Question

2 blocks of mass m and M are connected by a string of constant k. the blocks are kept on a constant horizontal plane and are at rest. the spring is unstretched when a constant force F starts acting on the block of mass M ti pull it. Find the maximum extension of the spring

Answer 1

Answer on Question#49848 - Physics - Mechanics - Kinematics - Dynamics

2 blocks of mass $m$ and $M$ are connected by a string of constant $k$ . The blocks are kept on a constant horizontal plane and are at rest. The spring is unstretched when a constant force $F$ starts acting on the block of mass $M$ to pull it. Find the maximum extension of the spring.

Solution:

Let $x_{m}$ be the displacement of mass $m$ and $x_{M}$ be the displacement of mass $M$ , then the extension of the spring is $x_{M} - x_{m}$ . According to the 2 Newton's law the equation of motion of the mass $M$ can be written as follows

$$M \ddot {x} _ {M} = F - k \left(x _ {M} - x _ {m}\right)$$

And for the mass $m$ :

$$m \ddot {x} _ {m} = k \left(x _ {M} - x _ {m}\right)$$

Dividing equation (1) by $M$ and equation (2) by $m$ we obtain

$$\left\{ \begin{array}{c} \ddot {x} _ {M} = \frac {F}{M} - \frac {k}{M} (x _ {M} - x _ {m}) \\ \ddot {x} _ {m} = \frac {k}{m} (x _ {M} - x _ {m}) \end{array} \right.$$

Subtracting the second equation of this system from the first we obtain

$$\frac {d ^ {2} (x _ {M} - x _ {m})}{d t ^ {2}} = \frac {F}{M} - k \left(\frac {1}{M} + \frac {1}{m}\right) (x _ {M} - x _ {m}) = \frac {F}{M} - \frac {k}{\frac {m M}{m + M}} (x _ {M} - x _ {m})$$

Denoting $(x_{M} - x_{m})$ by $\Delta x$ and $\frac{mM}{m + M}$ by $\mu$ we obtain

$$\Delta \ddot {x} = \frac {F}{M} - \frac {k}{\mu} \Delta x$$

This equation has the following solution

$$\Delta x = A \sin \frac {k}{\mu} t + B \cos \frac {k}{\mu} t + \frac {F \mu}{M k}$$

where $A$ and $B$ are some constants. We can determine them from the initial conditions, which are

$$\Delta x (0) = 0, \quad \Delta x (0) = 0$$

Using the first condition we obtain

$$0 = B + \frac {F \mu}{M k}$$

It gives us the value of $B$:

$$B = - \frac {F \mu}{M k}$$

Taking the derivative of (3) we obtain

$$\dot {x} = A \frac {k}{\mu} \cos \frac {k}{\mu} t - B \frac {k}{\mu} \sin \frac {k}{\mu} t$$

Using second condition we obtain

$$\dot {x} (0) = A \frac {k}{M} = 0$$

It gives us the value of $A$:

$$A = 0$$

Therefore, the extension of the spring at any time is given by

$$\Delta x = \frac {F \mu}{M k} \left(1 - \cos \frac {k}{\mu} t\right)$$

It is easy to see that extension reaches its maximum when cosine equals -1. Therefore, the maximum extension of the spring has the following value

$$\Delta x _ {m a x} = 2 \frac {F \mu}{M k}$$

Answer: $2\frac{F\mu}{Mk}$

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