Q: Question 13The diagram shown below depicts a displacement time graph of particle a wave travelling with speed 20 m/s.The frequency of the wave is :50 Hz33.3 Hz100 Hz1000 Hz

50 HzReason — We know,Frequency = 1Time period\dfrac{1}{\text{Time period}}Time period1Given, time period = 0.02 secSubstituting we get,f = 10.02\dfrac{1}{0.02}0.021 = 1002\dfrac{100}{2}2100 = 50 HzHence, f = 50 Hz

Question 1

Question 13The diagram shown below depicts a displacement time graph of particle a wave travelling with speed 20 m/s.The frequency of the wave is :50 Hz33.3 Hz100 Hz1000 Hz

Accepted Answer

50 HzReason — We know,Frequency = 1Time period\dfrac{1}{	ext{Time period}}Time period1​Given, time period = 0.02 secSubstituting we get,f = 10.02\dfrac{1}{0.02}0.021​ = 1002\dfrac{100}{2}2100​ = 50 HzHence, f = 50 Hz

Question 2

Question 16Assertion (A): The flash of lightening is seen before the sound of thunder is heard.Reason (R): The speed of sound is greater than the speed of light.Both A and R are true and R is the correct explanation of ABoth A and R are true and R is not the correct explanation of Aassertion is false but reason is true.assertion is true but reason is false.

Accepted Answer

assertion is true but reason is false.ExplanationAssertion (A) is true. The flash of lightning is seen before the sound of thunder is heard due to the difference in the speeds of light and sound. Light travels much faster than sound, so the flash of lightning is observed almost instantaneously, while the sound of thunder takes some time to travel to the observer.Reason (R) is false. The speed of sound is not greater than the speed of light. Light travels faster than sound with a speed of 3 x 108 m/s whereas speed of sound is 330 m/s.

Question 3

Question 1The wavelength of waves produced on the surface of water is 20 cm. If the wave velocity is 24 ms-1, calculate (i) the number of waves produced in one second and (ii) the time in which one wave is produced.

Accepted Answer

(i) As we know,Velocity (V) = frequency (f) x wavelength (λ)Given,V = 24 ms-1λ = 20 cmConverting cm to m, we get,100 cm = 1 m,1 cm = 1100\dfrac{1}{100}1001​ m = 0.01 m,Therefore, 20 cm = (20 x 0.01) m = 0.2 mHence, λ = 0.2 mSubstituting the values in the formula above, we get,24=f×0.2⇒f=240.2⇒f=2402⇒f=120s−124 = f \times 0.2 \[0.5em] \Rightarrow f = \dfrac{24}{0.2} \[0.5em] \Rightarrow f = \dfrac{240}{2} \[0.5em] \Rightarrow f = 120 s ^{-1}\[0.5em]24=f×0.2⇒f=0.224​⇒f=2240​⇒f=120s−1Therefore, the number of waves produced in one second = 120 s-1(ii) As we know,Time in which one wave is produced (t) = 1f\dfrac{1}{f}f1​Substituting the values in the formula above, we get,t=1120⇒t=0.0083⇒t=0.0083st = \dfrac{1}{120} \[0.5em] \Rightarrow t = 0.0083 \[0.5em] \Rightarrow t = 0.0083 s \[0.5em]t=1201​⇒t=0.0083⇒t=0.0083sHence,Time in which one wave is produced = 8.3 x 10-3 s

Question 4

Question 4A man standing 25 m away from a wall produces a sound and receives the reflected sound. (a) Calculate the time after which he receives the reflected sound if the speed of sound in air is 350 ms-1. (b) Will the man be able to hear a distinct echo? Explain the answer.

Accepted Answer

(a) As we know,Speed of sound (V)=Total distance travelled (2d)time interval (t)\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \[0.5em]Speed of sound (V)=time interval (t)Total distance travelled (2d)​Total distance travelled by the sound in going and then coming back = 2dGiven,d = 25 mV = 350 ms-1Substituting the values in the formula above, we get,350=2×25t⇒t=50350⇒t=17⇒t=0.143s350 = \dfrac{2 \times 25}{t} \[0.5em] \Rightarrow t = \dfrac{50}{350} \[0.5em] \Rightarrow t = \dfrac{1}{7} \[0.5em] \Rightarrow t = 0.143 s \[0.5em]350=t2×25​⇒t=35050​⇒t=71​⇒t=0.143sHence, the time after which he receives the reflected sound = 0.143 seconds(b) Yes, the man will be able to hear a distinct echo.As an important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 s because original sound persists only for 0.1 s. Hence, a distinct echo is heard because the man receives the reflected sound after 0.143 seconds.

Question 5

Question 5A RADAR sends a signal to an aeroplane at a distance 300 km away, with a speed of 3 x 108 ms-1. After how much time is the signal received back after reflecting from the aeroplane?

Accepted Answer

(a) As we know,Speed of sound (V)=Total distance travelled (2d)time interval (t)\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \[0.5em]Speed of sound (V)=time interval (t)Total distance travelled (2d)​Total distance travelled by the sound in going and then coming back = 2dGiven,d = 300 kmConverting km to m, we get,1 km = 1000 mTherefore, 300 km = 3,00,000 mHence, d = 3 x 105 mV = 3 x 108 ms-1Substituting the values in the formula above, we get,3×108=2×(3×105)tt=2×(3×105)3×108t=2×10−33 \times 10^8 = \dfrac{2 \times (3 \times 10^5) }{t} \[0.5em] t = \dfrac{2 \times (3 \times 10^5)}{3 \times 10^8} \[0.5em] t = 2 \times 10^{-3} \[0.5em]3×108=t2×(3×105)​t=3×1082×(3×105)​t=2×10−3Hence, the signal is received back after 2 x 10-3 s

Question 6

Question 6A man standing 48 m away from a wall fires a gun. Calculate the time after which an echo is heard. (The speed of sound in air is 320 ms-1).

Accepted Answer

As we know,Speed of sound (V)=Total distance travelled (2d)time interval (t)\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \[0.5em]Speed of sound (V)=time interval (t)Total distance travelled (2d)​Total distance travelled by the sound in going and then coming back = 2dGiven,d = 48 mV = 320 ms-1Substituting the values in the formula above, we get,320=2×48tt=2×48320t=0.30s320 = \dfrac{2 \times 48}{t} \[0.5em] t = \dfrac{2 \times 48}{320} \[0.5em] t = 0.30 s \[0.5em]320=t2×48​t=3202×48​t=0.30sHence, time after which echo is heard = 0.30 s.

Question 7

Question 8A pendulum has a frequency of 5 vibrations per second. An observer starts the pendulum and fires a gun simultaneously. He hears an echo from the cliff after 8 vibrations of the pendulum. If the velocity of sound in air is 340 ms-1, find the distance between the cliff and the observer.

Accepted Answer

As we know,Speed of sound (V)=Total distance travelled (2d)time interval (t)\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \[0.5em]Speed of sound (V)=time interval (t)Total distance travelled (2d)​5 vibrations are produced in 1 second.Therefore, time for 8 vibrations = (15\dfrac{1}{5}51​ x 8) s= 85\dfrac{8}{5}58​ s = 1.6 sHence, t = 1.6 sV = 340 ms-1Substituting the values in the formula above, we get,340=2d1.6⇒d=340×1.62⇒d=340×0.8⇒d=272m340 = \dfrac{2d}{1.6} \[0.5em] \Rightarrow d = \dfrac{340 \times 1.6}{2} \[0.5em] \Rightarrow d = 340 \times 0.8 \[0.5em] \Rightarrow d = 272 m340=1.62d​⇒d=2340×1.6​⇒d=340×0.8⇒d=272mTherefore, the distance between the cliff and the observer = 272 m

Question 8

Question 9A person standing between two vertical cliffs produces a sound. Two successive echoes are heard at 4s and 6s. Calculate the distance between the cliffs.(Speed of sound in air = 320 ms-1)[HINT — First echo will be heard from the nearer cliff and the second echo from the farther cliff.]

Accepted Answer

As we know,Speed of sound (V)=Total distance travelled (2d)time interval (t)\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \[0.5em]Speed of sound (V)=time interval (t)Total distance travelled (2d)​Total distance travelled by the sound in going and then coming back = 2dGiven,t1 = 4 st2 = 6 sV = 320 ms-1 andFirst echo will be heard from the nearer cliff and the second echo from the farther cliff.Hence, Distance between cliffs = d1 + d2Therefore, substituting the values in the formula above, we get,For t1320=2d14⇒d1=320×42⇒d1=640m320 = \dfrac{2d_1 }{4} \[0.5em] \Rightarrow d_1 = \dfrac{320 \times 4}{2} \[0.5em] \Rightarrow d_1 = 640 m320=42d1​​⇒d1​=2320×4​⇒d1​=640mSo, For t2320=2d26⇒d2=320×62⇒d2=960m320 = \dfrac{2d_2}{6} \[0.5em] \Rightarrow d_2 = \dfrac{320 \times 6}{2} \[0.5em] \Rightarrow d_2 = 960 m320=62d2​​⇒d2​=2320×6​⇒d2​=960mDistance between cliffs = d1 + d2= 640 + 960= 1600 mHence, distance between cliffs = 1600 m

Question 9

Question 10A person standing at a distance x in front of a cliff fires a gun. Another person B standing behind the person A at a distance y from the cliff hears two sounds of the fired shot after 2 s and 3 s respectively. Calculate x and y (take speed of sound 320 ms-1)

Accepted Answer

Given,speed of sound 320 ms-1As we know,Speed=distancetime\text {Speed} = \dfrac{\text {distance}}{\text{time}}Speed=timedistance​The person B hears two sounds of the fired shot, the first one is direct from the gun while another sound comes after reflection from the cliffIn the case of first sound for person B, when t = 2s (i.e., when the gun was fired)Substituting the values in the formula above, we get,320=y - x2y - x=640320 = \dfrac{\text {y - x}}{\text{2}} \[0.5em] \text {y - x} = 640 \[0.5em]320=2y - x​y - x=640Hence, y - x = 640 m     [Equation 1]In the case of second sound for person B, when t = 3s (i.e. after reflection from the cliff)Substituting the values in the formula above, we get,320=y + x3y + x=960320 = \dfrac{\text {y + x}}{\text{3}} \[0.5em] \text {y + x} = 960 \[0.5em]320=3y + x​y + x=960Hence, y + x = 960 m     [Equation 2]Adding equation (1) and (2), we get,y - x = 640     [Equation 1]y + x = 960     [Equation 2]2y=1600y=16002⇒y=800m2y = 1600 \[0.5em] y = \dfrac{1600}{2} \[0.5em] \Rightarrow y = 800 m \[0.5em]2y=1600y=21600​⇒y=800mTherefore, y = 800 mSubstituting the value of y in eqn 1 we get the value of x,800−x=640⇒x=800−640⇒x=160800 - x = 640 \[0.5em] \Rightarrow x = 800 - 640 \[0.5em] \Rightarrow x = 160 \[0.5em]800−x=640⇒x=800−640⇒x=160Therefore, x = 160 mHence,x = 160 m and y = 800 m

Question 10

Question 11On sending an ultrasonic wave from a ship towards the bottom of a sea, the time interval between sending the wave and receiving it back is found to be 1.5 s. If the velocity of the wave in sea water is 1400 ms-1, find the depth of sea.

Accepted Answer

As we know,Speed of sound (V)=Total distance travelled (2d)time interval (t)\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \[0.5em]Speed of sound (V)=time interval (t)Total distance travelled (2d)​Total distance travelled by the sound in going and then coming back = 2dGiven,t = 1.5 sV = 1400 ms-1Substituting the values in the formula above, we get,1400=2d1.5⇒d=1400×1.52⇒d=700×1.5⇒d=1050m1400 = \dfrac{2d}{1.5} \[0.5em] \Rightarrow d = \dfrac{1400 \times 1.5}{2} \[0.5em] \Rightarrow d = 700 \times 1.5 \[0.5em] \Rightarrow d = 1050 m1400=1.52d​⇒d=21400×1.5​⇒d=700×1.5⇒d=1050mTherefore, the depth of the sea = 1050 m

Question 11

Question 12Figure below shows the distance-displacement graph of two waves A and B. Compare (i) the amplitude, (ii) the wavelength of the two waves.

Accepted Answer

(i) Amplitude = maximum displacement from the mean position.From graph,For A, Amplitude = 10 cm andFor B, Amplitude = 5 cmRatio between the two amplitudes —105\dfrac{10}{5}510​ = 21\dfrac{2}{1}12​ = 2 : 1Therefore, the ratio of the two amplitudes = 2 : 1(ii) For A, Wavelength λ1 = 8 cmFor B, Wavelength λ2 = 16 cmRatio between the two Wavelengths —816\dfrac{8}{16}168​ = 12\dfrac{1}{2}21​ = 1 : 2Therefore, the ratio of the two wavelengths = 1 : 2

Question 12

Question 14What is 'SONAR'? State the principle on which it is based.

Accepted Answer

The word 'SONAR' stands for Sound navigation and ranging.The figure given below shows the principle of a SONAR in which, ultrasonic waves are sent in sea water in all directions, from the ship. These waves are received after reflection from an obstacle such as the enemy submarine, iceberg, sunken ship etc. Hence, the distance between the ship and the obstacle can be determined.To find the distance of obstacle from the ship the time interval (t) between the instant when waves are sent and the instant when waves are received, after reflection from the obstacle is measured.The distance d of the obstacle from the source is then —d=Vt2\text{d} = \dfrac{Vt}{2} \[0.5em]d=2Vt​where, V is the speed of ultrasonic wave in water.The depth of sea can also be found by this method. The process is then called echo depth sounding.

Question 13

Question 2The natural frequency of a simple pendulum of length 1.0 m on earth's surface is :[Take g = 9.8 m/s2]0.5 Hz100 Hz50 Hz5 Hz

Accepted Answer

0.5 HzReason — Given,length = 1.0 mg = 9.8 m/s2We know,frequency (f) = 12πgL\dfrac{1}{2π} \sqrt\dfrac{g}{L}2π1​Lg​​Substituting we get,frequency (f) = 12π9.81\dfrac{1}{2π} \sqrt\dfrac{9.8}{1}2π1​19.8​​frequency (f) = 12π×3.13\dfrac{1}{2π} 	imes 3.132π1​×3.13 = 0.49 ≈ 0.5 Hz

Question 14

Question 7If l is the length of the string stretched between its ends, the wavelength of different modes in Figure (1), (2) and (3) will be :2l/2, 2l, 2l/32l/3, 2l/2, 2l2l, 2l/2, 2l/32l, 21/3, 2l/2

Accepted Answer

2l, 2l/2, 2l/3Reason — In fig. 1, we know l = λ2\dfrac{λ}{2}2λ​. Hence, λ = 2lIn fig. 2, l = 2λ2\dfrac{2λ}{2}22λ​ Hence, λ = 2l2\dfrac{2l}{2}22l​In fig. 3, l = 3λ2\dfrac{3λ}{2}23λ​. Hence, λ = 2l3\dfrac{2l}{3}32l​So, we get, the wavelength of different modes in Figure (1), (2) and (3) as 2l, 2l2\dfrac{2l}{2}22l​, 2l3\dfrac{2l}{3}32l​

Question 15

Question 15How do you tune your radio set to a particular station? Name the phenomenon involved in doing so and define it.

Accepted Answer

When we want to tune a radio (or TV) receiver, we merely adjust the values of the electronic components to produce vibrations of frequency equal to that of the radio waves which we want to receive.When both the frequencies match, resonance occurs and only the energy of signal of that particular frequency is received from the waves present in space, leaving the signals of other frequency of the receiver circuit. The signal received is then amplified in the receiver set.The phenomenon involved is Resonance.Resonance is a special case of forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude.

Question 16

Question 3Two waves of the same pitch have amplitudes in the ratio 1 : 3. What will be the ratio of their (i) loudness and (ii) frequencies?

Accepted Answer

Given,a1 : a2 = 1 : 3(i) As we know, that loudness is proportional to the square of the amplitudeLoudness ∝ (amplitude) 2Therefore,L1L2=(a1a2)2\dfrac{L_1}{L_2} = \Big(\dfrac{a_1}{a_2}\Big)^2 \[0.5em]L2​L1​​=(a2​a1​​)2Substituting the values, we getL1L2=(13)2L1L2=19\dfrac{L_1}{L_2} = \Big(\dfrac{1}{3}\Big)^2 \[0.5em] \dfrac{L_1}{L_2} = \dfrac{1}{9}L2​L1​​=(31​)2L2​L1​​=91​Hence, ratio of loudness between the two waves = 1 : 9(ii) Pitch of a wave depends on its frequency and if two waves have same frequencies then their pitch will be same.Hence, ratio of pitch between the two waves = 1 : 1.

Sound

Exercise 7A Multiple Choice Questions

Exercise 7A Numericals

Exercise 7A Short Questions

Exercise 7B Multiple Choice Type

Exercise 7B Short Questions

Exercise 7C Very Short Questions