Examples of Sound and Light Wave Problems and their Discussion
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Example Problems of Sound and Light Waves [+Pembahasan] – Light Wave is electromagnetic wave radiation that can be detected by the human eye and felt its effect.
Characteristics of Light Waves:
- Dispersion – The event of the decomposition of white light (polychromatic) into its components due to refraction.
- Light Interference – Interference is the blending of two or more waves into one new wave.
- Light Diffraction – the event of scattering or bending of waves by a narrow slit as an obstacle.
- Polarization of Light – the process of limiting vector waves that form a transverse wave so that it becomes one direction.
sound wave Longitudinal waves are waves that consist of particles that oscillate in the direction of the wave forming high and low pressure areas.
There are 3 conditions for sound to occur, namely:
- sound source
- medium and
1 – 10 Examples of Sound and Light Wave Problems
1. Determine the speed of sound in air at 17o! Match it if it is considered that sound propagates in air at a temperature of 150C is 340 m/s! What’s your conclusion? (𝛾 = 1.4; R = 8.31 /𝑚𝑜𝑙 ; = 29 × 103 /𝑚𝑜𝑙)
𝑇1 = 17𝑂= 290
𝑇2 = 15𝑂= 288
R = 8.31 /𝑚𝑜𝑙
= 29 × 103 /𝑚𝑜𝑙
Wanted: = 17𝑂= ?
So, the speed of sound in air at 170C is 341.1 m/s, while at a temperature of 150C of 339.9 m/s
2. Train A with a speed of 20 m/s and train B with a speed of 60 m/s are moving toward each other. The speed of sound in air is 340 m/s. If the train driver A whistles with a frequency of 1500 Hz, what is the frequency of the whistle heard by the train driver B?
𝑣𝑠 = 20 /𝑠
𝑣𝑝 = 60 /𝑠
= 340 /𝑠
𝑓𝑠 = 1500
Asked:𝑝 = ?
So, the frequency of the whistle heard by the train driver B is 1875 Hz.
3. A string 3 m long is tied at both ends. The resonant frequency of the first overtone of the string is 80 vibrations/second. If the mass per unit length of the string is 0.01 g/cm, what is the tension in the wire?
𝑓1 = 80
= 0.01 /𝑐𝑚 = 103/𝑚
Asked: = ?
So, the magnitude of the tension in the wire is 57.6 N
4. The length of a closed organ pipe is 4 m. If the speed of sound in air at 15 suhu0C is 340 m/s, what is the frequency of the first overtone?
L = 4 m
= 340 /𝑠
Asked:1 = ?
So, the frequency of the first overtone is 63.75 Hz.
5. A sound source transmits sound with an output power of 0.16𝜋 watts. Assuming the sound wavefront is spherical, determine: (𝐼0 = 1012 /𝑚2 )
- (a) Sound intensity at a distance of 2 m from the source
- (b) The level of sound intensity at a distance of 2 m from the source
𝐼0 = 1012 /𝑚2
- a) at 2𝑚 =…?
- b) at 2𝑚 =…?
So, at a distance of 2 m from the source the sound intensity is 0.01 /𝑚2 and the sound intensity level is 100 dB.
See also: Examples of Mechanical Wave Problems
6. Monochromatic light falling on two slits 0.06 mm apart produces a fourth bright band at an angle of 8o . What wavelength of light is used?
= 0.06 = 0.06 × 103𝑚
Asked: = ?
So, the wavelength of light used is 2.1 × 106 m
7. Light with a wavelength of 600 nm passes through a slit 2.5 × 103 mm. Determine the third dark band diffraction angle (𝜃3)!
= 600 = 6 × 104𝑚𝑚
= 2.5 × 103 𝑚𝑚
Asked:3 = ?
So, the third dark band diffraction angle is 460
8. A sound source has an intensity level of 60 dB. When 100 sources of the same sound sound simultaneously, the resulting intensity level is…
Use the equation:
TIn = TI1 + 10 log n
TI100 = 60 dB + 10 log 100
TI100 = 60 dB + 10 . 2 dB
TI 100 = 80 dB
9. The intensity level of one bee buzzing is 10 dB. If the buzzing sound of each bee is considered identical and the intensity of the human hearing threshold is 10−12 Wm2, then the intensity of the buzzing sound of 1000 bees is….
Count in advance TI 1000 bees
TI = TI 1 + 10 log n = 10 dB + 10 log 1000 = 40 dB
Counting I 1000 bees
TI = 10 logs (I / Io)
40 dB = 10 logs (I / 10-12)
log(I / 10-12) = 4
I / 10-12 = 104
I = 104 . 10-12 = 10-8 Wb
10. The intensity level of a sound source at a distance of 9 m from the observer is 50 dB. If nine identical sound sources are added together, the total intensity level for the observer becomes….
TI = TI 1 + 10 log n = 50 dB + 10 log 10 = 60 dB
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