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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 506: Lesson Check

Exercise 28

Step 1

1 of 2

A standing wave in pipe that is closed at one end is expected to have an antinode located at the open end of the pipe and a node at the closed end.

Result

2 of 2

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Exercise 29

Step 1

1 of 2

For a standing wave in a pipe open that is at both ends, the antinode must be located at each end of the pipe.

Result

2 of 2

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Exercise 30

Step 1

1 of 2

The first harmonic standing wave for a bottle has a node at the bottom and an antinode located at the open end. However, the first harmonic standing wave in a pipe that is open at both ends has antinodes at each end and a node located at the middle of the pipe. Consequently, the first harmonic of a pipe that is open at both ends must have more antinodes than the first harmonic of a bottle (closed at one end).

Result

2 of 2

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Exercise 31

Step 1

1 of 2

The fundamental frequency (n=1) of a pipe open at both ends is given by the equation

$f_1=dfrac{v}{2L}$

When the pipe is cut into half, the fundamental frequency is expected to double. Therefore, the fundamental frequency of each half would be greater than $200Hz$

Result

2 of 2

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Exercise 32

Step 1

1 of 2

The frequency of standing waves in a pipe that is open at one end (closed at the other) is given by the equation

$f_n=dfrac{nv}{4L}$

But for pipes that are open at both ends, the equation is

$f_n=dfrac{2v}{2L}$

Using this principle

$f_A=dfrac{1(343m/s)}{4(1m)}=86Hz$

$f_B=dfrac{(1)(343/s)}{2(1m)}=172;Hz$

$f_C=dfrac{2(343m/s)}{2(3m)}=114;Hz$

$f_D=dfrac{3(343m/s)}{4(3m)}=86;Hz$

Therefore, the correct arrangement is $A=D<C<B$

Result

2 of 2

$$
A=D<C<B
$$

Exercise 33

Step 1

1 of 2

Compute the fundamental frequency (n=1) using the following equation

$f_n=dfrac{nv}{4L}$

$$
f_1=dfrac{v}{4(H-h)}=dfrac{343;m/s}{4(0.260m-0.070m)}=451;Hz
$$

Result

2 of 2

$$
451;Hz
$$

Exercise 34

Step 1

1 of 2

Calculate the third harmonic (n=3) using the equation

$$
lambda_3=dfrac{4L}{n}=dfrac{4(2.7m)}{3}=3.6m
$$

Result

2 of 2

$$
3.6m
$$

Exercise 35

Step 1

1 of 2

Take the difference of the adjacent harmonics

$difference=625-375=250;Hz$

Since this fundamental frequency is twice the given fundamental frequency, we can say that

$f_1=dfrac{250;Hz}{2}=125;Hz$

Therefore, the fundamental frequency for the bottle is not 375 Hz because from the result of our calculation, it is 125 Hz. The answer is NO.

The next standing wave (second harmonic) that occurs after 625 Hz will occur at $2f_1=250Hz$ higher than 625 Hz.

Thus, the next standing wave will have a frequency

$650+250=875Hz$

Result

2 of 2

$$
875Hz
$$

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