Question

Review: A large block P attached to a light spring executes horizontal, simple harmonic motion as it slides across a frictionless surface with a frequency $\mathit{f}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{\text{Hz}}$. Block B rests on it as shown in Figure and the coefficient of static friction between the two is${\mathit{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{600}$. What maximum amplitude of oscillation can the system have if block B is not to slip?

Short answer

The maximum amplitude of oscillation can the system have if block B is not to slip A = 6.62 cm.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The frequency is, f = 1.50 Hz.
• The coefficient of static fraction is, ${\mu }_s=0.600$.

Significance of the Newton’s law of motion

Newton’s law of motion states that, the force acting on a body is equal to the mass m of the body multiplied by the acceleration of its center of mass,

F = ma

Determination of the maximum amplitude

When acceleration of block B exceeds the maximum possible value of acceleration of the block B that can be generated by the force of friction

UsingNewton’slawofmotionforblockP:

$$\begin{gathered} kx=Ma \\ a=\frac{kx}{M}...\left(1\right) \end{gathered}$$

This acceleration must be equal to acceleration produced in block B as:

$$\begin{gathered} \mu mg=ma \\ a=\mu g \end{gathered}$$

From equation (1) and (2), we get

$\frac{kx}{M}=\mu g$

For maximum amplitude x = A

role="math" localid="1663737780116" $\frac{kA}{M}=\mu g...\left(3\right)$

The expression of natural frequency is expressed as,

$$\begin{gathered} {{\omega }_0}^2=\left[\frac{k}{M}\right] \\ \frac{k}{M}={\left[2\pi f\right]}^2 \end{gathered}$$

Substitute all the value in the above equation,

$$\begin{gathered} \frac{k}{M}={\left[2\pi \times 1.50 \text{Hz}\right]}^2 \\ \frac{k}{M}=88.8 {\text{Hz}}^2 \end{gathered}$$

Substitute the value of $\frac{k}{M}$in equation (3)

$\left(88.8 \text{Hz}\right)A=\mu g$

$$\begin{gathered} \left(88.8 \text{Hz}\right)A=0.600\times 9.81 {\text{ms}}^{-2} \\ A=\frac{0.600\times 9.81 {\text{ms}}^{-2}}{88.8 \text{Hz}} \\ A=0.0662 \text{m} \\ A=6.62 \text{cm} \end{gathered}$$

Hence the maximum amplitude of oscillation can the system have if block B is not to slip is, A = 6.62 cm.