91影视

1. Frequency of oscillation of the rock$f=2.2\text{Hz}.$
2. Maximum displacement of the rock$x_m=1.0\text{cm}=0.01\text{m}$

A precariously balanced rock oscillates harmonically parallel to the ground when earthquakes occur. Hence, we can applytheequations of SHM to studythemotion of the rock.

The angular frequency of oscillation is given as-

$蝇=2蟺\text{f}$

The maximum acceleration is given as-

${\text{a}}_{max}={蝇}^2{\text{x}}_m$

The frequency of oscillation$\left(f\right)$ of diaphragm is related to angular frequency (蝇) by the relation,

$蝇=2蟺\text{f}$

Putting the values, we get

$\begin{array}{rcl}蝇& =& 2蟺\text{f}\\ & =& 2脳3.14脳2.2\text{Hz}\\ & =& 13.8\text{rad/s}\\ & & \end{array}$

For SHM, the magnitude of maximum acceleration is given by${\text{a}}_{max}={蝇}^2{\text{x}}_m$

Putting the values,

$$\begin{gathered} \begin{array}{rcl}{\text{a}}_{max}& =& {蝇}^2{\text{x}}_m{ \\ }^{\mathit{\text{=}}{\left(13.8\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^{\mathit{2}}}脳\left(0.01\text{m}\right)\\ & =& 1.90{\text{m/s}}^{\text{2}}\end{array} \end{gathered}$$

In terms of g, the acceleration is,

$$\begin{gathered} {\text{a}}_{max}=\frac{1.90{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}{9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}g \\ =0.19\text{g} \end{gathered}$$

The maximum acceleration of the oscillations is $0.19\text{g}$.