## College Physics (4th Edition)

The two possible frequencies are $106~Hz$ and $137~Hz$
We can find the wavelength for a frequency of $100~Hz$: $\lambda = \frac{v}{f} = \frac{343~m/s}{100~Hz} = 3.43~m$ We can find the wavelength for a frequency of $150~Hz$: $\lambda = \frac{v}{f} = \frac{343~m/s}{150~Hz} = 2.29~m$ If the sound is very soft compared to other places, then the waves from the two speakers are interfering destructively at this point. Then the path difference must have the form $(\frac{2n+1}{2}~\lambda)$, for some integer $n$. We can find the path difference to this point from the two speakers: $37.1~m-25.8~m = 11.3~m$ We can find an expression for the wavelength: $\frac{2n+1}{2}~\lambda = 11.3~m$ $\lambda = \frac{(2)(11.3~m)}{2n+1}$ $\lambda = \frac{22.6~m}{2n+1}$ We can find the wavelength for various integers $n$: $\lambda_2 = \frac{22.6~m}{(2)(2)+1} = 4.52~m$ $\lambda_3 = \frac{22.6~m}{(2)(3)+1} = 3.23~m$ $\lambda_4 = \frac{22.6~m}{(2)(4)+1} = 2.51~m$ $\lambda_5 = \frac{22.6~m}{(2)(5)+1} = 2.05~m$ The wavelengths which fall into the acceptable range between $2.29~m$ and $3.43~m$ are $2.51~m$ and $3.23~m$. We can find the frequency when the wavelength is $2.51~m$: $f = \frac{v}{\lambda} = \frac{343~m/s}{2.51~m} = 137~Hz$ We can find the frequency when the wavelength is $3.23~m$: $f = \frac{v}{\lambda} = \frac{343~m/s}{3.23~m} = 106~Hz$ The two possible frequencies are $106~Hz$ and $137~Hz$