Question

Question 5
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?

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.

Answer 2

The wavelength will be "1.5".

Frequency and Wavelength:

The proportion of waves that thus passes through a specific location in a particular length of time is a frequency.

The distance together in the wave's advancement connection from any one position towards the next position of the comparable phases, is considered as wavelength.

According to the question,

Frequency, f₁ = 266 Hz

                   f₂ = 400 Hz

Speed of sound, v = 340 m/s

We know the formula,

→ Wavelength, λ = $\frac{v}{f}$

Now,

λ₁ = $\frac{v}{f_1}$

By substituting the values,

   = $\frac{340}{266}$

   = 1.2782 m

and,

λ₂ = $\frac{v}{f_2}$

By substituting the values,

    = $\frac{340}{400}$

    = 0.85 m

hence,

The ratio of wavelength be:

→ $\frac{\lambda_1}{\lambda_2} = \frac{1.2782}{0.85}$

       = 1.5

Thus the above answer is correct.

Find out more information about frequency here:

https://brainly.com/question/10728818