Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q100P

Expert-verified
Fundamentals Of Physics
Found in: Page 513

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

Pipe has only one open end; pipe B is four times as long and has two open ends. Of the lowest 10 harmonic numbers nB of pipe B , what are the (a) smallest, (b) second smallest, and (c) third smallest values at which a harmonic frequency of matches one of the harmonic frequencies of A ?

  1. The smallest value of nB at which a harmonic frequency of B matches that of A is 2 .
  2. The second smallest value of nB at which a harmonic frequency of matches that of A is 6 .
  3. The third smallest value of nB at which a harmonic frequency of B matches that of A is 10.
See the step by step solution

Step by Step Solution

Step 1: Write the given data

The length of pipe B is LB=4LA

Step 2: Determine the concept of the Doppler Effect

Use the concept of two open-ended and one open-ended pipe. With both ends of the pipe open, any harmonic can be set up in the pipe but with only one end open, only odd harmonic can be set up.

Formulae:

  1. Resonant frequency for pipe with both ends open is f=nv2L
  2. Resonant frequency for pipe with both ends open isf=nv4L

Step 3: Calculate the smallest value of  nB

(a)

The resonant frequency for a pipe of length LB with two open ends using equation (i) is

f=nBv2LB

For, n=1,2,3

LB=nBv2f …… (1)

The resonant frequency for a pipe of length with only one open end is

f=nAv4LA

For, n=1,3,5

LA=nAv4f ….. (2)

The frequency of matches that of and the length of pipe B is LB=4LA. Hence, from equation (1) and equation (2), we get

nBv2f=4(nAv4f)nB2=(nA)oddnB=2(nA)odd …… (a)

Using equation (a), the smallest value of nB at which a harmonic frequency of B matches that of A is given as:

nB=2(1)=2

Hence, the smallest value of nBis 2.

Step 4: b) Calculate the second smallest value of  nB

Using equation (a), the second smallest value of nB at which a harmonic frequency of B matches that of A is given as:

nB=2(3)=6

Hence, the second smallest value of nB is 6.

Step 5: c) Calculate the third smallest value of  nB

Using equation (a), the third smallest value of nB at which a harmonic frequency of B matches that of A is given as:

nB=2(5)=10

Hence, the third smallest value of nB is 10.

Most popular questions for Physics Textbooks

Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.