Question

When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amplitude of 7.5 x10-7 m and a maximum speed of 3.3 x10-3 m/s. (a) What is the frequency (in Hz) of the eardrum's vibrations

Answer 1

Answer:

f = 700 Hz

Explanation:

given

amplitude =$a= 7.5 \times 10^{-7} m$

speed v =$3.3 \times 10^{-3}$ m/s

for (a)

we know $v_{max} = a\omega$

and $\omega = 2\pi f$

we get

$v_{max} = a( 2 \pi f)$

$f = \frac{v_{max}}{2 \pi a}$

$f = \frac{ 3.3 \times 10^{-3}}{ 2 \times 3.14 \times 7.5 \times 10^{-7}}$

$f = 700 Hz$