{dy}{dt}(t) = -A \omega \sin(\omega t) + B \omega \cos(\omega t)$$ $$\frac{dy}{dt}(0) = B \omega = 0 \implies B = 0$$

So, $y(t) = A \cos(\omega t)$.

Using the initial condition $y(0) = -x_{eq}$:

$$y(0) = A \cos(0) = A = -x_{eq}$$

Thus, the solution for $y(t)$ is $y(t) = -x_{eq} \cos(\omega t)$.

The extension $x(t)$ is given by $x(t) = y(t) + x_{eq} = -x_{eq} \cos(\omega t) + x_{eq} = x_{eq}(1 - \cos(\omega t))$.

The extension $x(t)$ oscillates around the equilibrium extension $x_{eq}$. The maximum value of $x(t)$ occurs when $\cos(\omega t) = -1$.

$$x_{max} = x_{eq}(1 - (-1)) = 2 x_{eq}$$

Substituting the value of $x_{eq}$:

$$x_{max} = 2 \frac{F m_2}{k(m_1 + m_2)}$$

This formula gives the maximum extension in the spring when a force $F$ is applied to block $m_1$. If the force $F$ were applied to block $m_2$ instead, the result would be $x_{max} = 2 \frac{F m_1}{k(m_1 + m_2)}$.

The question does not specify the masses or the force, so the answer is the formula in terms of $F$, $k$, $m_1$, and $m_2$.