b. 460.8 m/s
Explanation:
The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is
[tex]f=\frac{v}{2L}[/tex]
where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:
[tex]v=2Lf=2(0.90 m)(256 Hz)=460.8 m/s[/tex]
c. 18,000 m
Explanation:
The relationship between speed of the wave, distance travelled and time taken is
[tex]v=\frac{d}{t}[/tex]
where
v = 6,000 m/s is the speed of the wave
d = ? is the distance travelled
t = 3 s is the time taken
Re-arranging the formula and substituting the numbers into it, we find:
[tex]d=vt=(6,000 m/s)(3 s)=18,000 m[/tex]
