Respuesta :
Answer:
[tex] \lambda_1 = \frac{340m/s}{100 Hz}=3.4m[/tex]
[tex] \lambda_2 = \frac{340m/s}{1000 Hz}=0.34m[/tex]
[tex] \lambda_3 = \frac{340m/s}{10000 Hz}=0.034m[/tex]
So then the rank for this case would be:
[tex] \lambda_1 > \lambda_2 > \lambda_3 [/tex]
And makes sense since if we have a higher frequency we expect a lower wavelength by the inversely proportional relationship between the frequency and the wavelength.
Explanation:
For this case we can use the property that the speed of the sound is:
[tex] v_s = 343 \frac{m}{s}[/tex]
By definition the frequency is given by this formula:
[tex] f = \frac{v}{\lambda}[/tex]
Where f represent the frequency, v the velocity and [tex]\lambda[/tex] the waelength. If we solve for the wavelength we got:
[tex] \lambda = \frac{v}{f}[/tex]
Now if we find the wavlengths for each of the following cases we have:
[tex] \lambda_1 = \frac{340m/s}{100 Hz}=3.4m[/tex]
[tex] \lambda_2 = \frac{340m/s}{1000 Hz}=0.34m[/tex]
[tex] \lambda_3 = \frac{340m/s}{10000 Hz}=0.034m[/tex]
So then the rank for this case would be:
[tex] \lambda_1 > \lambda_2 > \lambda_3 [/tex]
And makes sense since if we have a higher frequency we expect a lower wavelength by the inversely proportional relationship between the frequency and the wavelength.
