(a) 0.0021 s, 2926.5 rad/s
The frequency of the B note is
![f= 466 Hz](https://tex.z-dn.net/?f=f%3D%20466%20Hz)
The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:
![T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7B466%20Hz%7D%3D0.0021%20s)
The angular frequency instead is given by
![\omega = 2\pi f](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20f)
And substituting
f = 466 Hz
We find
![\omega = 2\pi (466 Hz)=2926.5 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20%28466%20Hz%29%3D2926.5%20rad%2Fs)
(b) 20 Hz, 125.6 rad/s
In this case, the period of the sound wave is
T = 50.0 ms = 0.050 s
So the frequency is equal to the reciprocal of the period:
![f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B1%7D%7BT%7D%3D%5Cfrac%7B1%7D%7B0.050%20s%7D%3D20%20Hz)
While the angular frequency is given by:
![\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20f%20%3D%202%20%5Cpi%20%2820%20Hz%29%3D125.6%20rad%2Fs)
(c) ![4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s](https://tex.z-dn.net/?f=4.30%5Ccdot%2010%5E%7B14%7D%20Hz%2C%207.48%5Ccdot%201%5E%7B14%7D%20Hz%2C%202.33%5Ccdot%2010%5E%7B-15%7D%20s%2C%201.34%5Ccdot%2010%5E%7B-15%7Ds)
The minimum angular frequency of the light wave is
![\omega_1 = 2.7\cdot 10^{15}rad/s](https://tex.z-dn.net/?f=%5Comega_1%20%3D%202.7%5Ccdot%2010%5E%7B15%7Drad%2Fs)
so the corresponding frequency is
![f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B%5Comega%7D%7B2%20%5Cpi%7D%3D%5Cfrac%7B2.7%5Ccdot%2010%5E%7B15%7Drad%2Fs%7D%7B2%5Cpi%7D%3D4.30%5Ccdot%2010%5E%7B14%7D%20Hz)
and the period is the reciprocal of the frequency:
![T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7B4.30%5Ccdot%2010%5E%7B14%7DHz%7D%3D2.33%5Ccdot%2010%5E%7B-15%7Ds)
The maximum angular frequency of the light wave is
![\omega_2 = 4.7\cdot 10^{15}rad/s](https://tex.z-dn.net/?f=%5Comega_2%20%3D%204.7%5Ccdot%2010%5E%7B15%7Drad%2Fs)
so the corresponding frequency is
![f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B%5Comega%7D%7B2%20%5Cpi%7D%3D%5Cfrac%7B4.7%5Ccdot%2010%5E%7B15%7Drad%2Fs%7D%7B2%5Cpi%7D%3D7.48%5Ccdot%2010%5E%7B14%7D%20Hz)
and the period is the reciprocal of the frequency:
![T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7B7.48%5Ccdot%2010%5E%7B14%7DHz%7D%3D1.34%5Ccdot%2010%5E%7B-15%7Ds)
(d) ![2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s](https://tex.z-dn.net/?f=2.0%5Ccdot%2010%5E%7B-7%7Ds%2C%203.14%5Ccdot%2010%5E%7B7%7D%20rad%2Fs)
In this case, the frequency is
![f=5.0 MHz = 5.0 \cdot 10^6 Hz](https://tex.z-dn.net/?f=f%3D5.0%20MHz%20%3D%205.0%20%5Ccdot%2010%5E6%20Hz)
So the period in this case is
![T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7B5.0%5Ccdot%2010%5E6%20%20Hz%7D%3D2.0%20%5Ccdot%2010%5E%7B-7%7D%20s)
While the angular frequency is given by
![\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20f%3D2%20%5Cpi%20%285.0%5Ccdot%2010%5E%7B6%7DHz%29%3D3.14%5Ccdot%2010%5E%7B7%7D%20rad%2Fs)