Latihan Soal Fisika-Gelombang

1.    Guitar string guitar strings to resonate with the other when the two strings are the same ... 
a. Voltage. 
b. Length. 
c. Frequency. 
d. Broad cross-section. 

Discussion: 
Resonance is an event participated pulsate an object due to the influence of other objects the same frequency. 

2.    Propagation of sound waves in the air have a pattern ... 
a. Long and short wave. 
b. Density and renggangan 
c. Peak and base 
d. Mountains and valleys 

Discussion: 
The sound is one example of longitudinal waves. The characteristics of longitudinal waves: 
- Waves that have a direction parallel resistance (direction) to the direction of vibration. 
- Consists of renggangan and density. 
- These waves can occur in solid, liquid or gas but not in a vacuum. 
- Not having amplitude. 
Thus, sound waves propagate with renggangan pattern and density. 

3.    An insect prey can hear sound frequencies in the range 15Hz. The sound on these frequencies including ... 
a. Supersonic 
b. Infrasound 
c. Untrasonic 
d. Audiosonic 

Discussion: 
- Supersonic: planes that have a high speed. 
- Infrasound: sounds that have a frequency of less than 20Hz 
- Ultrasound: sound that has a frequency of more than 20.000Hz 
- Audiosonic: sounds that have frequencies between 20Hz to 20.000Hz.sound that can be heard by humans. 

4.    Apollo aircraft noise can not be heard on the moon the astronauts on the moon because of ... 
a. No steam. 
b. Vacuum. 
c. Temperature is too high. 
d. Pressure is very low. 
Discussion: 

Month vacuum so that no particles that deliver sound. Aircraft noise on the moon can not be heard at all the astronauts. 

5.    Pitch a string greater if ... 
a. The length of wire and string enlarged section area stringed minimized. 
b. Voltage and string minimized. 
c. Voltage and wire cross-sectional area is enlarged. 
d. Long strings of string tension is reduced and enlarged. 

Discussion: 

Pitch four factors influenced strings. Pitch / frequency of the larger wire if: 
- The length of wire strings shortened / reduced. 
- Wire cross-sectional area is minimized. 
- Style tense wire string enlarged. 
- The density of strings is minimized.

6.    A source emits with intensity 10^-5 watt/m2. If the intensity threshold value 10^-12 watt/m2, determine the level of intensity of the sound!

Discussion :

To calculate the intensity of a sound level using the following equation :

of the data obtained about :

7.       A child is at a distance of 100 m from a source of sound power 12.56 watts. Determine the level of intensity of the sound is heard the child if Π is 3.14 and the hearing threshold intensity I0 = 10^-12 watt/m2!

Discussion :

To determine the intensity of the sound level using the following equation :

The sound intensity level

 

Sound Intensity

Where P is power in watts, and A is the area in units of m2, assuming the sound spreads evenly to form an area in the form of a ball so that A = 4 Πr2, where r is the radius of the ball or equal to the distance of the child from the sound source.

So:

 

8.       A firebomb exploded at a distance of 20 meters from a child.

If the child is heard a blast with the intensity level of 120 dB, determine the level of intensity is heard another child who was at a distance of 180 m from the first child!

Discussion :

To determine the intensity level of a sound source heard from two different places used the following equation:

 

r1 is the distance to the first child of the sound source (20 m) and r2 is the distance of the second of the sound source (180 +20 = 200 m)

9.       A factory has 100 identical machines. If a machine has a sound intensity level of 70 dB, determine the value of the intensity level of the sound is heard when all the machines in the factory turned on simultaneously!

Discussion :

To determine the intensity of the sound by Taraf many identical sources using the following equation:

 where TI1 is IT a source of sound and n is the number of sound sources. So that :

10.   Determine the value of comparative intensity of a sound source from two places, namely A, which is 4 m from the source and from B within 9 m from the source!

Discussion :

The intensity of a sound source from two different places:

11. A spectator at the race car heard the sound (roar of the cars) were different, when a car approaching and away. Average race car emits 800 Hz. If the speed of sound in air 340 ms-1 speed and 20 ms-1, the frequency of which is heard when the car is approaching ....

A. 805 Hz

B. 810 Hz

C. 815 Hz

D. 850 Hz

E. 875 Hz

Discussion :

The application of the Doppler effect, the listener in a stationary position means Vp = ZERO, sources close to the listener is a sign for negative Vs.

12. A 1 kHz sound source moves directly toward the listener a break at a pace of 0.9 times the speed of sound. Frequency in kHz is received ....

A. 10.0

B. 1.9

C. 1.1

D. 0.5

E. 0.1

Discussion :

13. Piano string length of 0.5 m and a mass of 10^-2 kg stretched 200 N, then the basic piano tone is frequency ....

A. 100 Hz

B. 200 Hz

C. 400 Hz

D. 600 Hz

E. 800 Hz

Discussion :

Wave velocity in the string is:

Base notes on the strings occurs when:

Frequency string:

Formula:

14. An open organ pipe produces tones at a frequency of 150 Hz. Determine the frequency of the second tone of the pipe organ!

Discussion :

Comparison of frequency tones, the first tone, the tone on the second and so on of a closed organ pipe is:

F0: F1: F2: F3: ..... = 1: 3: 5: 7: .....

If you take the comparison between f2 and f0 obtained:

15. Rows of railroad car that is pulled by a locomotive leaving the station Tanjung Karang moving at a pace of 36 km / h. At that time, an officer at the station blew the whistle with a frequency 1700 Hz. If the speed of propagation of sound waves in the air 340 ms-1, find the frequency whistle is heard by an observer in the train.

Discussion:

Given: vp = 36 km / hr = 10ms^-1 ; vs = 340 ms^-1; fs = 1700 Hz

So the frequency of the whistle is heard by an observer on the train at 1650 Hz.

16. The noise of a typewriter equal to 70 dB. How dB noise due to 100 pieces an office typewriter?

Discussion:

Consider the intensity of the typewriter = I1

the intensity of 100 typewriters I1 = I2 = 100,

Find additional noise due to 100 typewriter:

∆TI = 10 log I2/I1 = 10 log 100 = 20 dB I1/I1

So noise is 100 typewriters: TI1 + TI2 = ∆TI = 90 Db

17. Sound waves from a source having the propagation of 340 m / s. If the frequency of sound waves is 500 Hz, determine the wavelength!

Discussion :

Data about:

ν = 340 m / s

f = 500 Hz

λ = ?

Relations wavelength, wave propagation and frequency:

λ = ν / f

λ = 340/500

λ = 0.68 m

18. A child hear the sound that has a wavelength of 5 meters. If the propagation of sound in air is 340 m / s, find:

a) the frequency of the sound source

b) the period of the sound source

Discussion :

Data about:

ν = 340 m / s

λ = 5 m

f = .......... Hz

Relations wavelength, wave propagation and frequency:

f = ν / λ

f = 340/5

f = 68 Hz

19. A ship of a sea depth gauge using the sound device. Beep fired into the bottom waters and 5 seconds later reflected sound arriving back at the ship. If the propagation of sound in water is 1500 m / s, determine the depth of the waters!

Discussion :

Determine the distance of two points (depth) with the reflection of sound:

S = (ν x t) / 2

S = (1500 x 5) / 2

S = 3750 meters

20. When the weather was cloudy a child heard the thunder of 1.5 seconds after the visible lightning. If the propagation of sound in air is 320 m / s, determine the distance of the source of the child's thunder!

Discussion :

Determine the distance of two points with no sound reflections:

S = ν x t

S = 320 x 1.5

S = 480 m

 21. Sound waves with a frequency of 5 kHz propagating in air with a temperature of 30 ° C. If the propagation of sound in air at 0 ° C is 330 m / s, find:

a) the propagation of sound

b) the length of the sound wave

Discussion :

The difference in the propagation of sound due to the differences / changes in air temperature:

ν = ν0 + 0.6 t

ν = 330 + (0.6 x 30)

ν = 348 m / s

22. Sound with a wavelength of 1.5 m has a propagation speed of 330 m / s. Can sound be heard by the human ear normal?

Discussion :

Looking frequency first:

f = ν / λ

f = 330 / 1.5

f = 220 Hz

Sounds with frequencies between 20 and 20,000 Hz audiosonik considered, can be heard by humans.

Learn more:

infrasound: sound frequencies less than 20 Hz

ultrasonic: sound frequencies greater than 20000 Hz

23. A first resonance occurs when the air column tube tube length of 15 cm. Define:

a) the wavelength of the sound

b) the length of the air column during the second resonance

c) the length of the air column during the third resonasi

d) the length of the air column resonance during the fourth

e) the frequency of the sound, if the propagation of sound is 340 m / s

Discussion  :

a) the wavelength of the sound

The first resonance → L = (1/4) x λ

15 = (1/4) x λ

λ = 4 x 15

λ = 60 cm

b) the length of the air column during the second resonance

The second resonance → L = (3/4) x λ

L = (3/4) x 60 cm

L = 45 cm

c) the length of the air column during the third resonasi

The third resonance → L = (5/4) x λ

L = (5/4) x 60 cm

L = 75 cm

d) the length of the air column resonance during the fourth

Fourth resonance → L = (7/4) x λ

L = (7/4) x 60 cm

L = 105 cm

e) the frequency of the sound, if the propagation of sound is 339 m / s

λ = 60 cm = 0.6 meters

ν = 339 m / s

f = ?

f = ν / λ

f = 339 / 0.6

f = 565 Hz

24. Determine the ratio of the frequency of which is owned by a string A string length of 100 cm and a length of 50 cm B if the two strings are made of the same material

Discussion :

fA / fB = LB / LA

fA / fB = 50/100

fA: fB = 1: 2

25. Strings I and II strings are the same length. If the cross-section area stringed I was three times the size of the string section II, specify:

a) comparison of the frequency of the strings I and II strings, consider the strings have the same voltage

b) the frequency of the strings II if the frequency of the strings I was 500 Hz

Discussion:

a) f1 / f2 = A2 / A1

f1 / f2 = A2 / 3A2

f1: f2 = 1: 3

b) f1: f2 = 1: 3

f2 = 3 x f1

f2 = 3 x 500

f2 = 1500 Hz