An instrument plays a frequency of 266 Hz. Another identical instrument plays a frequency of 400 Hz. How do the wavelength of each of these sound waves compare?

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.