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3 years ago

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Sound waves are a longitudinal wave that have a speed of about 340 m/s when traveling through room temperature air. What is the

wavelength of a sound wave that has a frequency of 6,191 Hz?

1 answer:

AlexFokin [52]3 years ago

8 0

The wavelength of the wave is 0.055 m

Explanation:

The relationship between speed, frequency and wavelength of a wave is given by the wave equation:

$v=f\lambda$

where

v is the speed

f is the frequency

$\lambda$ is the wavelength

For the sound wave in this problem we have

v = 340 m/s is the speed

f = 6,191 Hz is the frequency

Solving for $\lambda$ , we find the wavelength:

$\lambda=\frac{v}{f}=\frac{340}{6191}=0.055 m$

Learn more about waves and wavelength:

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In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

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Making x the subject of the formula

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$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )

Also at equilibrium the moment about the point where the string holding the bar (where ( $m_1 +m_2$ ) and $m_3$ are hanged ) is attached is zero

So basically

$(m_1 + m_2 ) * 20 = m_3 * 30$

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3 years ago

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3 years ago

Read 2 more answers

A microphone is located on the line connecting two speakers that are 0.513 m apart and oscillating in phase. The microphone is 1

pantera1 [17]

Answer:

frequency 1 = 334.30 Hz

frequency 2 = 1002.92 Hz

Explanation:

Given data

speaker distance y = 0.513 m

microphone distance D = 1.80 m

to find out

lowest two frequencies

solution

we know velocity of sound is 343 m/s

so we consider point x

so at 1st speaker distance from x = D + (y/2)

1st speaker distance from x = 1.80 + (0.513/2) = 2.0565 m .....1

and

at 2nd speaker distance from x = D - (y/2)

2nd speaker distance from x = 1.80 - (0.513/2) = 1.5435 m .........2

so destructive interference from 1 and 2 we know

1st - 2nd = ( m + 0.5 ) wavelength

2.0565 m - 1.5435 m = ( 0+ 0.5) wavelength

wavelength = 1.026 m

so here 1st min frequency will be

frequency 1 = velocity of sound / wavelength

frequency 1 = 343 / 1.026 =334.30 Hz

and

2nd min frequency will be

frequency 2 =

2.0565 m - 1.5435 m = ( 1 + 0.5) wavelength

wavelength = 0.342 m

frequency 2 = velocity of sound / wavelength

frequency 2 = 343 / 0.342 = 1002.92 Hz

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3 years ago

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Answer:

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