Question

the velocity of sound in gas in which two waves of wavelength 1m and 1.01m produce 10 beats in 3 seconds.

1) 336.7 ms^-1
2)337.6
3)363.7
4)333.3

Answer 1

As usual, let’s start with what we are given, assign some names and do some quick calculations that follow immediately.

λ0=1.00mλ0=1.00m

λ1=1.01m=λ0+.01mλ1=1.01m=λ0+.01m

We will use f0f0 and f1f1 for the frequencies corresponding to the two wavelength. Since the two frequencies superimposed produce a variation of 10 beats in 3 seconds we also have:

f0−f1=103≈3.3f0−f1=103≈3.3 so

f1=f0−3.3f1=f0−3.3

We know that f0>f1f0>f1 since it has a shorter wavelength.

Unless we have a very weird medium, we can ignore any variation is velocity due to frequency since the frequencies are so close.

In general, we have v=λfv=λf and in particular:

λ0f0=λ1f1λ0f0=λ1f1 . Substituting gives:

λ0f0=(λ0+.01)(f0−3.3)λ0f0=(λ0+.01)(f0−3.3) . Simplifying gives:

0=−3.3λ0+.01f0−0.0330=−3.3λ0+.01f0−0.033

I’ve dropped the units since they don’t add any particular understanding. Since λ0=1.0λ0=1.0 we have:

0=.01f0−3.330=.01f0−3.33 so

f0=333f0=333 . Since λ0=1.0λ0=1.0 we have:

v=333msv=333ms . Double checking with our f1f1 and λ1λ1 values gives:

v=329.7×1.01=332.997v=329.7×1.01= 333.3 which is in agreement.

The answer is 4) 333.3

(Please mark it brainliest.)