Question

A block is on a horizontal surface (a shake table) that is moving back and forth horizontally with simple harmonic motion of frequency $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{H}\mathit{z}$. The coefficient of static friction between block and surface is0.50. How great can the amplitude of the SHM be if the block is not to slip along the surface?

Short answer

The greatest value for amplitude of SHM will be0.031m, if the block is not to slip along the surface.

Step-by-step solution

The given data

1. Frequency of block in SHM,$f=2.0\mathrm{Hz}$
2. Coefficient of static friction,${\mu }_s=0.5$

Understanding the concept of Newton’s law of motion

Using Newton’s 2nd law, we can equate the frictional force to the product of mass and acceleration. Hence, we can find the maximum amplitude for SHM.

Formula:

The normal force applied to a body due to its weight, $F_N=mg$(i)

The angular frequency of an oscillation, $\omega =2\pi f$(ii)

The force on a body according to Newton’s second law, $F=ma$(iii)

Acceleration of block in SHM,$a_m=X_m{\omega }^2$ (iv)

Calculation of amplitude of SHM

The maximum force that can be exerted by the surface must be less than ${\mu }_s{F}_N$,

So, we can say that,

$F={\mu }_s{F}_N$ (V)

Substituting equations (i), (iii) & (iv) in equation (v), we get

$$\begin{gathered} {\mu }_{3}mg=m\times \left(X_m{\omega }^2\right) \\ {X}_m=\frac{{\mu }_{s}g}{{\omega }^2} \\ =\frac{{\mu }_{s}g}{{\left(2\pi f\right)}^2} \end{gathered}$$

Substitute the values to find the value of amplitude,

$$\begin{gathered} {\mathrm{X}}_{\mathrm{m}}=\frac{0.58\times 9.8}{{\left(2\mathrm{\pi }\times 2\right)}^2} \\ =0.031\mathrm{m} \end{gathered}$$

Hence, the value of amplitude is$0.031\mathrm{m}$