{"id":31543,"date":"2026-05-21T17:04:13","date_gmt":"2026-05-21T11:34:13","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31543"},"modified":"2026-05-21T17:04:55","modified_gmt":"2026-05-21T11:34:55","slug":"sound-waves-ncert-solutions-class-9-science-exploration","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/","title":{"rendered":"Sound Waves &#8211; NCERT Solutions Class 9 Science Exploration"},"content":{"rendered":"\n<p><strong><strong>\u00a0Sound Waves: Characteristics and Applications<\/strong><\/strong> &#8211; NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9 <a href=\"https:\/\/ncert.nic.in\/textbook.php?iesc1=0-13\">Science Exploration<\/a> textbook.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">NCERT Solutions Class 9<\/h2>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-english-kaveri\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >English Kaveri<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-hindi-ganga\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Hindi Ganga<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-sanskrit-sharada\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Sanskrit Sharada<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-maths-ganita-manjari\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Maths Ganita Manjari<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-science-exploration\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Science Exploration<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-social-understanding-society\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Social Understanding Society<\/a>\n\n\n\n<h2 class=\"wp-block-heading\">Sound Waves \u2013 NCERT Solutions<\/h2>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q<strong>.1: Two astronauts are repairing the arm of a space station together during a spacewalk. Can they talk to each other and hear the sounds of metal clanking as they do on the Earth?<\/strong><\/p>\n\n\n\n<p>Solution: No, the astronauts cannot talk to each other directly in space or hear the sound of metal clanking.<\/p>\n\n\n\n<p>Sound needs a material medium (like air, water, or solids) to travel. In space, there is no atmosphere (vacuum), so sound waves cannot propagate. Therefore, even though they may speak or the metal may collide, the sound cannot reach the other astronaut.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.2: How do most bats use sound to locate their prey in the dark at night?<\/strong><\/p>\n\n\n\n<p>Solution: Bats use a method called <strong>echolocation<\/strong> to locate their prey in the dark.<\/p>\n\n\n\n<p>They produce <strong>ultrasonic sound waves<\/strong> (high-frequency sounds not audible to humans). These sound waves travel, hit objects (like insects), and reflect back as echoes. By detecting the reflected sound, bats can determine the location, distance, and even the size of their prey.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.3: Explore various ways of producing sound.<\/strong><\/p>\n\n\n\n<p>Solution: Sound is produced when an object <strong>vibrates<\/strong>. There are several ways of producing sound:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Plucking:\u00a0<\/strong>Vibrations are produced when strings are plucked, e.g., guitar, sitar.<\/li>\n\n\n\n<li><strong>Striking:\u00a0<\/strong>Sound is produced when objects are struck, e.g., drum, bell.<\/li>\n\n\n\n<li><strong>Blowing:\u00a0<\/strong>Vibrations of air column produce sound, e.g., flute, whistle.<\/li>\n\n\n\n<li><strong>Friction (rubbing):\u00a0<\/strong>Sound is produced due to rubbing of surfaces, e.g., violin bow on strings.<\/li>\n<\/ol>\n\n\n\n<p>Thus, in all cases, sound is produced due to <strong>vibrations of objects or air columns<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.4: Make a list of different types of musical instruments and identify their vibrating parts which produce sound.<\/strong><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td><strong>Type of musical instrument<\/strong><\/td><td><strong>Examples<\/strong><\/td><td><strong>Vibrating part that produces sound<\/strong><\/td><\/tr><tr><td>String instruments<\/td><td>Guitar, Sitar, Violin<\/td><td>Stretched strings<\/td><\/tr><tr><td>Wind instruments<\/td><td>Flute, Trumpet, Clarinet<\/td><td>Air column inside the instrument<\/td><\/tr><tr><td>Membrane instruments<\/td><td>Drum, Tabla<\/td><td>Stretched membrane (skin)<\/td><\/tr><tr><td>Solid instruments (Percussion)<\/td><td>Bell, Cymbals<\/td><td>Whole body of the instrument<\/td><\/tr><tr><td>Keyboard instruments<\/td><td>Piano<\/td><td>Stretched strings inside (struck by hammers)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.5: <strong>Assertion (A):<\/strong> We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out.<br><strong>Reason (R):<\/strong> Sound requires a medium to travel.<\/p>\n\n\n\n<p>Options:<br>(1) Both A and R are true and R is the correct explanation of A. \u2705<br>(2) Both A and R are true but R is not the correct explanation of A.<br>(3) A is true but R is false.<br>(4) A is false but R is true.<\/p>\n\n\n\n<p>Explanation: Sound cannot travel in vacuum. When air is removed from the jar, there is no medium for sound to propagate, so the sound of the bell cannot be heard.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.6: <strong>Assertion (A):<\/strong> Compressions and rarefactions move through the medium.<br><strong>Reason (R):<\/strong> Individual particles of the medium continuously move forward with the wave.<\/p>\n\n\n\n<p>Options:<br>(1) Both A and R are true and R is the correct explanation of A.<br>(2) Both A and R are true but R is not the correct explanation of A.<br>(3) A is true but R is false. \u2705<br>(4) A is false but R is true.<\/p>\n\n\n\n<p>Explanation: <strong>Assertion (A): True<\/strong><br>Compressions and rarefactions move through the medium as a sound wave propagates.<\/p>\n\n\n\n<p><strong>Reason (R): False<\/strong><br>The particles of the medium do <strong>not move forward with the wave<\/strong>; they only oscillate back and forth about their mean position.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.7: When sound travels from a tuning fork to your ear, which of the following actually reaches your ear?<\/strong><\/p>\n\n\n\n<p>Options:<br>(1) Air particles near the tuning fork<br>(2) Energy carried by sound waves \u2705<br>(3) The tuning fork material<br>(4) A continuous stream of compressed air<\/p>\n\n\n\n<p>Explanation: Sound travels through a medium as a wave. The particles of air only vibrate back and forth and do not move from the source to the ear. What is actually transferred is energy, not matter.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.8: If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20 Hz, then how many oscillations does the piston complete per minute?<\/strong><\/p>\n\n\n\n<p>Solution: Given:<br>Frequency of sound wave = 20 Hz<\/p>\n\n\n\n<p><strong>Step 1:<\/strong> Meaning of frequency<\/p>\n\n\n\n<p>20 Hz means the piston completes 20 oscillations per second.<\/p>\n\n\n\n<p><strong>Step 2: <\/strong>Oscillations in 1 minute<\/p>\n\n\n\n<p>1 minute = 60 seconds<\/p>\n\n\n\n<p>Total&nbsp;oscillations&nbsp;in&nbsp;1&nbsp;minute = 20 {tex}\\times{\/tex} 60 = 1200<br>The piston completes 1200 oscillations per minute.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.9: For the sound wave represented by the graph shown in the figure, what is half of its wavelength?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777025090-skchhg.jpg\"><\/p>\n\n\n\n<p>Solution: From the graph:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One complete wavelength is the distance between two consecutive compressions (or rarefactions).<\/li>\n\n\n\n<li>From the figure, this distance is 3.0 cm.<\/li>\n<\/ul>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>{tex} \\lambda=3.0 {~cm}{\/tex}<\/p>\n\n\n\n<p>Half of wavelength:<\/p>\n\n\n\n<p>{tex} \\frac{\\lambda}{2}=\\frac{3.0}{2}=1.5 {~cm}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.10: Table shows the speed of sound in a few media at atmospheric pressure.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td><strong>State<\/strong><\/td><td><strong>Substance\/Medium<\/strong><\/td><td><strong>Approximate speed<\/strong><\/td><\/tr><tr><td>Solid<\/td><td>Steel<\/td><td>{tex}5000 {~m} {s}^{-1}{\/tex}<\/td><\/tr><tr><td>Liquid<\/td><td>Water<\/td><td>{tex}1500 {~m} {s}^{-1}{\/tex}<\/td><\/tr><tr><td>Gas<\/td><td>Air<\/td><td>{tex}340 {~m} {s}^{-1}{\/tex}<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Compare the speeds in different media by finding the ratio of<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>the speed of sound in water with respect to the speed in the air.<\/li>\n\n\n\n<li>the speed of sound in steel with respect to the speed in the water.<\/li>\n<\/ol>\n\n\n\n<p>Solution: Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Speed in water {tex}=1500 {~m} \/ {s}{\/tex}<\/li>\n\n\n\n<li>Speed in air {tex}=340 {~m} \/ {s}{\/tex}<\/li>\n\n\n\n<li>Speed in steel {tex}=5000 {~m} \/ {s}{\/tex}<\/li>\n<\/ul>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Ratio of speed in water to speed in air<br>{tex} \\frac{1500}{340}=\\frac{150}{34}=\\frac{75}{17} \\approx 4.41{\/tex}<br>{tex} 75: 17 \\text { (or approximately } 4.41: 1 \\text { ) }{\/tex}<\/li>\n\n\n\n<li>Ratio of speed in steel to speed in water {tex} \\frac{5000}{1500}=\\frac{50}{15}=\\frac{10}{3} \\approx 3.33{\/tex} {tex} 10: 3 \\text { (or approximately } 3.33: 1 \\text {)}{\/tex}<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.11: An experiment is being set up that requires echoes to arrive at least 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 m s<sup>-1<\/sup>.<\/strong><\/p>\n\n\n\n<p>Solution: For an echo, the sound has to travel to the reflecting surface and back, so the total distance covered is twice the distance to the surface.<\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Speed of sound {tex}v=343 {~m} \/ {s}{\/tex}<\/li>\n\n\n\n<li>Time delay {tex}t=0.2 {~s}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Step:<\/strong><br>Total distance traveled by sound:<\/p>\n\n\n\n<p>{tex} \\text {Total distance }=v \\times t=343 \\times 0.2=68.6 {~m}{\/tex}<br>Since this is a round trip, the one-way distance (actual distance to the reflecting surface) is:<\/p>\n\n\n\n<p>{tex} \\frac{68.6}{2}=34.3 {~m}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.12: 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 s to return. What is the depth of the ocean at that location if the speed of sound in seawater is 1500 m s<sup>-1<\/sup>?<\/strong><\/p>\n\n\n\n<p>Solution: <strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Speed of sound in seawater {tex}v=1500 {~m} \/ {s}{\/tex}<\/li>\n\n\n\n<li>Total time {tex}t=4 {~s}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Step:<\/strong><br>Total distance traveled by the sound:<\/p>\n\n\n\n<p>{tex} \\text {Distance}=v \\times t=1500 \\times 4=6000 {~m}{\/tex}<br>Since this is a round trip, the actual depth is half of this:<\/p>\n\n\n\n<p>{tex} \\frac{6000}{2}=3000 {~m}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.13: If there are 10 density oscillations in 2 seconds at a given position, then calculate the (i) frequency of sound wave, and (ii) its time period.<\/strong><\/p>\n\n\n\n<p>Solution: Given:<br>Number of oscillations = 10<br>Time = 2 s<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Frequency of sound wave<br>We know that frequency is the number of oscillations per second.<br>{tex}\\text { Frequency }({f}) =\\frac{\\text { Number of oscillations }}{\\text { Time taken }}{\/tex}<br>{tex}f =\\frac{10}{2}=5 {~Hz}{\/tex}<\/li>\n\n\n\n<li>Time period of sound wave<br>Time period is the time taken for one oscillation.<br>{tex}T=\\frac{1}{f}{\/tex}<br>{tex}T=\\frac{1}{5}=0.2 {~s}{\/tex}<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.14: Human hearing roughly spans 20 Hz to 20 kHz. What are the corresponding wavelengths in air for these two frequencies? Use the speed of sound in air as 344 m s<sup>-1<\/sup>.<\/strong><\/p>\n\n\n\n<p>Solution: Using the relation between wavelength {tex}(\\lambda){\/tex}, frequency {tex}(\\nu){\/tex}, and speed ({tex}v{\/tex}),<br>speed of the wave {tex}={\/tex} frequency \u00d7 wavelength<br>Therefore, wavelength {tex}\\lambda=\\frac{\\text { speed of the wave }}{\\text { frequency }}{\/tex}<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>For {tex}\\nu =20 {~Hz}, \\lambda=\\frac{344 {~m} {~s}^{-1}}{20 {~s}^{-1}}=17.2 {~m}{\/tex}<\/li>\n\n\n\n<li>For {tex}\\nu =20 {kHz}=20000 {~Hz}, {\/tex}\u00a0{tex}\\lambda=\\frac{344 {~m} {~s}^{-1}}{20000 {~s}^{-1}}=0.0172 {~m}=1.72 {~cm}{\/tex}<\/li>\n<\/ol>\n\n\n\n<p>The wavelength of sound in air corresponding to the frequency (i) 20 Hz is 17.2 m, and (ii) 20000 Hz is 1.72 cm.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.15: During a thunderstorm, lightning is seen before thunder is heard because sound travels much slower than light. If the time delay between seeing the lightning flash and hearing the thunder is measured to be 5 s, estimate the distance to the lightning strike. Use the speed of sound in air as {tex}340 {m} {s}^{-1}{\/tex}. Assume that light (speed {tex}=300000 {~km} {s}^{-1}{\/tex}) reaches you almost instantaneously.<\/strong><\/p>\n\n\n\n<p>Solution: Distance {tex}=v \\times t=340 {~m} {s}^{-1} \\times 5 {~s}=1700 {~m}{\/tex}<br>Lightning struck about 1.7 km away.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.16: From the graphical representation of a sound wave propagating in steel, find its wavelength. Calculate its frequency and time period if the speed of sound in steel is 5000 m s<sup>-1<\/sup>.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777026966-wwhu8w.jpg\"><\/p>\n\n\n\n<p>Solution: From the graph, the wavelength {tex}\\lambda=50 {~m}{\/tex}<br>Using Eq. ({tex}v=\\lambda \\times \\nu{\/tex}), the frequency of the sound wave is<\/p>\n\n\n\n<p>{tex} \\nu=\\frac{v}{\\lambda}=\\frac{5000 {~m} {~s}^{-1}}{50 {~m}}=100 {~Hz}{\/tex}<\/p>\n\n\n\n<p>Using Eq. ({tex}v=\\frac{1}{T}{\/tex}), the time period of the sound wave is<\/p>\n\n\n\n<p>{tex} T=\\frac{1}{\\nu}=\\frac{1}{100 {~Hz}}=0.01 {~s}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.17: You clap in an empty corridor and hear an echo after 0.5 s. If the speed of sound in air is 340 m s<sup>-1<\/sup>, calculate your distance from the wall.<\/strong><\/p>\n\n\n\n<p>Solution: Sound travels to the wall and back, thus,<br>distance from wall {tex}=\\frac{v \\times t}{2}=\\frac{340 {~m} {s}^{-1} \\times 0.5 {~s}}{2}=85 {~m}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.18: A naval sonar signal sent into seawater returns after 0.90 s. The speed of sound in seawater is 1530 m s<sup>-1<\/sup>. How far is the object?<\/strong><\/p>\n\n\n\n<p>Solution: Time taken for the signal to reach the object and travel back = 0.90 s<br>Time taken to reach the object is half of above time {tex}=\\frac{0.90 {~s}}{2}=0.45 {~s}{\/tex}<br>Thus, distance {tex}={\/tex} speed \u00d7 time {tex}=1530 {~m} {s}^{-1} \\times 0.45 {~s}=688.5 {~m}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q<strong>.19: Which observation best supports the idea that sound is a mechanical wave?<\/strong><\/p>\n\n\n\n<p>Options:<br>(1) Sound shows reflection<br>(2) Sound needs a medium to propagate \u2705<br>(3) Sound has frequency<br>(4) Sound carries energy<\/p>\n\n\n\n<p>Explanation: Mechanical waves are those waves which require a material medium for propagation. Sound is a mechanical wave because it cannot travel in vacuum and needs particles of a medium (like air, water, or solids) to move and transfer energy.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Reflection, frequency, and energy are properties common to many types of waves (including light, which is not mechanical).<\/li>\n\n\n\n<li>The need for a medium is the distinguishing feature of mechanical waves.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.20: For a sound wave propagating in a medium, increasing its frequency will increase its<\/strong><\/p>\n\n\n\n<p>Options:<br>(1) wavelength<br>(2) speed<br>(3) number of compressions per second \u2705<br>(4) time period<\/p>\n\n\n\n<p>Explanation: Frequency is defined as the number of oscillations (or compressions and rarefactions) per second.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If frequency increases \u2192 number of compressions per second increases<\/li>\n\n\n\n<li>Wavelength decreases (inverse relation)<\/li>\n\n\n\n<li>Speed remains constant in the same medium<\/li>\n\n\n\n<li>Time period decreases {tex}\\left(T=\\frac{1}{f}\\right){\/tex}<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.21: If 20 compressions pass a point in 4 seconds, the frequency is<\/strong><\/p>\n\n\n\n<p>Options:<br>(1) 80 Hz<br>(2) 5 Hz \u2705<br>(3) 10 Hz<br>(4) 0.2 Hz<\/p>\n\n\n\n<p>Explanation: Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Number of compressions {tex}=20{\/tex}<\/li>\n\n\n\n<li>Time {tex}=4{\/tex} seconds<\/li>\n<\/ul>\n\n\n\n<p>Concept:<br>Frequency is defined as the number of vibrations (or compressions) per second.<\/p>\n\n\n\n<p>{tex} \\text { Frequency }=\\frac{\\text { Number of compressions }}{\\text { Time }}{\/tex}<br>{tex} f=\\frac{N}{t}{\/tex}<br><strong>Calculation:<\/strong><\/p>\n\n\n\n<p>{tex} f=\\frac{20}{4}=5 {~Hz}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.22: In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify your answer.<\/strong><\/p>\n\n\n\n<p>Solution: The human brain possesses a property called <strong>persistence of hearing<\/strong>. When we hear a sound, the sensation remains in our brain for approximately <strong>0.1 seconds<\/strong>.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>The Threshold for Echo:<\/strong> To hear a distinct echo, the reflected sound must reach our ears <strong>after<\/strong> the first sound has faded- meaning the time interval between the original sound and the reflected sound must be <strong>at least 0.1 s<\/strong>.<\/li>\n\n\n\n<li><strong>The Given Scenario:<\/strong> In this case, the reflected sound arrives in only <strong>0.05 s<\/strong>.<\/li>\n\n\n\n<li><strong>The Result:<\/strong> Since 0.05 s is less than 0.1 s ({tex}0.05\\text{ s} &lt; 0.1\\text{ s}{\/tex}), the original sound and the reflected sound will overlap and &#8220;fused&#8221; together in the brain.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.23: Graphs representing two sound waves are given in the figure. If the scales on the X and Y axes of the two graphs are the same, which of the two sound waves has (i) greater wavelength, and (ii) smaller amplitude?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777029753-s4gm82.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Greater Wavelength<br><strong>Wave (a) has the greater wavelength.<\/strong><br><strong>Reasoning:<\/strong> Wavelength ({tex}\\lambda{\/tex}) is the horizontal distance between two consecutive peaks (crests) or two consecutive troughs. Looking at the X-axis (horizontal), the distance between the peaks in graph (a) is significantly larger than the distance between the peaks in graph (b). Since the scales are the same, a longer physical distance on the graph represents a greater wavelength.<\/li>\n\n\n\n<li>Smaller Amplitude<br><strong>Wave (a) has the smaller amplitude.<\/strong><br><strong>Reasoning:<\/strong> Amplitude is the maximum vertical displacement of the wave from its central equilibrium position (the X-axis). Looking at the Y-axis (vertical), the &#8220;height&#8221; of the peaks in wave (a) is much shorter compared to the peaks in wave (b). Therefore, wave (a) represents a softer sound with smaller amplitude, while wave (b) represents a louder sound with greater amplitude.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.24: The sound waves emitted by three sources A, B and C are represented in the figure. If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777029942-pax32k.jpg\"><\/p>\n\n\n\n<p>Solution: From the figure:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The wave with maximum number of oscillations (shortest wavelength) has maximum frequency \u2192 A<\/li>\n\n\n\n<li>The wave with minimum oscillations (longest wavelength) has minimum frequency \u2192 C<\/li>\n\n\n\n<li>The remaining wave is B<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.25: Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.<\/strong><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777279328-mjq8as.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.26: In a movie, while showing the explosion of a spacecraft in space, a flash of light is shown along with sound at the same time. What are the errors in this depiction?<\/strong><\/p>\n\n\n\n<p>Solution: In space, there is vacuum, so sound waves cannot propagate. Therefore, during a spacecraft explosion, only light should be seen, not heard.<\/p>\n\n\n\n<p>Also, even if sound were possible, light travels much faster than sound, so they cannot be observed simultaneously.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.27: A source produces a sound wave of wavelength 3.44\u2009m. If the wave travels with a speed of 344\u2009m s<sup>-1<\/sup> find its time period.<\/strong><\/p>\n\n\n\n<p>Solution: Given:<br>Wavelength {tex}(\\lambda)=3.44 {~m}{\/tex}<br>Wave speed {tex}(v)=344 {~m} \/ {s}{\/tex}<\/p>\n\n\n\n<p>Concept:<\/p>\n\n\n\n<p>{tex} v=\\lambda \\nu \\text { and } T=\\frac{1}{\\nu}{\/tex}<\/p>\n\n\n\n<p>Calculation:<\/p>\n\n\n\n<p>{tex}\\nu =\\frac{v}{\\lambda}=\\frac{344}{3.44}=100 {~Hz}{\/tex}<br>{tex}T =\\frac{1}{\\nu}=\\frac{1}{100}=0.01 {~s}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.28: A ship searching for a sunken ship sent a sonar signal and detected an echo after 5\u2009s. If ultrasonic wave travels at 1525 m s<sup>-1<\/sup> in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?<\/strong><\/p>\n\n\n\n<p>Solution: Given:<br>Time for echo {tex}({t})=5 {~s}{\/tex}<br>Speed of sound in seawater {tex}(v)=1525 {~m} \/ {s}{\/tex}<\/p>\n\n\n\n<p>Concept:<br>Sound travels to the object and back, so total distance {tex}=2 {~d}{\/tex}<\/p>\n\n\n\n<p>{tex} d=\\frac{v \\times t}{2}{\/tex}<\/p>\n\n\n\n<p>Calculation:<\/p>\n\n\n\n<p>{tex} d=\\frac{1525 \\times 5}{2}=\\frac{7625}{2}=3812.5 {~m}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.29: A vehicle is fitted with an ultrasonic distance sensor as part of parking assistance system which provides echolocation, while the driver is reversing the vehicle. It emits ultrasonic wave (about 40 kHz) which is reflected by the obstacle. When the warning beep starts sounding at a distance of 1.2 m from the obstacle, how much time is taken by ultrasonic wave to travel to the obstacle and come back? Assume the speed of ultrasonic wave in air to be 345 m s<sup>-1<\/sup>.<\/strong><\/p>\n\n\n\n<p>Solution: Given:<br>Distance from obstacle (d) = 1.2 m<br>Speed of sound {tex}(v)=345 {~m} \/ {s}{\/tex}<\/p>\n\n\n\n<p>Concept:<br>The ultrasonic wave travels to the obstacle and back, so total distance {tex}=2 {~d}{\/tex}<\/p>\n\n\n\n<p>{tex} t=\\frac{2 d}{v}{\/tex}<\/p>\n\n\n\n<p>Calculation:<\/p>\n\n\n\n<p>{tex}t=\\frac{2 \\times 1.2}{345}=\\frac{2.4}{345}{\/tex}<br>{tex}t \\approx 0.007 {~s}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.30: The speed of sound in air is about {tex}331 {~m} {s}^{-1}{\/tex} at {tex}0^{\\circ} {C}{\/tex} and nearly {tex}344 {~m} {s}^{-1}{\/tex} at {tex}22^{\\circ} {C}{\/tex}. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m, if the air temperature changes from {tex}22^{\\circ} {C}{\/tex} to {tex}0^{\\circ} {C}{\/tex}? Assume that all other conditions remain unchanged.<\/strong><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<br>Distance (d) = 1720 m<br>Speed at {tex}22^{\\circ} {C}\\left({v}_1\\right)=344 {~m} \/ {s}{\/tex}<br>Speed at {tex}0^{\\circ} {C}\\left(v_2\\right)=331 {~m} \/ {s}{\/tex}<\/p>\n\n\n\n<p>Concept:&nbsp;{tex} t=\\frac{d}{v}{\/tex}<\/p>\n\n\n\n<p>Calculation:<br>Time at {tex}22^{\\circ} {C}{\/tex}:<\/p>\n\n\n\n<p>{tex} t_1=\\frac{1720}{344}=5 {~s}{\/tex}<\/p>\n\n\n\n<p>Time at {tex}0^{\\circ} {C}{\/tex}:<\/p>\n\n\n\n<p>{tex} t_2=\\frac{1720}{331} \\approx 5.19 {~s}{\/tex}<\/p>\n\n\n\n<p>Extra time taken:<\/p>\n\n\n\n<p>{tex} \\Delta t=t_2-t_1=5.19-5=0.19 {~s}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.31: The variation of density of medium for a sound wave propagating with a speed of 340 m s<sup>-1<\/sup> is shown in figure. Calculate the wavelength and frequency of the sound wave.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777031570-ec7gw7.jpg\"><\/p>\n\n\n\n<p>Solution: From the given density-distance graph of a sound wave:<br>The distance between two consecutive compressions (or rarefactions) = one wavelength ({tex}\\lambda{\/tex}).<\/p>\n\n\n\n<p>From the figure, this distance is {tex}{4 0 ~ c m}={0 . 4 0 ~ m}{\/tex}.<\/p>\n\n\n\n<p>So,<br>wavelength {tex}\\lambda=0.40 {~m}{\/tex}<\/p>\n\n\n\n<p>Now using the wave relation from NCERT:<\/p>\n\n\n\n<p>{tex} v=f \\lambda{\/tex}<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>{tex}f=v \/ \\lambda{\/tex}<br>{tex}f=340 \/ 0.40{\/tex}<br>{tex}f=850 {~Hz}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.32: The graphical representation of two sound waves A and B propagating at the same speed of 345 m s<sup>-1<\/sup> is shown in the figure. What is the wavelength of each of them? Also, calculate their frequencies.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777031827-xvapcy.jpg\"><\/p>\n\n\n\n<p>Solution: Given:<br>Speed of sound {tex}(v)=345 {~m} \/ {s}{\/tex}<\/p>\n\n\n\n<p>Concept:<br>Wavelength is the distance between two consecutive crests.<\/p>\n\n\n\n<p>{tex} v=\\lambda f{\/tex}<br>From the graph:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Wave A has wavelength {tex}2 {~cm}(0.02 {~m}){\/tex}<\/li>\n\n\n\n<li>Wave {tex}B{\/tex} has wavelength {tex}4 {~cm}(0.04 {~m}){\/tex}<\/li>\n<\/ul>\n\n\n\n<p>For A:<\/p>\n\n\n\n<p>{tex} f_A=\\frac{345}{0.02}=17250 {~Hz}{\/tex}<\/p>\n\n\n\n<p>For B:<\/p>\n\n\n\n<p>{tex} f_B=\\frac{345}{0.04}=8625 {~Hz}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.33: Two identical sound sources are placed at A and B\u2009-\u2009one in air and one submerged in water. Both produce sounds at the same time, which travel horizontally to the vertical side of the cliff and come back. If the time taken by the sound to return to A is 4.5 times that of B, what is the ratio between the speeds of sound in air and water?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777032127-y3feb7.jpg\"><\/p>\n\n\n\n<p>Solution: Given: Time taken by sound in air to return to {tex}{A}\\left(t_A\\right)\\ is\\ 4.5{\/tex} times that of {tex}{B}\\left(t_B\\right){\/tex} i.e., {tex}t_A=4.5 t_B{\/tex}<br>To Find: Ratio of speeds of sound in air and water {tex}\\left(\\frac{v_{\\text {air }}}{v_{\\text {water }}}\\right){\/tex}<br>Concept: For the same distance, time taken is inversely proportional to speed.<\/p>\n\n\n\n<p>{tex}\\text {Time taken} \\ t=\\frac{\\text { distance }}{\\text { speed }}{\/tex}<\/p>\n\n\n\n<p>Let the distance between the source and the cliff be {tex}d{\/tex}.<br>For sound in air: {tex}t_A=\\frac{2 d}{v_{\\text {air }}}{\/tex}<br>For sound in water: {tex}t_B=\\frac{2 d}{v_{\\text {water }}}{\/tex}<br>Given: {tex}t_A=4.5 t_B{\/tex}<br>So, {tex}\\frac{2 d}{v_{\\text {air }}}=4.5 \\times\\left(\\frac{2 d}{v_{\\text {water }}}\\right){\/tex}<\/p>\n\n\n\n<p>{tex} \\frac{1}{v_{\\text {air }}}=\\frac{4.5}{v_{\\text {water }}}{\/tex}<\/p>\n\n\n\n<p>{tex}\\frac{v_{\\text {water }}}{v_{\\text {air }}} =4.5{\/tex}<br>{tex}\\frac{v_{\\text {air }}}{v_{\\text {water }}} =\\frac{1}{4.5}=\\frac{2}{9}{\/tex}<\/p>\n\n\n\n<p>The ratio between the speeds of sound in air and water is {tex}2: 9{\/tex}.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0Sound Waves: Characteristics and Applications &#8211; NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9 Science Exploration textbook. NCERT Solutions Class 9 Sound Waves \u2013 NCERT Solutions Q.1: Two astronauts are repairing the arm of a space station together during a spacewalk. Can they talk to each &#8230; <a title=\"Sound Waves &#8211; NCERT Solutions Class 9 Science Exploration\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/\" aria-label=\"More on Sound Waves &#8211; NCERT Solutions Class 9 Science Exploration\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[281,2071,2072],"tags":[216],"class_list":["post-31543","post","type-post","status-publish","format-standard","hentry","category-ncert-solutions","category-ncert-solutions-class-9","category-ncert-solutions-class-9-science-exploration","tag-ncert-solutions"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Sound Waves - NCERT Solutions Class 9 Science Exploration | myCBSEguide<\/title>\n<meta name=\"description\" content=\"Sound Waves - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Sound Waves - NCERT Solutions Class 9 Science Exploration | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Sound Waves - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/\" \/>\n<meta property=\"og:site_name\" content=\"myCBSEguide\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/mycbseguide\/\" \/>\n<meta property=\"article:published_time\" content=\"2026-05-21T11:34:13+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2026-05-21T11:34:55+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777025090-skchhg.jpg\" \/>\n<meta name=\"author\" content=\"myCBSEguide\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:site\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"myCBSEguide\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"16 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/\"},\"author\":{\"name\":\"myCBSEguide\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65\"},\"headline\":\"Sound Waves &#8211; NCERT Solutions Class 9 Science Exploration\",\"datePublished\":\"2026-05-21T11:34:13+00:00\",\"dateModified\":\"2026-05-21T11:34:55+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/\"},\"wordCount\":3422,\"publisher\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777025090-skchhg.jpg\",\"keywords\":[\"NCERT Solutions\"],\"articleSection\":[\"NCERT Solutions\",\"NCERT Solutions Class 9\",\"NCERT Solutions Class 9 Science Exploration\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/\",\"url\":\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/\",\"name\":\"Sound Waves - NCERT Solutions Class 9 Science Exploration | myCBSEguide\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#primaryimage\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777025090-skchhg.jpg\",\"datePublished\":\"2026-05-21T11:34:13+00:00\",\"dateModified\":\"2026-05-21T11:34:55+00:00\",\"description\":\"Sound Waves - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9\",\"breadcrumb\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#primaryimage\",\"url\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777025090-skchhg.jpg\",\"contentUrl\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777025090-skchhg.jpg\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/mycbseguide.com\/blog\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"NCERT Solutions\",\"item\":\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"NCERT Solutions Class 9\",\"item\":\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/\"},{\"@type\":\"ListItem\",\"position\":4,\"name\":\"NCERT Solutions Class 9 Science Exploration\",\"item\":\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-science-exploration\/\"},{\"@type\":\"ListItem\",\"position\":5,\"name\":\"Sound Waves &#8211; NCERT Solutions Class 9 Science Exploration\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#website\",\"url\":\"https:\/\/mycbseguide.com\/blog\/\",\"name\":\"myCBSEguide\",\"description\":\"\",\"publisher\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/mycbseguide.com\/blog\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\",\"name\":\"myCBSEguide\",\"url\":\"https:\/\/mycbseguide.com\/blog\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png\",\"contentUrl\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png\",\"width\":180,\"height\":180,\"caption\":\"myCBSEguide\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/\"},\"sameAs\":[\"https:\/\/www.facebook.com\/mycbseguide\/\",\"https:\/\/x.com\/mycbseguide\",\"https:\/\/www.linkedin.com\/company\/mycbseguide\/\",\"http:\/\/in.pinterest.com\/mycbseguide\/\",\"https:\/\/www.youtube.com\/channel\/UCxuqSnnygFzwJG0pwogCNEQ\"]},{\"@type\":\"Person\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65\",\"name\":\"myCBSEguide\",\"sameAs\":[\"http:\/\/mycbseguide.com\"]}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"Sound Waves - NCERT Solutions Class 9 Science Exploration | myCBSEguide","description":"Sound Waves - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/","og_locale":"en_US","og_type":"article","og_title":"Sound Waves - NCERT Solutions Class 9 Science Exploration | myCBSEguide","og_description":"Sound Waves - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9","og_url":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/","og_site_name":"myCBSEguide","article_publisher":"https:\/\/www.facebook.com\/mycbseguide\/","article_published_time":"2026-05-21T11:34:13+00:00","article_modified_time":"2026-05-21T11:34:55+00:00","og_image":[{"url":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777025090-skchhg.jpg","type":"","width":"","height":""}],"author":"myCBSEguide","twitter_card":"summary_large_image","twitter_creator":"@mycbseguide","twitter_site":"@mycbseguide","twitter_misc":{"Written by":"myCBSEguide","Est. reading time":"16 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#article","isPartOf":{"@id":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/"},"author":{"name":"myCBSEguide","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65"},"headline":"Sound Waves &#8211; NCERT Solutions Class 9 Science Exploration","datePublished":"2026-05-21T11:34:13+00:00","dateModified":"2026-05-21T11:34:55+00:00","mainEntityOfPage":{"@id":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/"},"wordCount":3422,"publisher":{"@id":"https:\/\/mycbseguide.com\/blog\/#organization"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#primaryimage"},"thumbnailUrl":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777025090-skchhg.jpg","keywords":["NCERT Solutions"],"articleSection":["NCERT Solutions","NCERT Solutions Class 9","NCERT Solutions Class 9 Science Exploration"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/","url":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/","name":"Sound Waves - NCERT Solutions Class 9 Science Exploration | myCBSEguide","isPartOf":{"@id":"https:\/\/mycbseguide.com\/blog\/#website"},"primaryImageOfPage":{"@id":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#primaryimage"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#primaryimage"},"thumbnailUrl":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777025090-skchhg.jpg","datePublished":"2026-05-21T11:34:13+00:00","dateModified":"2026-05-21T11:34:55+00:00","description":"Sound Waves - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9","breadcrumb":{"@id":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#primaryimage","url":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777025090-skchhg.jpg","contentUrl":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777025090-skchhg.jpg"},{"@type":"BreadcrumbList","@id":"https:\/\/mycbseguide.com\/blog\/sound-waves-ncert-solutions-class-9-science-exploration\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/mycbseguide.com\/blog\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions","item":"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/"},{"@type":"ListItem","position":3,"name":"NCERT Solutions Class 9","item":"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/"},{"@type":"ListItem","position":4,"name":"NCERT Solutions Class 9 Science Exploration","item":"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-science-exploration\/"},{"@type":"ListItem","position":5,"name":"Sound Waves &#8211; NCERT Solutions Class 9 Science Exploration"}]},{"@type":"WebSite","@id":"https:\/\/mycbseguide.com\/blog\/#website","url":"https:\/\/mycbseguide.com\/blog\/","name":"myCBSEguide","description":"","publisher":{"@id":"https:\/\/mycbseguide.com\/blog\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/mycbseguide.com\/blog\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/mycbseguide.com\/blog\/#organization","name":"myCBSEguide","url":"https:\/\/mycbseguide.com\/blog\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/","url":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png","contentUrl":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png","width":180,"height":180,"caption":"myCBSEguide"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/mycbseguide\/","https:\/\/x.com\/mycbseguide","https:\/\/www.linkedin.com\/company\/mycbseguide\/","http:\/\/in.pinterest.com\/mycbseguide\/","https:\/\/www.youtube.com\/channel\/UCxuqSnnygFzwJG0pwogCNEQ"]},{"@type":"Person","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65","name":"myCBSEguide","sameAs":["http:\/\/mycbseguide.com"]}]}},"_links":{"self":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/31543","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/comments?post=31543"}],"version-history":[{"count":1,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/31543\/revisions"}],"predecessor-version":[{"id":31544,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/31543\/revisions\/31544"}],"wp:attachment":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/media?parent=31543"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/categories?post=31543"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/tags?post=31543"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}