Question

For shown situation find the maximum elongation in the spring. Neglect friction everywhere. Initially, the blocks are at rest and spring is unstretched.

• A. $\frac{4F}{3K}$
• B. $\frac{3F}{4K}$
• C. $\frac{4F}{K}$
• D. $\frac{2F}{K}$

Answer 1

Answer: (A)

$\frac{4F}{3K}$

Answer 2

The problem asks for the maximum elongation in the spring for the given system.

The system consists of two blocks with masses $m_1 = 3m$ and $m_2 = 6m$, connected by a spring with spring constant $K$. An external force $F_1 = F/2$ is applied to the left on block $m_1$, and an external force $F_2 = F$ is applied to the right on block $m_2$. Initially, the blocks are at rest and the spring is unstretched. Friction is negligible.

We need to analyze the relative motion of the two blocks. Let $x$ be the elongation of the spring, so $x = x_2 - x_1$, where $x_1$ and $x_2$ are the displacements of block $m_1$ and $m_2$ respectively from their initial positions.

Let's write down the equations of motion for each block:

For block $m_1$ (mass $3m$): The forces acting on $m_1$ are the applied force $F/2$ (to the left) and the spring force $Kx$ (to the right, assuming $x$ is positive for elongation).

$3m a_1 = Kx - F/2$ (Equation 1)

For block $m_2$ (mass $6m$): The forces acting on $m_2$ are the applied force $F$ (to the right) and the spring force $Kx$ (to the left).

$6m a_2 = F - Kx$ (Equation 2)

The relative acceleration $a_{rel} = a_2 - a_1$. Substitute $a_1 = \frac{Kx - F/2}{3m}$ and $a_2 = \frac{F - Kx}{6m}$ into the expression for $a_{rel}$:

$a_{rel} = \frac{F - Kx}{6m} - \frac{Kx - F/2}{3m}$

To combine the terms, find a common denominator (which is $6m$):

$a_{rel} = \frac{(F - Kx) - 2(Kx - F/2)}{6m}$

$a_{rel} = \frac{F - Kx - 2Kx + F}{6m}$

$a_{rel} = \frac{2F - 3Kx}{6m}$

We know that $a_{rel} = \frac{d^2x}{dt^2}$. So, the equation of motion for the relative displacement $x$ is:

$\frac{d^2x}{dt^2} = \frac{2F - 3Kx}{6m}$

This can be rewritten as:

$6m \frac{d^2x}{dt^2} = 2F - 3Kx$

This equation describes the motion of the relative coordinate $x$. It is a form of simple harmonic motion (SHM). To make it more explicit, divide by 3:

$2m \frac{d^2x}{dt^2} = \frac{2F}{3} - Kx$

Let $\mu = 2m$ be the reduced mass for this effective single-body oscillation.

$\mu \frac{d^2x}{dt^2} = \frac{2F}{3} - Kx$

This equation can be written as $\mu \frac{d^2x}{dt^2} = -K(x - x_{eq})$, where $x_{eq}$ is the equilibrium position for the oscillation. Comparing this with our equation:

$-K(x - x_{eq}) = \frac{2F}{3} - Kx$

$-Kx + Kx_{eq} = \frac{2F}{3} - Kx$

$Kx_{eq} = \frac{2F}{3}$

$x_{eq} = \frac{2F}{3K}$

The system starts from rest with the spring unstretched. This means at $t=0$, $x=0$ and the relative velocity $\frac{dx}{dt} = 0$.

For a simple harmonic motion, if the initial velocity is zero, the starting position is an extreme position (amplitude) relative to the equilibrium position.

The initial displacement from the equilibrium position is $x(0) - x_{eq} = 0 - \frac{2F}{3K} = -\frac{2F}{3K}$.

The amplitude of oscillation $A$ is the magnitude of this initial displacement from equilibrium:

$A = \left|-\frac{2F}{3K}\right| = \frac{2F}{3K}$.

The maximum elongation in the spring occurs at the other extreme of the oscillation, which is $x_{eq} + A$.

$x_{max} = x_{eq} + A = \frac{2F}{3K} + \frac{2F}{3K}$

$x_{max} = \frac{4F}{3K}$