Question 14.4: Loudspeaker. The cone of a loudspeaker (Fig. 14-9) oscillate......
Loudspeaker. The cone of a loudspeaker (Fig. 14-9) oscillates in SHM at a frequency of 262 \mathrm{~Hz} (“middle C”). The amplitude at the center of the cone is A=1.5 \times 10^{-4} \mathrm{~m}, and at t=0, x=A. (a) What equation describes the motion of the center of the cone? (b) What are the velocity and acceleration as a function of time? (c) What is the position of the cone at t=1.00 \mathrm{~ms}\left(=1.00 \times 10^{-3} \mathrm{~s}\right) ?
APPROACH The motion begins (t=0) with the cone at its maximum displacement (x=A at t=0). So we use the cosine function, x=A \cos \omega t, with \phi=0.
(a) The amplitude A=1.5 \times 10^{-4} \mathrm{~m} and
\omega=2 \pi f=(6.28~ \mathrm{rad})\left(262 \mathrm{~s}^{-1}\right)=1650~ \mathrm{rad} / \mathrm{s} .
The motion is described as
x=A \cos \omega t=\left(1.5 \times 10^{-4} \mathrm{~m}\right) \cos (1650~t)
where t is in seconds.
(b) The maximum velocity, from Eq. 14-9a (v_{\max }=\omega A=\sqrt{\frac{k}{m}} A), is
v_{\max }=\omega A=(1650~ \mathrm{rad} / \mathrm{s})\left(1.5 \times 10^{-4} \mathrm{~m}\right)=0.25 \mathrm{~m} / \mathrm{s} \text {, }
so
v=-(0.25 \mathrm{~m} / \mathrm{s}) \sin (1650 t)
From Eq. 14-9b (a_{\max }=\omega^{2} A=\frac{k}{m} A) the maximum acceleration is a_{\max }=\omega^{2} A=1650 \,\mathrm{rad} / \mathrm{s})^{2}\left(1.5 \times 10^{-4} \mathrm{~m}\right)=410 \mathrm{~m} / \mathrm{s}^{2}, which is more than 40 \mathrm{~g} ‘s. Then
a=-\left(410 \mathrm{~m} / \mathrm{s}^{2}\right) \cos (1650 t)
(c) At t=1.00 \times 10^{-3} \mathrm{~s},
\begin{aligned}x=A \cos \omega t & =\left(1.5 \times 10^{-4} \mathrm{~m}\right) \cos \left[(1650 \,\mathrm{rad} / \mathrm{s})\left(1.00 \times 10^{-3} \mathrm{~s}\right)\right] \\& =\left(1.5 \times 10^{-4} \mathrm{~m}\right) \cos (1.65 \,\mathrm{rad})=-1.2 \times 10^{-5} \mathrm{~m}\end{aligned}
NOTE Be sure your calculator is set in RAD mode, not DEG mode