Respuesta :
Answer:
Explanation:
Frequency of sound n = 5.1 MHz = 5.1 x 10⁶ Hz
angular frequency ω = 2π x n
= 2 x 3.14 x 5.1 x 10⁶ rad /s
= 32.028 x 10⁶ rad /s
period of vibration = 1 / n
1 / 5.1 x 10⁶
= .196 x 10⁻⁶ s
T = .196 μs
The period and angular frequency of the molecular vibrations is 1.96 * 10⁻⁷s and 3.202 * 10⁷ rad/s.
Based on the given question,
• The frequency used to detect tumors is 5.10 MHz or 5.10 × 10⁶ Hz.
Now the period and angular frequency of the molecular vibrations caused by the given pulse of sound is,
Period,
[tex]= T = \frac{1}{f}[/tex]
[tex]= \frac{1}{5.10 * 10^{6} } Hz\\= 1.96 * 10^{-7} }s[/tex]
Angular frequency = 2πf
[tex]= 2*3.14 rad* (5.10 * 10^{6} Hz)\\= 3.202 * 10^{7} \frac{rad}{s}[/tex]
Thus, the period is 1.96 * 10⁻⁷ s and the angular frequency is 3.202 * 10⁷ rad/s.
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