# Ch 10.1 Help

## Ch 10.1 Help

Question 1: (1 point)

1) The speed of sound in air is 343 m/s. What is the wavelength of a sound wave that has a frequency of 483.5636 Hz (see sheet 3,3') ? Indicate with a positive (negative) sign whether the formula you use is only valid for sound waves (is valid for many waves, e.g. waves on a string, radio waves,...).

(How to solve):

Question 2: (1 point)

2) A sound source assumed to be a point emits a power of 10.001 W. What is the sound intensity at a distance of 3.87 m from the source (see sheet 9.2)?

(How to solve):

Question 3: (1 point)

3) You are at a distance of 8.208 m from a point-like sound source and experience a sound intensity of 0.0096 W/m2. How close to the source can you move before you experience a sound intensity of 0.8873 W/m2, the pain threshold (see sheet 9.2-9.4)? Indicate with a positive (negative) sign whether for a large wall full of loudspeakers as the sound source you would experience the same (smaller increase) of the sound intensity upon approach.

(How to solve):

Question 4: (1 point)

4) A train approaches a station with 46.32 km/hr and emits a whistle sound at 748 Hz. What is the frequency registered at the station (see sheet 12,use 343 m/s for the speed of sound) ? Indicate with a negative (positive) sign whether the frequency registered at the station would be higher (lower) if the train moved away from the station.

(How to solve):

hwilson
Guest

## Re: Ch 10.1 Help

Question 1:
velocity (speed of sound in air) = frequency x wavelength
wavelength = velocity/frequency

The Answer is NEGATIVE because this formula is true for all waves

Question 2:
Intensity = Power/Area, where area is = 4pi*r^2 where r is your distance
And its positive because there is no other choice

Kathleen
Guest

## Re: Ch 10.1 Help

how do u do 3 and 4

abc
Guest

## Question 3 and 4

I have been working on 3 and 4 for a while and am still confused can someone help? please?

sedwards
Guest

## Re: Ch 10.1 Help

question 3 negative
equation 10.3
I2/I1=R1^2/R2^2
You are at a distance of 7.304 m from a point-like sound source and experience a sound intensity of 0.0082 W/m2. How close to the source can you move before you experience a sound intensity of 0.8421 W/m2, the pain threshold (see sheet 9.2-9.4) ? Indicate with a positive (negative) sign whether for a large wall full of loudspeakers as the sound source you would experience the same (smaller increase) of the sound intensity upon approach.
.8421/.0082=7.304^2/x^2
===-0.7208

1
Guest

## Question 3

I did question three a little differently but still got a correct answer nonetheless

First, I used the intensity formula to solve for power using the distance and intensity first given,
so power = intensity x 4pi x distance^2
power = X

Then, use the power "X" you just calculated and the intensity they want you to "experience" (second one listed) to find the distance

("X")/(intensity) = Y
Y/ 4pi = Z
sqrt z = the distance

taking into account then that area (aka 4pi*distance^2) = power/intensity, distance^2 = (power/intensity)/(4pi)

Get it?

And as already mentioned, it is negative

Kathleen
Guest

helpme
Guest

## Q4

can someone help with number 4 ?

angelbab
Guest

## Re: Ch 10.1 Help

4 positive
A train approaches a station with 57.41 km/hr and emits a whistle sound at 849.9 Hz. What is the frequency registered at the station (see sheet 12,use 343 m/s for the speed of sound) ? Indicate with a negative (positive) sign whether the frequency registered at the station would be higher (lower) if the train moved away from the station.
first change the train km/hr to m/s by: 57.41km/h x 1000m/3600s= 15.947m/s
then equation 10.5
fprime= (v of sound / v of sound - v of train) x Freq
(minus because the train approaches)

fprime=(343/343-15.947)849.9 = 891.3
god i am an idiot took me for ever to realize i had to change the train speed to si units

1
Guest

## Question 4

First, convert km/hr to m/s by multiplying value in km/hr x (1000 m/km)/(3600 s/hr)

A train approaches a station with 44.56 km/hr and emits a whistle sound at 862 Hz. What is the frequency registered at the station (see sheet 12,use 343 m/s for the speed of sound) ? Indicate with a negative (positive) sign whether the frequency registered at the station would be higher (lower) if the train moved away from the station.

In my example, 44.56 km/hr x (1000/3600) = 12.377
Then use the equation f' = ((v+v_0)/(v+v_s)) * frequency
v = sound (which they say to use 343 m/s
v_o is observer which in this case is the train
vs = source which is the station which isn't moving so it is 0

f = ((343 + 12.377)/(343 + 0 )) * 862 = 893.106

Kathleen
Guest

## Re: Ch 10.1 Help

sorry, i never refresh the page because I was working through the problems so I didn't realize you already posted how to do them. either way, i hope you guys find one of these useful and good luck!

Kathleen
Guest