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NCERT Class 9 Science Sound Waves: Characteristics and Applications: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Science book, chapter “Sound Waves: Characteristics and Applications”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   Which observation best supports that sound is a mechanical wave? (i) it reflects (ii) it needs a medium (iii) it has a frequency (iv) it carries energy

   1. The chapter defines mechanical waves as those that require a material medium for propagation.
   2. It explicitly states that sound waves cannot propagate without a medium, making this their defining characteristic.
   3. While sound does reflect, have frequency, and carry energy, these properties are shared by other types of waves like electromagnetic waves.
   4. Therefore, the need for a medium is the specific observation that identifies sound as a mechanical wave.

   Answer: (ii)

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2. Exercises Q2

   Raising the frequency of a sound wave in a medium raises its (i) wavelength (ii) speed (iii) number of compressions per second (iv) time period

   1. Frequency is defined as the number of density oscillations or compressions occurring at a fixed point in one second.
   2. Since frequency counts how many times the wave pattern repeats per second, a higher frequency means more compressions pass a point every second.
   3. The speed of sound in a medium depends on the medium's properties, not the source frequency, so the speed remains constant.
   4. Wavelength and time period are inversely related to frequency, meaning they decrease when frequency increases.

   Answer: (iii)

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3. Exercises Q3

   20 compressions pass a point in 4 s. The frequency is (i) 80 Hz (ii) 5 Hz (iii) 10 Hz (iv) 0.2 Hz

   1. The frequency of a wave is defined as the number of oscillations or compressions passing a point per unit time.
   2. Here, the number of compressions passing the point is 20.
   3. The total time taken for these compressions to pass is 4 seconds.
   4. Using the formula Frequency = Number of compressions ÷ Time, we substitute the values.
   5. Frequency = 20 ÷ 4.
   6. Calculating the division gives 5 Hz.

   Answer: (ii) 5 Hz

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4. Exercises Q4

   In a room, the reflected sound reaches the ear 0.05 s after it is made. Echo or reverberation? Justify.

   1. We hear two sounds separately only if they reach us at least 0.1 s apart; if the gap is less than 0.1 s, we cannot tell them apart.
   2. Here the reflected sound arrives only 0.05 s after the original, which is less than 0.1 s.
   3. So it is not heard as a separate echo; it merges with the original sound and makes it last longer.
   4. Sound persisting because of reflections that arrive too soon to be heard separately is reverberation.

   Answer: reverberation

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5. Exercises Q5

   Two density-distance graphs (Fig. 10.30) on the same scales: (a) has wider, taller waves; (b) narrower, shorter-spaced waves. Which has (i) the greater wavelength (ii) the smaller amplitude?

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) Wavelength is the distance between consecutive crests, so the graph with wider spacing has the greater wavelength.
   2. (i) Graph (a) shows wider, taller waves, meaning its peaks are farther apart than in graph (b).
   3. (ii) Amplitude is the height of the wave from the mean position, so the graph with shorter peaks has the smaller amplitude.
   4. (ii) Graph (b) shows narrower, shorter-spaced waves, meaning its peaks are lower than in graph (a).

   Answer: (i) (a); (ii) (b)

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6. Exercises Q6

   Fig. 10.31 shows three waves (green, red, blue curves). A has the highest frequency and C the lowest: match A, B and C to the curves.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) Frequency is the number of oscillations per second, so a higher frequency means more waves fit in the same time or distance.
   2. (ii) Curve A has the most compressions and rarefactions packed closely together, indicating the shortest wavelength and highest frequency.
   3. (iii) Curve C has the fewest oscillations with the widest spacing between peaks, indicating the longest wavelength and lowest frequency.
   4. (iv) Curve B has an intermediate number of oscillations, placing its frequency between A and C.

   Answer: (i) A is the green curve with the highest frequency; (ii) B is the red curve with medium frequency; (iii) C is the blue curve with the lowest frequency.

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7. Exercises Q7

   Draw a sound wave graph with density amplitude 3 units and wavelength 4 cm.

   1. Draw the x-axis to represent distance in centimetres and the y-axis to represent density.
   2. Mark the average density as a horizontal dashed line on the y-axis.
   3. Set the amplitude to 3 units, so the crest is at +3 and the trough is at -3 relative to the average density.
   4. Set the wavelength to 4 cm, meaning the distance between two consecutive crests or troughs is 4 cm.
   5. Draw a smooth sinusoidal curve starting from the average density, rising to a crest, falling to a trough, and returning to the average density over 4 cm.

   Answer: A sinusoidal graph with density amplitude 3 units and wavelength 4 cm, showing crests above and troughs below the average density line.

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8. Exercises Q8

   A film shows a spacecraft exploding in space with its flash and its bang at the same moment. What is wrong?

   1. Outer space is a near vacuum where there is no medium for sound to travel through.
   2. Sound waves require a medium such as a solid, liquid, or gas to propagate.
   3. Since there is no medium in space, the explosion cannot produce a sound that is heard.
   4. Light travels much faster than sound and does not need a medium, so the flash is visible immediately.
   5. The depiction is wrong because it shows sound occurring simultaneously with the light in a vacuum.

   Answer: The error is that sound cannot propagate in the vacuum of space, so the explosion would be silent while the light flash is visible.

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9. Exercises Q9

   A sound of wavelength 3.44 m travels at 344 m s⁻¹. Find its time period.

   1. Write the formula relating speed, wavelength, and frequency: v = f × λ.
   2. Substitute the given values into the formula: 344 = f × 3.44.
   3. Solve for frequency f: f = 344 ÷ 3.44.
   4. Calculate the frequency: f = 100 Hz.
   5. Write the formula for time period: T = 1/f.
   6. Substitute the frequency into the formula: T = 1/100.
   7. Calculate the time period: T = 0.01 s.

   Answer: 0.01 s

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10. Exercises Q10

    A ship's sonar echo returns after 5 s. Ultrasound travels at 1525 m s⁻¹ in seawater. About how deep is the wreck?

    1. The total time for the echo to travel to the wreck and back is 5 s.
    2. The time taken to travel only to the wreck is half of the total time, which is 5 ÷ 2 = 2.5 s.
    3. The speed of ultrasound in seawater is given as 1525 m s⁻¹.
    4. The depth of the wreck is calculated by multiplying the speed by the one-way time: 1525 × 2.5.
    5. The calculation gives 3812.5 m.

    Answer: 3812.5 m

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11. Exercises Q11

    A car's parking sensor (about 40 kHz) starts beeping 1.2 m from an obstacle. With ultrasound at 345 m s⁻¹ in air, how long does the wave take to reach the obstacle and come back?

    1. The total distance the sound wave travels is twice the distance to the obstacle.
    2. Total distance = 2 × 1.2 m = 2.4 m.
    3. Time is calculated by dividing the total distance by the speed of sound.
    4. Time = 2.4 m ÷ 345 m s⁻¹.
    5. Time ≈ 0.0069565 s.
    6. Rounding to three significant figures gives 0.007 s.

    Answer: 0.007 s

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12. Exercises Q12

    Sound travels at about 331 m s⁻¹ at 0 °C and 344 m s⁻¹ at 22 °C. Roughly how much longer does thunder take to cover 1720 m if the air cools from 22 °C to 0 °C?

    1. Time taken at 22 °C is 1720 ÷ 344 = 5 s.
    2. Time taken at 0 °C is 1720 ÷ 331 s.
    3. The extra time is (1720 ÷ 331) − 5 = (1720 − 5 × 331) ÷ 331 = 65/331 s.
    4. This is approximately 0.196 s.

    Answer: 65/331 s or 0.196 s

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13. Exercises Q13

    A sound wave moves at 340 m s⁻¹; in Fig. 10.32 two full wavelengths span 8 cm. Find its wavelength and frequency.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. Two full wavelengths span a distance of 8 cm.
    2. Convert the distance to metres: 8 cm = 0.08 m.
    3. Since two wavelengths equal 0.08 m, one wavelength is 0.08 m ÷ 2.
    4. The wavelength is 0.04 m.
    5. Use the wave speed formula: speed = frequency × wavelength.
    6. Rearrange the formula to find frequency: frequency = speed ÷ wavelength.
    7. Substitute the values: frequency = 340 m/s ÷ 0.04 m.
    8. The frequency is 8500 Hz.

    Answer: wavelength = 0.04 m; frequency = 8500 Hz

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14. Exercises Q14

    Waves A and B both move at 345 m s⁻¹; in Fig. 10.33 one wavelength of A spans 2.5 cm and one of B spans 5.0 cm. Find each wavelength and frequency.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. Wavelength of wave A is given as 2.5 cm.
    2. Convert 2.5 cm to metres: 2.5 ÷ 100 = 0.025 m.
    3. Wavelength of wave B is given as 5.0 cm.
    4. Convert 5.0 cm to metres: 5.0 ÷ 100 = 0.05 m.
    5. Use the formula v = f × λ to find frequency, so f = v ÷ λ.
    6. Frequency of wave A is 345 ÷ 0.025 = 13800 Hz.
    7. Frequency of wave B is 345 ÷ 0.05 = 6900 Hz.

    Answer: Wavelength of A is 0.025 m and frequency is 13800 Hz; Wavelength of B is 0.05 m and frequency is 6900 Hz.

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15. Exercises Q15

    Identical sources at A (in air) and B (under water) send sound the same horizontal distance to a cliff and back; the sound takes 4.5 times as long to return to A as to B. Find the ratio of the speed of sound in air to that in water.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. Let the horizontal distance to the cliff be d, so the total path length for both sources is 2d.
    2. Let the speed of sound in air be v_air and in water be v_water.
    3. The time taken for sound to travel to the cliff and back in air is t_air = 2d / v_air.
    4. The time taken for sound to travel to the cliff and back in water is t_water = 2d / v_water.
    5. According to the question, the time in air is 4.5 times the time in water, so t_air = 4.5 × t_water.
    6. Substituting the expressions for time gives 2d / v_air = 4.5 × (2d / v_water).
    7. Canceling 2d from both sides gives 1 / v_air = 4.5 / v_water.
    8. Rearranging to find the ratio v_air / v_water gives v_air / v_water = 1 / 4.5.
    9. Converting 4.5 to a fraction gives 4.5 = 9/2, so 1 / 4.5 = 2/9.
    10. Thus, the ratio of the speed of sound in air to that in water is 2:9.

    Answer: 2:9

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