Chapter 11, Problem 84GP

To determine

To Calculate: The maximum possible amplitude of the oscillations.

Answer to Problem 84GP

The maximum possible amplitude of the oscillations is $\text{0}\text{.14m}$ when block with mass m is not slip off on the block with mass M.

Explanation of Solution

Given info:

The mass of the first block is, $\text{M=5}\text{.0kg}$ and the mass of the second block is, $m\text{=1}\text{.25kg}$

The spring constant is, $\text{k}\text{=}\text{130}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.$

The static friction coefficient is 0.30.

Formula used:

The acceleration in the simple harmonic motion is given as

  ${\text{a}}_{\max }\text{=}{\text{ω}}^{\text{2}}\text{A}$

Where, the angular frequency is $\text{ω}$ with the amplitude A.

The angular frequency is given as,

  $\begin{array}{l}\text{ω}\text{=}\sqrt{\frac{\text{k}}{{\text{m}}_{total}}}\\ \Rightarrow {\text{ω}}^{\text{2}}\text{=}\frac{\text{k}}{{\text{m}}_{total}}\end{array}$

Where, k is the spring constant and ${\text{m}}_{total}$ is the total mass

Calculation:

Due to the static friction, the block with mass m is stayed on the block with mass M without slipping. So, they are executing the simple harmonic motion with respect to the ground. So, a maximum static frictional force is exerted on the mass ‘m’which is given as,

  ${\text{F}}_{\text{fric}}{}_{{}_{\text{max}}}\text{=}{\text{μ}}_{\text{s}}\text{mg}$

The mass m block gets accelerated due to the frictional force.

  $\begin{array}{l}\therefore {\text{ma}}_{\max }={\text{μ}}_{\text{s}}\text{mg}\\ \Rightarrow {\text{a}}_{\max }={\text{μ}}_{\text{s}}\text{g}\end{array}$

But according to the simple harmonic oscillations, the acceleration is given as,

  ${\text{a}}_{\max }\text{=}{\text{ω}}^{\text{2}}\text{A=}\frac{\text{k}}{{\text{m}}_{total}}A$

Now, by comparing these two expressions of the acceleration with each other, the results equation is

  ${\text{a}}_{\max }=\frac{\text{k}}{{\text{m}}_{total}}A={\text{μ}}_{\text{s}}\text{g}$

From this equation, here the amplitude of the wave is determined in such a way that,

  $\begin{array}{l}\Rightarrow \text{A}\text{=}{\text{μ}}_{\text{s}}\text{g}\frac{{\text{m}}_{\text{total}}}{\text{k}}\text{=}\frac{{\text{μ}}_{\text{s}}\text{g}\left(\text{M+m}\right)}{\text{k}}\\ \text{=}\frac{\left(\text{0}\text{.30}\right)\text{×}\left(\text{9}\text{.80}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\right)\text{×}\left(\text{5}\text{.0}\text{kg+1}\text{.25}\text{kg}\right)}{\text{130}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.}\\ \text{A}\text{=}\text{0}\text{.14}\text{m}\end{array}$

Conclusion:

The amplitude is $\text{0}\text{.14}\text{m}$ .