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

28 questions · 16 still being checked

Pause and Ponder 10.1–10.13 (part 1 of 3)

  1. Exercise 10.1

    Explore various ways of producing sound.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Sound is produced whenever something vibrates — every method below is just a different way of setting an object vibrating.
    NCERT_Solution_Class9_Science_Ch10_PP_Q10-1
    Plucking a stretched band or string — stretch a rubber band across an open cardboard box and pluck it (Activity $\displaystyle 10.1$); the band is seen vibrating, and the moment the vibration stops, so does the sound.
    Striking a metal object — a taal or any sonorous metal rings when struck.
    Blowing across an air column — in a bansuri the air inside the hollow pipe vibrates.
    Striking a tuning fork on a rubber pad — the prongs vibrate; touch a prong to water and ripples spread out, proving the vibration (Activity $\displaystyle 10.2$).
    Using the vocal cords — talking or singing vibrates the stretched flaps in the larynx; touch your throat while you speak and feel it.
    Rubbing body parts — grasshoppers and crickets rub their wings or legs together.
    Knocking, scratching or tapping — knock a desk (Activity $\displaystyle 10.3$), or tap two spoons together in air and again under water (Activity $\displaystyle 10.4$).
    Rapid expansion of heated gas — a firecracker or a clap of thunder heats gas so fast that the sudden density disturbance is heard as a loud pulse.
    Common thread: in every case the vibrating object is the source, and the sound ends when the vibration ends.

  2. Exercise 10.2

    Make a list of different types of musical instruments and identify their vibrating parts which produce sound.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Every instrument sounds because some part of it vibrates — the instruments named in this chapter, with that part, are:
    NCERT_Solution_Class9_Science_Ch10_PP_Q10-2
    Bansuri (flute) — the air column inside the hollow pipe, set vibrating by blowing.
    Tabla and mridangam — the stretched membrane of the drum head, whose vibration is shaped by the black syaahi patch at its centre.
    Sitar, sarangi, veena, tanpura, ektara — the stretched strings; the sitar, sarangi and veena add extra strings to enrich the sound.
    Taal — the metal discs themselves, which ring when struck because metals are sonorous.
    Tuning fork — the two prongs (tines) of the U, struck against a rubber pad.
    Rubber band stretched on an open box — the band, the simplest instrument you can make yourself.
    Most instruments have more than one vibrating part — a plucked string also sets the body and the air inside it vibrating.
    That is why a flute, an ektara and a tabla playing the same note at the same loudness still sound different: their timbre comes from the pattern of overtones their shape and material produce.

  3. Exercise 10.3

    Assertion (A): We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out. Reason (R): Sound requires a medium to travel. Choose the correct statement: (i) Both A and R are true, but R is not the correct explanation of A. (ii) Both A and R are true, and R is the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT’s answer
    (ii)
    (ii) Both A and R are true, and R is the correct explanation of A.
    A is true: in the vacuum bell jar experiment the bell grows fainter as air is pumped out, and at near vacuum almost nothing is heard even though the bell is clearly seen ringing.
    R is true: sound is a mechanical wave — it travels as compressions and rarefactions of the particles of a medium, so with no particles there is nothing to carry it.
    R explains A: pumping out the air removes the medium, and removing the medium is exactly why the sound dies away.
    The proof that this is the cause: let the air back in and the sound returns, gradually becoming as loud as before.

  4. Exercise 10.4

    Assertion (A): Compressions and rarefactions move through the medium. Reason (R): Individual particles of the medium continuously move forward with the wave. Choose the correct statement: (i) Both A and R are true, but R is not the correct explanation of A. (ii) Both A and R are true, and R is the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT’s answer
    (iii)
    (iii) A is true, but R is false.
    A is true: compressions and rarefactions — the regions of higher and lower density — really do travel forward through the medium, each particle passing the disturbance to the next by collision.
    R is false: the particles themselves do not move forward with the wave; each only oscillates to and fro about its own mean position, parallel to the direction of propagation.
    The slinky shows it directly: the marked turn stays where it is and merely oscillates, while the closely spaced and spread-out regions run the whole length of the slinky.
    What travels forward is the disturbance and the energy it carries, not the matter of the medium.

  5. Exercise 10.5

    When sound travels from a tuning fork to your ear, which of the following actually reaches your ear? (i) Air particles near the tuning fork (ii) Energy carried by sound waves (iii) The tuning fork material (iv) A continuous stream of compressed air

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT’s answer
    (ii)
    (ii) Energy carried by sound waves.
    NCERT_Solution_Class9_Science_Ch10_PP_Q10-5
    The air particles near the fork only oscillate about their mean positions and never travel to your ear, so (i) and (iv) are wrong — there is no stream of air moving across the room.
    The fork's material stays in the fork, so (iii) is wrong.
    Sound is a form of energy: the vibrating prongs hand energy to the neighbouring air particles, whose collisions pass it on from particle to particle until it sets your eardrum vibrating.

  6. Exercise 10.6

    The variation of density of the medium for two sound waves is shown in Fig. 10.17\displaystyle 10.17 (a) and (b). Label compression and rarefaction by C and R on it. In the graph given in Fig. 10.17\displaystyle 10.17 (c) and (d), label the axes and draw the curves corresponding to Fig. 10.17\displaystyle 10.17 (a) and (b).NCERT_Question_Class9_Science_Ch10_PP_Q10-6

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Compressions (C) are the crowded regions and rarefactions (R) the thinly spread ones — mark them alternately, C R C R …, along both Fig. $\displaystyle 10.17$ (a) and (b).
    In (a) and (b): write C wherever the dots representing the particles are packed closer than average, and R wherever they are more spread out than average.
    For graphs (c) and (d): label the x-axis "Distance" and the y-axis "Density".
    Draw a horizontal dashed line across each graph for the average density — this is the level the curve oscillates about.
    On each graph draw a smooth wavy curve with a crest (maximum density) directly above every C and a trough (minimum density) directly above every R, the curve cutting the dashed line midway between them.
    Keep both graphs on the same distance scale: the strip whose C's are closer together must give the curve with crests closer together, i.e. the shorter wavelength, and the other strip the longer-wavelength curve.
    The height of a crest above the dashed line (and the depth of a trough below it) is the density amplitude — draw it the same on both unless one strip is visibly more crowded than the other.

  7. Exercise 10.7

    Conduct Activity 10.1\displaystyle 10.1 once again with a thick rubber band and then with a thin rubber band. Does the thin rubber band vibrate faster than the thick rubber band? If yes, how do the frequency and time period of the sound produced by the thin rubber band differ from that of the thick rubber band?

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Yes — the thin rubber band vibrates faster than the thick one.
    NCERT_Solution_Class9_Science_Ch10_PP_Q10-7
    Faster vibration means more density oscillations reach your ear each second, so the thin band gives the higher frequency and the shriller, higher-pitched sound.
    Its time period is shorter, because frequency and time period are inversely related: \(\displaystyle \nu = \dfrac{1}{T} \), so a larger \(\displaystyle \nu \) forces a smaller \(\displaystyle T \).
    The thick band is the opposite — lower frequency, longer time period, deeper sound.
    The chapter gives the same rule in the human voice: during adolescence boys' vocal cords lengthen and thicken, vibrate less frequently, and the voice 'deepens'.
    Tightening either band also raises its frequency, which is the change you hear in step $\displaystyle 6$ of Activity 10.1.

  8. Exercise 10.8

    If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20\displaystyle 20 Hz, then how many oscillations does the piston complete per minute?
    NCERT’s answer
    $\displaystyle 1200$ oscillations
    $\displaystyle 1200$ oscillations per minute.
    Frequency is the number of oscillations per unit time, so \(\displaystyle \nu = 20\ \text{Hz} \) means the piston completes $\displaystyle 20$ oscillations every second.
    One minute is $\displaystyle 60$ s, so \(\displaystyle \text{oscillations} = 20\ \text{s}^{-1} \times 60\ \text{s} = 1200 \).
    Each of those $\displaystyle 1200$ oscillations sends out one compression and one rarefaction into the tube.

  9. Exercise 10.9

    For the sound wave represented by the graph shown in Fig. 10.19\displaystyle 10.19, what is half of its wavelength?NCERT_Question_Class9_Science_Ch10_PP_Q10-9
    NCERT’s answer
    1.$\displaystyle 5$ cm
    Half the wavelength is $\displaystyle 1.5$ cm, so the full wavelength is \(\displaystyle 3.0\ \text{cm} \).
    On the distance axis of Fig. $\displaystyle 10.19$ the marks \(\displaystyle 0,\ 1.5,\ 3.0,\ 4.5\ \text{cm} \) fall alternately on a crest and a trough.
    Wavelength is the distance between two consecutive crests (or two consecutive troughs): \(\displaystyle \lambda = 3.0\ \text{cm} - 0 = 3.0\ \text{cm} \).
    \(\displaystyle \dfrac{\lambda}{2} = \dfrac{3.0\ \text{cm}}{2} = 1.5\ \text{cm} \) — which is exactly the crest-to-next-trough spacing you can read straight off the graph.

  10. Exercise 10.10

    Table 10.1\displaystyle 10.1 shows the speed of sound in a few media at atmospheric pressure.
    StateSubstance/MediumApproximate speed
    SolidSteel5000\displaystyle 5000 m s−1\displaystyle s^{-1}
    LiquidWater1500\displaystyle 1500 m s−1\displaystyle s^{-1}
    GasAir340\displaystyle 340 m s−1\displaystyle s^{-1}
    Compare the speeds in different media by finding the ratio of (i) the speed of sound in water with respect to the speed in the air. (ii) the speed of sound in steel with respect to the speed in the water.
    NCERT’s answer
    (i)
    $\displaystyle 75$:$\displaystyle 17$ (ii) $\displaystyle 10$:$\displaystyle 3$
    (i) water : air = $\displaystyle 75$ : 17. (ii) steel : water = $\displaystyle 10$ : 3.
    NCERT_Solution_Class9_Science_Ch10_PP_Q10-10
    From Table $\displaystyle 10.1$ at $\displaystyle 15$ ºC: steel \(\displaystyle 5000\ \text{m s}^{-1} \), water \(\displaystyle 1500\ \text{m s}^{-1} \), air \(\displaystyle 340\ \text{m s}^{-1} \).
    (i) \(\displaystyle \dfrac{v_{\text{water}}}{v_{\text{air}}} = \dfrac{1500}{340} = \dfrac{150}{34} = \dfrac{75}{17} \approx 4.4 \) — sound is about $\displaystyle 4$–$\displaystyle 5$ times faster in water than in air.
    (ii) \(\displaystyle \dfrac{v_{\text{steel}}}{v_{\text{water}}} = \dfrac{5000}{1500} = \dfrac{10}{3} \approx 3.3 \) — and steel is about $\displaystyle 15$ times faster than air.
    Both ratios confirm the chapter's ordering: sound is fastest in solids, slower in liquids and slowest in gases.

  11. Exercise 10.11

    Two friends are standing along a steel fence at a distance of 340\displaystyle 340 m from each other (Fig. 10.23\displaystyle 10.23). Gunjan places her ear over the fence and her friend knocks the fence with a metal object. Using the values of the speed of sound in steel and air given in Table 10.1\displaystyle 10.1, calculate the time difference between the sound that reached Gunjan through the air and the steel. Would it have been possible for her to distinguish between the two sounds? (The time interval between two sounds must be at least 0.1\displaystyle 0.1 s to be heard separately.)NCERT_Question_Class9_Science_Ch10_PP_Q10-11
    NCERT’s answer
    0.$\displaystyle 932$ s; Yes
    Time difference \(\displaystyle = 0.932\ \text{s} \), and yes — Gunjan could hear the two as separate sounds.
    Through air: \(\displaystyle t_{\text{air}} = \dfrac{d}{v_{\text{air}}} = \dfrac{340\ \text{m}}{340\ \text{m s}^{-1}} = 1\ \text{s} \).
    Through steel: \(\displaystyle t_{\text{steel}} = \dfrac{340\ \text{m}}{5000\ \text{m s}^{-1}} = 0.068\ \text{s} \).
    \(\displaystyle \Delta t = 1\ \text{s} - 0.068\ \text{s} = 0.932\ \text{s} \).
    \(\displaystyle 0.932\ \text{s} \) is far more than the \(\displaystyle 0.1\ \text{s} \) the ear needs, so the two arrivals are heard separately.
    The knock reaches her through the steel first (sound is fastest in solids), and the same knock arrives through the air nearly a second later.

  12. Exercise 10.12

    An experiment is being set up that requires echoes to arrive at least 0.2\displaystyle 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be 343\displaystyle 343 m s−1\displaystyle s^{-1}.
    NCERT’s answer
    34.$\displaystyle 3$ m
    The reflecting surface must be at least $\displaystyle 34.3$ m away.
    NCERT_Solution_Class9_Science_Ch10_PP_Q10-12
    The sound must travel to the surface and back, so in \(\displaystyle 0.2\ \text{s} \) it covers twice the distance \(\displaystyle d \).
    \(\displaystyle 2d = v \times t = 343\ \text{m s}^{-1} \times 0.2\ \text{s} = 68.6\ \text{m} \).
    \(\displaystyle d = \dfrac{68.6\ \text{m}}{2} = 34.3\ \text{m} \).
    Any nearer and the echo returns sooner than \(\displaystyle 0.2\ \text{s} \), which the experiment does not allow.

  13. Exercise 10.13

    Sound travels much farther in water than light, and thus, is used for various underwater applications. A sonar signal sent to find the depth of ocean takes 4\displaystyle 4 s to return. What is the depth of the ocean at that location if the speed of sound in seawater is 1500\displaystyle 1500 m s−1\displaystyle s^{-1}?
    NCERT’s answer
    $\displaystyle 3000$ m
    The ocean is $\displaystyle 3000$ m deep at that location.
    NCERT_Solution_Class9_Science_Ch10_PP_Q10-13
    The $\displaystyle 4$ s is the time for the signal to go down and come back, so the one-way time is \(\displaystyle \dfrac{4\ \text{s}}{2} = 2\ \text{s} \).
    \(\displaystyle \text{depth} = v \times t = 1500\ \text{m s}^{-1} \times 2\ \text{s} = 3000\ \text{m} \).
    This is sonar — ultrasonic pulses are used because sound travels far in water, while light does not.