Question

The block shown in the figure, is acted on by a spring with spring constant $k$ and a weak frictional force of constant magnitude $f$. The block is pulled a distance $x_0$ from equilibrium position and then released. It oscillates many times and ultimately comes to rest. Find the number of cycles the mass oscillates before coming to rest :

• A. $\frac{1}{4}[\frac{k{x}_0}{f}-1]$
• B. $\frac{1}{2}[\frac{k{x}_0}{f}-1]$
• C. $\frac{1}{3}[\frac{k{x}_0}{f}-1]$
• D. none of the above

Answer 1

The correct option is A $\frac{1}{4}[\frac{k{x}_0}{f}-1]$

Let us assume the that block compress spring up to $x$,

where $x_o>x$ because of loss in energy due to frictional force

Now work done against friction will be loss of spring energy.

So $f(x_o+x)=\frac{k(x_o^2-x^2)}{2}$

So $x_o-x=\frac{2f}{k}$ is loss of amplitude which is constant.

Now block will come to rest when $f=kx$

So $x_n=\frac{f}{k}$ ....(2)

From equation 1 we can write,

$x_o-x_1=\frac{2f}{k}$

$x_1-x_2=\frac{2f}{k}$

Similarly upto nth order $x_{n-1}-x_n=\frac{2f}{k}$

Adding all the equation, we get

$x_0-x_n=n\frac{2f}{k}$

$x_n=x_o-n\frac{2f}{k}$ ....(3)

On comparing 2 and 3

$n=\frac{x_o-k/f}{2f/k}$

Number of cycles

$m=n/2=\frac{1}{4}(\frac{k{x}_o}{f}-1)$