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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 501: Lesson Check

Exercise 10

Step 1

1 of 2

When an oscillating or vibrating object creates alternating regions of compressed and expanded air, sound wave is produced. These regions goes away from its source as longitudinal waves.

Result

2 of 2

Click here to see the explanation.

Exercise 11

Step 1

1 of 2

The pitch of the sound is dependent on the frequency of its sound wave. Therefore, 600 Hz sound is expected to have higher pitch than the 400 Hz sound.

Result

2 of 2

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Exercise 12

Step 1

1 of 2

When two similar frequencies are played simultaneously the sound waves interfere.

The manner of this interference changes with time, in some instants it is constructive (resulting in louder sound), and in some instants it is destructive (resulting in softer sound)

This varying from larger to softer sound is periodical, and can be heard by the listener. These changes in loudness are referred to as beats.

Result

2 of 2

Beats are formed by the interfering of two frequencies that are played simultaneously. They can be heard as a periodical change in loudness of the resulting sound.

Exercise 13

Step 1

1 of 2

The speed of a sound wave is highly dependent on the properties of the medium where it travels, and not on the frequency of the wave. Hence, the velocity of sound will remain constant when the frequency is doubled.

Result

2 of 2

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Exercise 14

Step 1

1 of 2

The speed of a wave is defined as the product of frequency and wavelength. However, the wave speed is constant for all frequencies. Hence, the wavelength of the sound wave must decreased to half if the frequency is doubled so that the speed remains constant.

Result

2 of 2

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Exercise 15

Step 1

1 of 2

Sound waves are energy carrier. Its properties can be characterized by wavelength, frequency, and speed. The sound wave and the wave on the string are similar in this manner. However, the difference lies on its mode of propagation. Sound waves are longitudinal while waves on strings are tranverse.

Result

2 of 2

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Exercise 16

Step 1

1 of 2

The sound wave must travel to the click and back.

Hence, multiply the distance by 2 and divide by the speed of sound to solve for the time.

$$
t=dfrac{d}{v}=dfrac{2(155m)}{343;m/s}=0.904;s
$$

Result

2 of 2

$$
=0.904;s
$$

Exercise 17

Step 1

1 of 2

Beat frequency is the absolute value of the difference of two frequencies.

$$
f_{beat}=|f_1-f_2|=|278; Hz-292;Hz|=14;Hz
$$

Result

2 of 2

14 Hz

Exercise 18

Step 1

1 of 2

The beat frequency is the absolute value of the difference of the two frequencies.

Hence, the beat frequencies are:

$f_A=|149-145| = 4;Hz$

$f_B=|12-22| = 10;Hz$

$f_C=|901-900|=1;Hz$

$f_D=|332-228|=6;Hz$

Therefore,

The correct arrangement is

$$
C<A<D<B
$$

Result

2 of 2

$$
C<A<D<B
$$

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