Question

A loudspeaker diaphragm is oscillating in simple harmonic motion with a frequency of$\mathbf{440}\mathbf{Hz}$and a maximum displacement of$\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{mm}$.

1. What is the angular frequency?
2. What is the maximum speed?
3. What is the magnitude of the maximum acceleration?

Short answer

1. Angular frequency$=2.8\times {10}^3\text{rad}/\text{s}$
2. Maximum speed$=2.1\text{m}/\text{s}$
3. Magnitude of maximum acceleration$=5.7\times {10}^3\text{m}/{\text{s}}^2$

Step-by-step solution

Given

1. Frequency of oscillation of diaphragm$f=440\text{\hspace{0.17em}Hz}$
2. Maximum displacement of diaphragm

$\begin{array}{l}{x}_m\text{=0.75\hspace{0.17em}mm}\\ {\text{=0.75×10}}^{\text{-3}}\text{m}\end{array}$

Understanding the concept

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

The angular frequency is given as-

$\omega =2\pi \text{f}$

The maximum velocity is given as-

${\text{v}}_{max}=\omega {\text{x}}_m$

The maximum acceleration is given as-

${\text{a}}_{max}={\omega }^2{\text{x}}_m$

(a) Calculate the angular frequency

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

$\omega =2\pi \text{f}$

Putting the values, we get

$\begin{array}{rcl}\omega & =& 2\pi \text{f}\\ & =& 2\times 3.14\times 440\text{Hz}\\ & =& 2.8\times {10}^3\text{rad/s}\\ & & \end{array}$

(b) Calculate the maximum speed

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

Putting the values, we get

$\begin{array}{l}{\text{v}}_{max}=\omega {\text{x}}_m\\ =2.8\times {10}^3\text{\hspace{0.17em}rad}/\text{s}\times 0.75\times {10}^{-3}\text{\hspace{0.17em}m}\\ =2.1\text{\hspace{0.17em}m}/\text{s}\end{array}$

The maximum velocity is$2.1\text{\hspace{0.17em}m}/\text{s}$.

(c) Calculate the magnitude of the maximum acceleration

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

Putting the values,

$\begin{array}{l}{\text{a}}_{max}={\omega }^2{\text{x}}_m\\ ={(2.8\times {10}^3\text{\hspace{0.17em}rad}/\text{s})}^2\times (0.75\times {10}^{-3}\text{\hspace{0.17em}m})\\ =5.7\times {10}^3\text{\hspace{0.17em}m}/{\text{s}}^2\end{array}$

The maximum acceleration is given as$5.7\times {10}^3\text{\hspace{0.17em}m}/{\text{s}}^2$-.