Chapter 12, Problem 60P

(a)

To determine

Prove that the maximum stretching of spring from mean position is $x=\frac{{\mu }_{s}mg}{k}$.

Answer to Problem 60P

It is proved that the maximum stretching of spring from mean position is $x=\frac{{\mu }_{s}mg}{k}$.

Explanation of Solution

Write the relation between force on spring and maximum static frictional force.

    $kx={\mu }_{s}n$

Here, $k$ is the spring force, $x$ is the displacement of spring, ${\mu }_s$ is the coefficient of static friction, and $n$ is the normal reaction force.

Write the expression for $n$.

  $n=mg$

Here, $m$ is the mass of block and $g$ is the gravitational acceleration.

Rewrite the first equation by substituting the above one.

    $kx={\mu }_{s}mg$

Conclusion:

Rewrite the expression for $x$.

    $x=\frac{{\mu }_{s}mg}{k}$

Therefore, it is proved that the maximum stretching of spring from mean position is $x=\frac{{\mu }_{s}mg}{k}$.

$x=\frac{{\mu }_{s}mg}{k}$.

(b)

To determine

Prove that block oscillates about mean point when spring is stretched by $x=\frac{{\mu }_{s}mg}{k}$.

Answer to Problem 60P

It is proved that block oscillates about mean point when spring is stretched by $x=\frac{{\mu }_{s}mg}{k}$.

Explanation of Solution

The block will be almost at rest at the break point because of the smaller value of velocity. The block starts to oscillate s soon when the force on block becomes $-kx$ and $+{\mu }_{k}mg$. Write the expression for net force on block while sliding.

    $F_{net}=-kx+{\mu }_{k}mg$

Here, $F_{net}$ is the net force on block,.

Rewrite the above equation by multiplying and dividing the second term with $k$.

    $\begin{array}{c}{F}_{net}=-kx+\frac{k{\mu }_{k}mg}{k}\\ =-k\left(x-\frac{{\mu }_{k}mg}{k}\right)\end{array}$

Express the term $\frac{{\mu }_{k}mg}{k}$ into a single variable.

    $x-\frac{{\mu }_{k}mg}{k}=x_{rel}$

Here, $x_{rel}$ is the displacement of block from the point $\frac{{\mu }_{k}mg}{k}$.

Rewrite the expression for $F_{net}$ by substituting the above relation.

    $F_{net}=-k\left(x_{rel}\right)$

Conclusion:

From the above calculation, it is evident that block starts to be in simple harmonic motion about mean position on displacing the spring

Therefore, it is proved that block oscillates about mean point when spring is stretched by $x=\frac{{\mu }_{s}mg}{k}$.

(c)

To determine

Plot the variation of position of block with time.

Answer to Problem 60P

The plot is

Explanation of Solution

The graph is plotted with time on x-axis and position on y-axis.

The dotted line along x-axis denotes the function $x_{rel}=\frac{{\mu }_{k}mg}{k}$. The peak on curve denotes $x=\frac{{\mu }_{s}mg}{k}$

Conclusion:

Therefore, the graph is plotted

(d)

To determine

Prove that the amplitude of oscillation is $A=\frac{\left({\mu }_s-{\mu }_k\right)mg}{k}$.

Answer to Problem 60P

It is proved that the amplitude of oscillation is $A=\frac{\left({\mu }_s-{\mu }_k\right)mg}{k}$.

Explanation of Solution

Write the expression for $x_{rel}$ from part (b).

    $x_{rel}=x-\frac{{\mu }_{k}mg}{k}$

Rewrite the above expression by substituting $\frac{{\mu }_{s}mg}{k}$ for $x$.

    $\begin{array}{c}{x}_{rel}=\frac{{\mu }_{s}mg}{k}-\frac{{\mu }_{k}mg}{k}\\ =\left({\mu }_s-{\mu }_k\right)\frac{mg}{k}\end{array}$

Amplitude is the maximum displacement from the mean position only. Rewrite the expression as amplitude.

    $A=\left({\mu }_s-{\mu }_k\right)\frac{mg}{k}$

Here, $A$ is the amplitude.

On pulling the block to distance $x=\frac{{\mu }_{s}mg}{k}$, it will start to oscillate about $x=\frac{{\mu }_{k}mg}{k}$.

Conclusion:

Therefore, it is proved that the amplitude of oscillation is $A=\frac{\left({\mu }_s-{\mu }_k\right)mg}{k}$.

(e)

To determine

Prove that the period of oscillation is $T=\frac{2\left({\mu }_s-{\mu }_k\right)mg}{vk}+\pi \sqrt{\frac{m}{k}}$.

Answer to Problem 60P

It is proved that the period of oscillation is $T=\frac{2\left({\mu }_s-{\mu }_k\right)mg}{vk}+\pi \sqrt{\frac{m}{k}}$.

Explanation of Solution

Write the expression for time taken by block to move with the board

    $T_1=\frac{2A}{v}$

Here, $T_1$ is the time period and $v$ is the speed of motion.

Rewrite the expression by substituting $\left({\mu }_s-{\mu }_k\right)\frac{mg}{k}$ for $A$.

    $T_1=\frac{2\left(\left({\mu }_s-{\mu }_k\right)\frac{mg}{k}\right)}{v}$

Time taken by block to move from mean position to an extreme position.

    $\begin{array}{c}{T}_{1/2}=\frac{1}{2}\left(2\pi \sqrt{\frac{m}{k}}\right)\\ =\pi \sqrt{\frac{m}{k}}\end{array}$

Here, $T_{1/2}$ is the time taken to complete the half of SHM (same as that of time needed to spring back the block to original position.

The period of oscillation is the sum of $T_1$ and $T_{1/2}$.

    $T=T_1+T_{1/2}$

Here, $T$ is the time period of oscillation.

Conclusion:

Rewrite the above equation by substituting the relations for $T_1$ and $T_{1/2}$.

    $\begin{array}{c}T=\frac{2\left(\left({\mu }_s-{\mu }_k\right)\frac{mg}{k}\right)}{v}+\pi \sqrt{\frac{m}{k}}\\ =\frac{2\left(\left({\mu }_s-{\mu }_k\right)mg\right)}{kv}+\pi \sqrt{\frac{m}{k}}\end{array}$

Therefore, It is proved that the period of oscillation is $T=\frac{2\left({\mu }_s-{\mu }_k\right)mg}{vk}+\pi \sqrt{\frac{m}{k}}$.