Explanation:
the wavelength of the first instrument is approximately 1.5 greater than that produced by the second instrument.
Explanation:
Given:
f₁ = frequency = 266 Hz
f₂ = 400 Hz
Question: How do the wavelength of each of these sound waves compare, λ₁ = ?, λ₂ = ?
The equation to solve this question is
\lambda =\frac{v}{f}λ=
f
v
Here
λ = wavelength of the sound
v = speed of sound = 340 m/s
f = frequency of each instrument
You need to calculate both wavelengths
\lambda _{1} =\frac{v}{f_{1} } =\frac{340}{266} =1.2782mλ
1
=
f
1
v
=
266
340
=1.2782m
\lambda _{2} =\frac{v}{f_{2} } =\frac{340}{400} =0.85mλ
2
=
f
2
v
=
400
340
=0.85m
Ratio=\frac{\lambda _{1}}{\lambda _{2} } =\frac{1.2782}{0.85} =1.5038Ratio=
λ
2
λ
1
=
0.85
1.2782
=1.5038
According to the results, the wavelength of the first instrument is approximately 1.5 greater than that produced by the second instrument.
