Answer
a.) Music.
T = 2.15 x 10^{-3} s
\omega = 2.93 x 10^(3) rad/s
b.) Hearing.
f = 2.0 x 10^{4} Hz
\omega = 1.26 x 10^{5} rad/s
c.) Vision.
f = 4.3 x 10^{14} Hz to f = 7.5 x 10^{14} Hz
T = 1.3 x 10^{-15} s to T = 2.5 x 10^{-15} s
d.) Ultrasound.
T = 2.0 x 10^{-7} s
\omega = 3.1 x 10^{7} rad/s
Work Step by Step
a.) Music.
required: T and \omega
given: f = 466 Hz
Solution:
The period T and frequency f are related by the equation: T = \frac{1}{f}
Therefore the period that it takes for the vocal chords to vibrate through one complete cycle is T = \frac{1}{466hz} = 2.15 x 10^{-3} s.
The angular frequency is given by the equation: \omega = 2\pi f
Therefore the angular frequency of the chords \omega = 2\pi (466Hz) = 2.93 x 10^(3) rad/s.
b.) Hearing.
required: f and \omega
given: 50.0 \mu s
Solution:
First, we convert 50.0 \mu s into SI units, which gives: 5 x 10^{-5} s.
The period T and frequency f are related by the equation: T = \frac{1}{f}. We can multiply both sides of the equation by \frac{f}{T} so we get the equation for the frequency: f = \frac{1}{T}
Solving for the frequency we get, f = \frac{1}{T} = \frac{1}{5 x 10^{-5}} = 2.0 x 10^{4} Hz
The angular frequency is given by the equation: \omega = 2\pi f
Therefore the angular frequency of the vibrating eardrum is \omega = 2\pi f = 2\pi (2.0 x 10^{4} Hz = 1.26 x 10^{5} rad/s.
c.) Vision.
required: Limits of the period T and frequency F
given: \omega = 2.7 x 10^{15} rad/s and \omega = 4.7 x 10^{15} rad/s
Solution:
The frequency f and angular frequency \omega is related by \omega = 2\pi f , and by dividing both sides by 2\pi we get f = \frac{\omega}{2\pi}.
The lower limit is f = \frac{2.7 x 10^{15} rad/s}{2\pi rad} = 4.3 x 10^{14} Hz.
The upper limit is f = \frac{4.7 x 10^{15} rad/s}{2\pi rad} = 7.5 x 10^{14} Hz.
The period T and frequency f is related by: T = \frac{1}{f}.
The lower limit for the T is T = \frac{1}{7.5 x 10^{14} Hz} = 1.3 x 10^{-15} s.
And the upper limit for the T is T = \frac{1}{4.3 x 10^{14} Hz } = 2.5 x 10^{-15} s.
d.) Ultrasound.
required: T and \omega
given: 5.0 MHz
Solution:
First, we convert 5.0 MHz to SI units, which gives: 5 x 10^{6} Hz.
The period T and frequency f is related by T = \frac{1}{f}
Therefore, T = \frac{1}{5 x 10^{6} Hz} = 2.0 x 10^{-7} s.
The frequency and angular frequency \omega is related by \omega = 2\pi f
Therefore, \omega = 2\pi (5 x 10^{6} Hz) = 3.1 x 10^{7} rad/s.