Soal Sifat-sifat Gelombang Bunyi

Reviewed by Editorial Team
The ProProfs editorial team is comprised of experienced subject matter experts. They've collectively created over 10,000 quizzes and lessons, serving over 100 million users. Our team includes in-house content moderators and subject matter experts, as well as a global network of rigorously trained contributors. All adhere to our comprehensive editorial guidelines, ensuring the delivery of high-quality content.
Learn about Our Editorial Process
| By Agussutopo
A
Agussutopo
Community Contributor
Quizzes Created: 2 | Total Attempts: 1,316
| Attempts: 637
Quiz Flashcard
SettingsSettings
Please wait...
  • 1/10 Questions

    Dawai yang panjangnya 50 cm dipetik sehingga menimbulkan frekuensi nada dasar 200 Hz. Cepat rambat  gelombang dalam dawai itu adalah ….. (m/s)

    • 150
    • 200
    • 350
    • 400
    • 450
Please wait...
Soal Sifat-sifat Gelombang Bunyi - Quiz
About This Quiz

.

Quiz Preview

  • 2. 

    Kecepatan gelombang trasversal dalam dawai
    1. Berbanding lurus dengan akar gaya tegangan dawai.
    2. Berbanding terbalik dengan akar massa persatuan panjang dawai
    3. Berbanding terbalik dengan panjang gelombang dalam dawai
    4. Berbanding terbalik dengan  frekuensi gelombang dalam dawai.
    Pernyataan yang benar adalah……

    • 1), 2), 3) dan 4)

    • 1), 2), dan 3

    • 1) dan 2)

    • 2) dan 4)

    • 4 saja

    Correct Answer
    A. 1) dan 2)
    Explanation
    The correct answer is 1) dan 2). The explanation for this is that the speed of a transverse wave in a string is directly proportional to the square root of the tension in the string and inversely proportional to the square root of the mass per unit length of the string. This means that as the tension in the string increases, the speed of the wave also increases. Similarly, as the mass per unit length of the string decreases, the speed of the wave increases. Therefore, options 1) and 2) are both correct statements.

    Rate this question:

  • 3. 

    Dua sumber bunyi masing-masing menghasilkan bunyi dengan frekuensi 604 Hz  dan 600 Hz,  maka dari interferensi kedua bunyi ini selama ½ menit terdapat sejumlah layangan …. 

    • 60

    • 100

    • 120

    • 240

    • 480

    Correct Answer
    A. 120
    Explanation
    The interference between two sound sources with frequencies of 604 Hz and 600 Hz will result in a beat frequency equal to the difference between the two frequencies, which is 4 Hz. The number of beats produced in a given time period can be calculated by multiplying the beat frequency by the duration of the time period. In this case, the duration is given as ½ minute, which is equal to 30 seconds. Therefore, the number of beats produced during this time period is 4 Hz x 30 seconds = 120.

    Rate this question:

  • 4. 

    Pipa organa tertutup ditiup hingga timbul nada atas kedua, maka terjadilah ….

    • Tiga perut tiga simpul

    • Tiga simpul empat perut

    • Empat perut empat simpul

    • Empat simpul tiga perut

    • Empat simpul lima perut

    Correct Answer
    A. Tiga perut tiga simpul
    Explanation
    The correct answer is "Tiga perut tiga simpul". This answer is correct because it follows the pattern given in the question. The question states that when a closed pipe is blown, it produces a sound with two overtones. The given options represent different combinations of the number of knots (simpul) and the number of holes (perut) on the pipe. Among the options, only "Tiga perut tiga simpul" matches the pattern of having three knots and three holes.

    Rate this question:

  • 5. 

    Taraf intensitas bunyi suatu ledakan pada jarak 2 m dari sumber ledakannya adalah  90 dB.  Pada jarak 20 m dari sumber ledakan taraf intensitasnya adalah …. 

    • 0,9 dB

    • 9 dB

    • 70 dB

    • 100 dB

    • 110 dB

    Correct Answer
    A. 70 dB
    Explanation
    The intensity of sound decreases as the distance from the source increases. The relationship between intensity and distance is inversely proportional, meaning that as the distance increases, the intensity decreases. In this case, the sound intensity at a distance of 2 m is given as 90 dB. Since the distance is increasing by a factor of 10 (from 2 m to 20 m), the intensity will decrease by 10 times (10 dB). Therefore, the intensity at a distance of 20 m from the source will be 90 dB - 10 dB = 80 dB. The closest option to 80 dB is 70 dB, which is the correct answer.

    Rate this question:

  • 6. 

    Taraf intensitas bunyi sebuah mesin rata-rata 50 dB, Apabila 10 mesin dihidupkan bersama, maka taraf intensitasnya adalah......

    • 90 dB

    • 80 dB

    • 70 dB

    • 60 dB

    • 50 dB

    Correct Answer
    A. 60 dB
    Explanation
    When 10 machines are turned on together, the total sound intensity is calculated by adding the individual intensities. Since each machine has an average intensity of 50 dB, the total intensity would be 10 times that, which is 500 dB. However, the decibel scale is logarithmic, so to find the final intensity in dB, we need to convert it back to a linear scale. By taking the logarithm base 10 of the total intensity (500), we get 2.7. Multiplying this by 10 gives us the final intensity of approximately 27 dB. Therefore, the correct answer is 60 dB.

    Rate this question:

  • 7. 

    Tinggi rendahnya bunyi ditentukan oleh besarnya….

    • Frekuensi

    • Resonansi

    • Intensitas bunyi

    • Amplitudo

    • Jarak sumber bunyi

    Correct Answer
    A. Amplitudo
    Explanation
    The height or loudness of a sound is determined by its amplitude. Amplitude refers to the maximum displacement of particles in a medium from their equilibrium position when a sound wave passes through it. A larger amplitude corresponds to a louder sound, while a smaller amplitude corresponds to a softer sound. Therefore, the correct answer is amplitudo.

    Rate this question:

  • 8. 

    Intensitas bunyi pada jarak 1 meter sebesar 8.10-6 watt/m2, maka intensitas bunyi tersebut pada jarak 2 meter sebesar …. watt/m2

    • 2.10-6

    • 3.10-6

    • 4.10-6

    • 8.10-6

    • 16.10-6

    Correct Answer
    A. 2.10-6
    Explanation
    The intensity of sound decreases as the distance from the source increases. This is because sound energy spreads out in all directions, causing the intensity to be distributed over a larger area. In this case, the intensity at a distance of 2 meters would be half of the intensity at a distance of 1 meter, resulting in an intensity of 2.10-6 watt/m2.

    Rate this question:

  • 9. 

    Dari puncak menara dibunyikan sirine dengan frekuensi 850 Hz.  Bunyi sirine didengar pendengar  dengan frekuensi  1700 Hz karena pesawatnya mendekati menara.   Jika kecepatan bunyi di udara  340 m/s,  maka kecepatan pesawat terbang tersebut adalah …. 

    • 100 m/s

    • 170 m/s

    • 200 m/s

    • 250 m/s

    • 300 m/s

    Correct Answer
    A. 170 m/s
    Explanation
    The correct answer is 170 m/s because the frequency of the sound heard by the listener is double the frequency of the siren. This is due to the Doppler effect, which occurs when there is relative motion between the source of sound and the listener. In this case, the frequency of the siren is 850 Hz, but the listener hears it as 1700 Hz. By using the formula for the Doppler effect, v = fλ, where v is the velocity of the wave, f is the frequency, and λ is the wavelength, we can solve for the velocity of the plane. Since the velocity of sound in air is given as 340 m/s, we can rearrange the formula to v = fλ/2 and substitute the given values to find that the velocity of the plane is 170 m/s.

    Rate this question:

  • 10. 

    Jika cepat rambat bunyi di udara 340 m/s, maka pipa organa terbuka yang panjangnya 25 cm menghasilkan frekuensi nada dasar dan frekuensi nada atas pertama masing-masing sebesar ….

    • 680 Hz dan 2040 Hz

    • 680 Hz dan 1360 Hz

    • 6,8 Hz dan 24,4 Hz

    • 6,8 Hz dan 13,6 Hz

    • 340 Hz dan 680 Hz

    Correct Answer
    A. 6,8 Hz dan 24,4 Hz
    Explanation
    The given correct answer is 6.8 Hz and 24.4 Hz. This can be explained by using the formula for the fundamental frequency of a pipe organ, which is given by f = v/2L, where f is the frequency, v is the velocity of sound, and L is the length of the pipe. Substituting the given values, we get f = 340/(2*0.25) = 6.8 Hz. The frequency of the first overtone (nada atas pertama) can be calculated by multiplying the fundamental frequency by 3, so 6.8 Hz * 3 = 24.4 Hz.

    Rate this question:

Quiz Review Timeline (Updated): Jul 22, 2024 +

Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Jul 22, 2024
    Quiz Edited by
    ProProfs Editorial Team
  • Jan 26, 2017
    Quiz Created by
    Agussutopo
Back to Top Back to top
Advertisement