1) The speed of a wave is given by
[tex]v=\lambda f[/tex]
where
[tex]\lambda[/tex] is the wavelength
f is the frequency of the wave
For the wave in this problem, the distance between two successive crests is 15 m, this means its wavelength is [tex]\lambda=15 m[/tex] (because the wavelength is the distance between two successive crests). The period of the wave is 6 s, and the frequency is the reciprocal of the period:
[tex]f= \frac{1}{T}= \frac{1}{6 s}=0.167 Hz [/tex]
Therefore, the speed of the waves is
[tex]v= \lambda f=(15 m)(0.167 Hz)=2.51 m/s[/tex]
2) The speed of sound in air is [tex]v=343 m/s[/tex]. The frequency of the note is f=400 Hz, so we can find its wavelength by using the same relationship we used at point 1)
[tex]\lambda = \frac{v}{f}= \frac{343 m/s}{400 Hz}=0.86 m [/tex]
Since you are sitting at a distance of d=184 m from the source of the sound wave, the number of wavelengths between you and Barry is
[tex]N= \frac{d}{\lambda}= \frac{184 m}{0.86 m}=213.9 \sim 214 [/tex]
So, there are approximately 214 fully wavelengths between you and Barry.
2b) The time delay between the time the note is played and the time you hear it corresponds to the time the sound takes to arrive to your ear, therefore the distance divided by the speed of sound:
[tex]t= \frac{d}{v}= \frac{184 m}{343 m/s}=0.54 s [/tex]
3) The wavelength of the ultrasound wave is [tex]\lambda=0.0085 m[/tex] while its speed is the speed of sound: [tex]v=343 m/s[/tex], therefore its frequency is
[tex]f= \frac{v}{\lambda}= \frac{343 m/s}{0.0085 m}= 4.04 \cdot 10^4 Hz=40.4 kHz[/tex]
4) The period of the wave is T=2 s, so the frequency is
[tex]f= \frac{1}{2 s}=0.5 Hz [/tex]
and the spacing between the crests, which corresponds to the wavelength, is
[tex]\lambda=20 m[/tex]
Therefore, the speed of the wave is
[tex]v=\lambda f=(20 m)(0.5 Hz)=10 m/s[/tex]
