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?

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

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)

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$0.031\mathrm{m}$