91Ó°ÊÓ

1. Frequency of oscillation of diaphragm$f=440\text{ H³ú}$
2. Maximum displacement of diaphragm

$\begin{array}{l}{x}_m\text{=0.75″¾³¾}\\ {\text{=0.75×10}}^{\text{-3}}\text{m}\end{array}$

Use the fact that an oscillating loudspeaker diaphragm executes Simple Harmonic Motion.

The angular frequency is given as-

$\mathrm{Ó\neg }=2\mathrm{Ï€}\text{f}$

The maximum velocity is given as-

${\text{v}}_{max}=\mathrm{Ó\neg }{\text{x}}_m$

The maximum acceleration is given as-

${\text{a}}_{max}={\mathrm{Ó\neg }}^2{\text{x}}_m$

The frequency of oscillation (f) of diaphragm is related to angular frequency (Ó¬) by the relation,

$\mathrm{Ó\neg }=2\mathrm{Ï€}\text{f}$

Putting the values, we get

$\begin{array}{rcl}\mathrm{Ó\neg }& =& 2\mathrm{Ï€}\text{f}\\ & =& 2×3.14×440\text{Hz}\\ & =& 2.8×{10}^3\text{rad/s}\\ & & \end{array}$

For an SHM, maximum speed of oscillations is given by${\text{v}}_{max}=\mathrm{Ó\neg }{\text{x}}_m$

Putting the values, we get

$\begin{array}{l}{\text{v}}_{max}=\mathrm{Ó\neg }{\text{x}}_m\\ =2.8×{10}^3\text{ r²¹»å}/\text{s}×0.75×{10}^{−3}\text{″¾}\\ =2.1\text{″¾}/\text{s}\end{array}$

The maximum velocity is$2.1\text{″¾}/\text{s}$.

For SHM, the magnitude of maximum acceleration is given by$a_{max}={\mathrm{Ó\neg }}^2{x}_m$

Putting the values,

$\begin{array}{l}{\text{a}}_{max}={\mathrm{Ó\neg }}^2{\text{x}}_m\\ ={(2.8×{10}^{3\text{ }}\text{ r²¹»å}/\text{s})}^2×(0.75×{10}^{−3}\text{″¾})\\ =5.7×{10}^3\text{″¾}/{\text{s}}^2\end{array}$

The maximum acceleration is given as$5.7×{10}^3\text{″¾}/{\text{s}}^2$-.