Question

Question 5

Answer 1

Answer:

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

$\lambda =\frac{v}{f}$

Here

λ = wavelength of the sound

v = speed of sound = 340 m/s

f = frequency of each instrument

You need to calculate both wavelengths

$\lambda _{1} =\frac{v}{f_{1} } =\frac{340}{266} =1.2782m$

$\lambda _{2} =\frac{v}{f_{2} } =\frac{340}{400} =0.85m$

$Ratio=\frac{\lambda _{1}}{\lambda _{2} } =\frac{1.2782}{0.85} =1.5038$

According to the results, the wavelength of the first instrument is approximately 1.5 greater than that produced by the second instrument.