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A group of boys hear a guitar playing. The sound reaches the first boy after 3 seconds and the second boy after 4 seconds. The boys notice that the colour of the audience's clothes change with the fla

Physics

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Answer

d

0.75 m0.75 \text{ m}

Step-by-step solution

3 steps

Here's the explanation for the boys:

Part a

**Why they heard the sound at different intervals.

The boys heard the sound at different intervals because they were standing at different distances from the guitar player on the floating stage. Sound travels at a finite speed (330 ms−1330 \text{ ms}^{-1}). Boy X, hearing the sound after 3 seconds, was 330 ms−1×3 s=990 m330 \text{ ms}^{-1} \times 3 \text{ s} = 990 \text{ m} away. Boy Y, hearing the sound after 4 seconds, was 330 ms−1×4 s=1320 m330 \text{ ms}^{-1} \times 4 \text{ s} = 1320 \text{ m} away. Since Boy Y was further away, the sound took longer to reach him.

Part b

**Why the colour of the clothes of the audience kept on changing when coloured lights flashed on them.

The apparent color of an object depends on the colors of light it reflects and the colors of light illuminating it. The disco lights were flashing red, blue, and green. When a specific colored light (e.g., red) shines on clothes, the clothes will only reflect the red light if they contain pigments that can reflect red. If they absorb red light, they will appear dark. As the illuminating colored lights change, the wavelengths of light reflected by the clothes also change, causing their perceived color to change.

Part c

**Why the laser source of light preferred to provide laser light that enhanced visibility late in the late hours of the night.

Laser light is preferred for enhancing visibility because it is highly collimated. This means the light rays are nearly parallel and do not spread out significantly over long distances. This property allows the laser light to remain intense and focused, providing clear and concentrated illumination over large areas or long ranges, which is crucial for visibility in the dark.

Part d

0.75 m0.75 \text{ m}

**Compare the wave length of the sound waves and laser light waves in air medium.

  1. Calculate the wavelength of the sound waves.

    The formula for wavelength is λ=vf\lambda = \frac{v}{f}.
    Given:
    Speed of sound in air (vsoundv_{\text{sound}}) = 330 ms−1330 \text{ ms}^{-1}
    Frequency of sound (fsoundf_{\text{sound}}) = 440 Hz440 \text{ Hz}

    λsound=330 ms−1440 Hz\lambda_{\text{sound}} = \frac{330 \text{ ms}^{-1}}{440 \text{ Hz}} λsound=0.75 m\lambda_{\text{sound}} = \boxed{0.75 \text{ m}}

  2. Calculate the wavelength of the laser light waves.

    Given:
    Speed of light in air (vlightv_{\text{light}}) = 3.0×108 ms−13.0 \times 10^8 \text{ ms}^{-1}
    Frequency of laser light (flaserf_{\text{laser}}) = 4.7×109 MHz4.7 \times 10^9 \text{ MHz}

    First, convert the frequency from MHz to Hz:

    flaser=4.7×109 MHz×106 Hz1 MHz=4.7×1015 Hzf_{\text{laser}} = 4.7 \times 10^9 \text{ MHz} \times \frac{10^6 \text{ Hz}}{1 \text{ MHz}} = 4.7 \times 10^{15} \text{ Hz}

    Now, calculate the wavelength:

    λlaser=3.0×108 ms−14.7×1015 Hz\lambda_{\text{laser}} = \frac{3.0 \times 10^8 \text{ ms}^{-1}}{4.7 \times 10^{15} \text{ Hz}} λlaser≈6.38×10−8 m\lambda_{\text{laser}} \approx 6.38 \times 10^{-8} \text{ m}

  3. Compare the wavelengths.

    The wavelength of sound waves is 0.75 m0.75 \text{ m}.
    The wavelength of laser light waves is approximately 6.38×10−8 m6.38 \times 10^{-8} \text{ m}.
    The wavelength of the sound waves is significantly larger than the wavelength of the laser light waves. Specifically, the sound wavelength is approximately 0.75 m/(6.38×10−8 m)≈1.17×1070.75 \text{ m} / (6.38 \times 10^{-8} \text{ m}) \approx 1.17 \times 10^7 times larger than the laser light wavelength.

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