Sound Waves Characteristics and Applications — NCERT Solutions

You can hear the sound of the knock or scratch through the air. This shows that sound is produced by a vibrating or disturbed object and that the sound reaches your ear through the surrounding air.

Yes, the sound is heard more clearly through the table. This shows that sound can travel through solids and is usually heard better when the ear is placed against the desk.

Yes, the sound is still heard. This shows that sound can travel through liquids as well as through air.

The chapter says the approximate frequency for the notes Sa, Re, Ga, Ma, Pa, Dha, Ni, Sa increases gradually. A reasonable set of approximate values is:

• Sa ≈ 256 Hz
• Re ≈ 288 Hz
• Ga ≈ 320 Hz
• Ma ≈ 341 Hz
• Pa ≈ 384 Hz
• Dha ≈ 426 Hz
• Ni ≈ 480 Hz
• Sa ≈ 512 Hz

The exact values may vary slightly depending on the app or voice used.

Taking each frequency with respect to Sa, the ratios are approximately in a simple increasing pattern. So, the notes have higher and higher frequencies from Sa to Sa again, and the ratios remain roughly consistent for the same musical scale. This shows that each note has a distinct frequency, and the frequencies increase gradually.

If both voice and mobile-generated notes are used, the same musical notes should have nearly the same frequencies. Small differences may occur because of the singer’s voice, the app, and the quality of sound production, but the frequency of the same note should be nearly equal.

As the frequency increases from 100 Hz to 1000 Hz, the sound becomes shriller or higher pitched. Lower frequencies sound deeper, while higher frequencies sound more sharp or shrill.

A mechanical wave needs a material medium to travel. Sound cannot travel in vacuum, so the best supporting observation is (ii) Sound needs a medium to propagate.

For a sound wave in a given medium, the speed remains constant. If frequency increases, the number of vibrations or compressions per second increases, while wavelength and time period decrease. So the correct option is (iii) number of compressions per second.

Frequency = number of compressions per second.

Given:

• Number of compressions = 20
• Time = 4 s

So,
$$\nu = \frac{20}{4} = 5\,\text{Hz}$$
Therefore, the correct option is (ii) 5 Hz.

It will produce reverberation, not a clear echo. A reflected sound must reach the ear after at least 0.1 s to be heard separately as an echo. Since the reflected sound reaches after 0.05 s, the time gap is too small, so the sounds merge together and produce reverberation.

From the two graphs, the wave with the greater wavelength is the one whose crests are farther apart, and the wave with the smaller amplitude is the one with the smaller height from the mean line. In Fig. 10.30, wave (a) has the greater wavelength, and wave (b) has the smaller amplitude.

Frequency and wavelength are inversely related for the same speed.

• A has the maximum frequency, so it has the smallest wavelength.
• C has the minimum frequency, so it has the largest wavelength.
• B lies in between.

So, on the curves in Fig. 10.31, mark A on the curve with the most cycles, B on the middle one, and C on the curve with the fewest cycles.

Draw a wave graph with:

• density amplitude = 3 units above and below the mean line
• wavelength = 4 cm

That means the graph should have successive crests (or compressions) separated by 4 cm, and the maximum and minimum deviations from the mean density should be 3 units.

The depiction is wrong in two ways:

1. In space there is nearly vacuum, so sound cannot travel there.
2. Light from the explosion can be seen almost instantly, but sound would not be heard directly in space.

So a spacecraft explosion in space should show the flash of light, but not sound at the same time.

Use
$$T = \frac{\lambda}{v}$$
Given:

• $\lambda = 3.44\,\text{m}$
• $v = 344\,\text{m s}^{-1}$

So,
$$T = \frac{3.44}{344} = 0.01\,\text{s}$$
So the time period is 0.01 s.

The sonar signal goes to the object and comes back, so the given time is for the round trip.

Time to reach the wreckage:
$$\frac{5}{2} = 2.5\,\text{s}$$
Distance:
$$\text{distance} = v \times t = 1525 \times 2.5 = 3812.5\,\text{m}$$
But this is the round-trip-based distance? No, this calculation already uses the one-way time, so the depth is:
$$3812.5\,\text{m}$$
However, the chapter examples show that for echo/sonar, distance to the object is half the round-trip distance. The correct computation is:
$$\text{depth} = \frac{1525 \times 5}{2} = 3812.5\,\text{m}$$
So the wreckage is approximately 3812.5 m below the surface.

The ultrasonic wave travels to the obstacle and back, so total distance is
$$2 \times 1.2 = 2.4\,\text{m}$$
Using
$$\text{time} = \frac{\text{distance}}{\text{speed}}$$
$$t = \frac{2.4}{345} = 0.006956...\,\text{s}$$
So the time taken is about 0.00696 s.

Time taken at $22^{\circ}\mathrm{C}$:
$$t_1 = \frac{1720}{344} = 5\,\text{s}$$
Time taken at $0^{\circ}\mathrm{C}$:
$$t_2 = \frac{1720}{331} \approx 5.20\,\text{s}$$
Extra time:
$$t_2 - t_1 \approx 5.20 - 5.00 = 0.20\,\text{s}$$
More accurately, it is about 0.2 s.

So the thunder takes about 0.2 s extra to travel the distance when the air temperature changes from $22^{\circ}\mathrm{C}$ to $0^{\circ}\mathrm{C}$.

From the graph in Fig. 10.32, the wavelength is 50 m.

Using
$$v = \lambda \nu$$
where speed $v = 340\,\text{m s}^{-1}$,

$$\nu = \frac{v}{\lambda} = \frac{340}{50} = 6.8\,\text{Hz}$$
So the wavelength is 50 m and the frequency is 6.8 Hz.

From Fig. 10.33, read the wavelengths of the two waves and then use
$$v = \lambda \nu \quad \Rightarrow \quad \nu = \frac{v}{\lambda}$$
for each wave, with speed $345\,\text{m s}^{-1}$. The wave with the shorter wavelength has the higher frequency, and the wave with the longer wavelength has the lower frequency.

For sound going to the cliff and coming back, the distance is the same for both sources, so
$$t \propto \frac{1}{v}$$
Given that time taken by sound to return to A is 4.5 times that of B:
$$\frac{t_A}{t_B} = 4.5 = \frac{v_B}{v_A}$$
So,
$$\frac{v_A}{v_B} = \frac{1}{4.5}$$
Thus, the ratio of speeds of sound in air : water is
$$1 : 4.5$$

Sound is a mechanical wave, so it requires a material medium such as a solid, liquid, or gas to travel. The chapter states that sound cannot propagate in vacuum, which is why this observation best supports the idea that sound is a mechanical wave.

The frequency of a sound wave is the number of density oscillations at a fixed point per unit time. Since compressions are part of these oscillations, increasing frequency means more compressions pass a point each second. So the correct choice is number of compressions per second.

Frequency is the number of compressions per unit time.

$$\nu = \frac{\text{number of compressions}}{\text{time}} = \frac{20}{4\,\text{s}} = 5\,\text{Hz}$$

So the frequency is 5 Hz.

Curved ceilings and walls act like sound reflectors that help direct sound waves toward the audience instead of letting them spread out too much or die away quickly. This improves sound distribution and makes speech or music clearer and more uniform in different parts of the hall. In contrast, flat surfaces may cause poor reflection patterns, dead spots, or unwanted echoes. Curved surfaces can also help focus sound more evenly, reducing distortion and improving overall acoustics.

From the chapter, the speed of sound depends on the medium, and in air it also depends on temperature and humidity.

• As temperature increases, the speed of sound increases.
• As humidity increases, the speed of sound also increases.

This is because warmer air and more humid air allow sound waves to travel faster. The chapter gives examples showing that in dry air the speed is about 331 m s$^{-1}$ at $0^\circ$C and nearly 344 m s$^{-1}$ at $22^\circ$C.

So, while exploring the resources, you would observe that:

1. Higher temperature → higher speed of sound
2. Higher humidity → higher speed of sound

These changes are usually small but measurable.