Chapter 11 (Sound) - Past Paper PDF

## Taleem City ### **Chapter # 11 (Sound)** - Choose the correct answer from the following choices: 1. Which is an example of a longitudinal wave? - Sound wave - Light wave - Radio wave - Water wave 2. How does sound travel from its source to your ear? - By ch...

## Taleem City ### **Chapter # 11 (Sound)** - Choose the correct answer from the following choices: 1. Which is an example of a longitudinal wave? - Sound wave - Light wave - Radio wave - Water wave 2. How does sound travel from its source to your ear? - By changes in air pressure - By vibrations in wires or strings - By electromagnetic wave - By infrared waves 3. Which form of energy is sound? - Electrical - Mechanical - Thermal - Chemical 4. Astronauts in space need to communicate with each other by radio links because: - Sound waves travel very slowly in space - Sound waves travel very fast in space - Sound waves cannot travel in space - Sound waves have low frequency in space 5. The loudness of a sound is most closely related to its - Period - Frequency - Amplitude - Wavelength 6. For a normal person, audible frequency range for sound wave lies between. - 10Hz to 10kHz - 20Hz to 20kHz - 25Hz to 25kHz - 30Hz to 30kHz 7. When the frequency of a sound wave is increased, which of the following decrease? - I only - III only - I and II only - I and III only ### **Answer:** 1. a 2. a 3. b 4. c 5. c 6. c 7. c ### **Review Questions** - Write short answers of the following questions: 1. What is the necessary condition for the production of sound? - Sound is produced by vibrating bodies. Due to vibration of bodies, the air around them also vibrates and the air vibrations produces sensation of sound in air. 2. What is the effect of the medium on the speed of sound? In which medium sound travels more faster; air, solid or liquid? Justify your answer. - Every medium has distinct density. Speed of sound increases with the increase in density. So, the speed of sound is maximum in solids due to their high density. Speed of sound in solids is about fifteen times than that in gases and speed of sound in liquid is five times than gases. 3. How can you prove the mechanical nature of sound by a simple experiment? - Sound waves require a material medium for their propagation, therefore, they are mechanical waves. - Experiment: Suspend an electric bell in a bell jar with a help of two wires connected to a power supply. When we switch ON the power supply, an electric bell will begin to ring. Now pump out air from the jar by a vacuum pump. Again switch ON, no sound is heard in the absence of air as a medium. Now start pumping out air from the jar, the sound of bell starts becoming more and more feeble and eventually dies out. It is concluded that sound waves can only propagate in the presence of air (medium). 4. What do you understand by the longitudinal wave? Describe the longitudinal nature of sound waves. - Longitudinal waves: “In a longitudinal wave, particles of the medium move back and forth along the direction of propagation of the wave.” - Explanation: Propagation of sound waves produced by a vibrating tuning fork can be understood by a vibrating tuning fork as shown in the figure. Before the vibration of tuning fork, the density of air molecules on the right side is uniform (Figure a). When the right prong of the tuning fork moves from the mean position O to B (Figure b), it exerts some pressure on the adjacent layer of air molecules and produces a compression and after some time rarefaction is also produced thus, this sound wave shows the longitudinal nature. 5. Sound is a form of wave. List at least three reasons to support the idea that sound is a wave. - Sound shows reflection like waves. - Sound shows refraction like waves. - Sound shows diffraction like waves. 6. We know that waves manifest the phenomenon of reflection, diffraction, and refraction. Does sound also manifest these characteristics? - Yes, sound waves also manifest these characteristics. 7. What is the difference between the loudness and intensity of sound? Derive the relationship between the two. - Loudness of sound:“It is the characteristics of sound by which loud and faint sound can be distinguished.” - Intensity of sound: “Sound energy passing per second through a unit area held perpendicular to the of sound wave is called the intensity of sound.” - Relationship between loudness and intensity of sound: Loudness (L) is directly proportional to the logarithm of intensity (I). - L α log(I). L = K log I. Here, K is constant. 8. On what factors does the loudness of sound depend? - Factors: Loudness of sound depends upon the number of factors. Some of them are given below: - Amplitude of vibrating body - Area of vibrating body - Distance from vibrating body 9. What do you mean by the term intensity level of the sound? Name and define the unit of intensity level of sound. - Intensity level of the sound: “The difference between the loudness L of unknown sound and loudness Lo is called intensity level of sound.” - Unit: The unit of intensity level of sound is a bel. - Bel is a bigger unit while decibel is a smaller unit. - 1 bel = 10dB. 10. What are the units of loudness? Why do we use a logarithmic scale to describe the range of the sound intensities we hear? - Unit of Loudness: Loudness depends upon the physical condition of the listener. It has no specific units. It is measured in terms of intensity level whose unit is (bel). - The use of a logarithmic scale is due to the following law i.e. Loudness is directly proportional to the logarithm of intensity. So, we use a logarithmic scale. 11. What is the difference between frequency and pitch? - Frequency: “The number of waves passing through a point in unit time is called frequency.” - Pitch: “It is the characteristics of sound by which we distinguish between a shrill an a grave sound.” - Relation between frequency and pitch: Frequency is directly proportional to pitch. 12. Describe the effect of change in amplitude on loudness and the effect of change in frequency on pitch of sound. - If the amplitude of a vibrating body increases, then loudness also increases, and vice versa. Similarly, if frequency increases pitch also increases, and vice versa. 13. If the pitch of sound is increased, what are the changes in the following: - The frequency - The wavelength - The wave velocity - The amplitude of a wave - If the pitch of sound is increased, frequency also increases. - Wavelength decreases when the pitch of sound increases. - Wave velocity remains the same. - Amplitude doesn’t change. 14. If we clap or speak in front of a building while standing at a particular distance, we rehear our sound after some time. Can you explain how does it happen? - This sound which we hear is called an echo and is a result of reflection of sound from the surface. When sound is incident on the surface of a medium, it bounces back into the first medium. This phenomenon is called echo or reflection of sound. 15. What is the audible frequency range for human ear? Does this range vary with the age of people? Explain. - Audible frequency range is from 20Hz to 20,000Hz and it varies with the age of people. 16. Explain that noise is a nuisance. - Noise is a nuisance: Some sounds produce unpleasant effects on our ears, such as the sound of machinery, the slamming of a door, and the sound of traffic in big cities. Sound which is jarring and unpleasant effect on our ears is called noise. Noise corresponds to irregular and sudden vibrations produced by some sounds. Noise has negative effects on human health as it can cause conditions such as hearing loss, sleep disturbances, aggression, hypertension, high stress levels. Noise can also cause accidents by interfering with communication and warning signals. 17. Describe the importance of acoustic protection. - Importance of Acoustics Protection: - Reflection of sound is less prominent if the surface is soft and irregular, but it is more prominent on rigid and smooth surfaces. - Soft porous material, such as draperies and rugs absorb a large amount of sound energy and, thus quit echoes and softening noises. - By using soft and sound insulating materials such as curtains, carpets, and glazed windows, we can reduce the level of noise pollution. - If the surface of the classrooms and public halls are too absorbent, then the sound level is low for the audience. 18. What are the uses of ultrasound in medicine? - Uses of Ultrasound: - In the medical field, ultrasonic waves are used to diagnose and treat different ailments. - Powerful ultrasound is now being used to remove blood clots formed in the arteries. - Ultrasound can also be used to get the pictures of thyroid gland for diagnostic purposes. ### **Conceptual Questions** - Why two tin cans with a string stretched between them could be better way to communicate than merely shouting through the air? - It is due to the fact that the speed of sound is 15 times higher in solids than air. So, it is easier to communicate through tin cans. - The other reason is that, it avoids spreading of the sound waves in air. - We can recognize persons speaking with the same loudness from their voice. How is this possible? - We can recognize persons due to a difference in the quality of their sounds because every person has unique quality of sound. - You can listen to your friend around a corner, but you cannot see him/her. Why? - Diffraction of sound is more prominent than diffraction of light, as light waves have smaller wavelength than sound waves. So, you can’t see your friend at a round corner but listen to him/her. - Why must the volume of a stereo in a room with wall-to-wall carpet be tuned higher than in a room with a wooden floor? - The reflection of sound waves in a wooden floor is maximum so, the sound will be higher. On the other hand, in a carpeted room reflection of sound waves is minimum so, the sound will be lower. - A student says that the two terms speed and frequency of the wave refer to the same thing. What is your response? - Speed is the distance covered by waves in unit time, while frequency is the number of waves passing from a point in unit times, so, they are two different quantities. But the time factor is similar in both quantities. - Two people are listening to the same music at the same distance. They disagree on its loudness. Explain how this could happen? - Loudness depends upon the physical conditions of the listener so, the sound appears louder to a person with a sensitive ear than a person with defective ears. - Is there any difference between echo and reflection of sound? Explain. - There is no difference between echo and reflection of sound. Because when sound falls on the surface of a medium then, it bounces back to the first medium, this is called reflection of sound or echo of sound. - Will two separate 50dB sounds together constitute a 100dB sound? Explain. - No, since the decibel scale is not a linear scale but a logarithmic scale, therefore, they cannot be added simply. Hence, two separate 50dB sounds together would not constitute a 100dB sound. - Why is ultrasound useful in medical fields? - Uses of Ultrasound: - In medical fields, ultrasonic waves are used to diagnose and treat different ailments. - Powerful ultrasound is now being used to remove blood clots formed in the arteries. - Ultrasound can also be used to get the pictures of thyroid gland for diagnostic purposes. ### **Important Formulas** - Intensity level of sound = 10 log(I/Io) (dB) - f = n/t - T = 1/f - V = d/t - S = vt ### **Values & Units** - Unit of intensity = I = Wm−2 - Unit of sound level = L − Lo = dB (decibel) - Or bel (larger unit) - Unit of wavelength = λ = m (meter) - Unit of frequency = f = Hz (Hertz) - log 10 = 1 - Intensity of Faintest sound = I = 10−12Wm−2 ### **Numericals** 1. A normal conversation involves sound intensities of about 3.0×10−6 Wm−2. What is the decibel level for this intensity? What is the intensity of the sound for 100 dB? **Given Data** - Intensity of faintest sound = Io = 10−12 Wm−2 - Intensiy of sound = I = 3.0×10−6 Wm−2 **To Find** - (a) Decibel level of normal conversation - (b) Intensity of sound **Solution:** - (a) Intensity level of sound: - Intensity level of normal conversation =10 log(I/Io) (dB) - Intensity level of normal conversation = 10 log(3.0×10−6/10−12) (dB) - Intensity level of normal conversation = 10 log (3.0 × 10−6+ 12) (dB) - Intensity level of normal conversation = 10 [log 3.0 × 106] (dB) - Intensity level of normal conversation = 10 [log 3.0 + log 106] (dB) - Intensity level of normal conversation = 10 [0.4771 + 6(1)] (dB) - Intensity level of normal conversation = 10 [0.4771 + 6] (dB) - Intensity level of normal conversation = 10 [6.4771] (dB) - Intenseity level of sound = 64.8dB - (b) Intensity level of a sound = 100dB 2. If at Anarkali bazaar Lahore, the sound intensity level is 80 dB, what will be the sound intensity there? **Given Data** - Sound level = 80dB **To Find** - Intensity = I = ? **Solution** - Sound level = 10log(I/Io) (dB) - 80 = 10 log(I/10−12) (dB) - 80 = 10 log I − 10log 10−12 - 80 = 10(log I − log 10−12) - 80/10 = log I − log 10−12 - 8 = log I − (-12)log 10 - 8 = log I + 12 - 8 − 12 = log I - -4 = log I - log I = 4.0000 - I = Anti log(4.0000) - I = 0.0001 Wm−2 - I = 10−4 Wm−2 3. At a particular temperature, the speed of sound in air is 330ms−1. If the wavelength of a note is 5cm, calculate the frequency of the sound wave. Is this frequency lie in the audible range of the human ear? **Given Data** - Speed of sound = v = 330ms−1 - Wavelength of note = λ = 5cm = (5/100) m = 0.05m **To Find** - Frequency of sound wave = f = ? **Solution:** - v = fλ - 330 = f(0.05) - 330 / 0.05 = f - f = 660 Hz - f = 6.6 × 103 Hz - Yes, this frequency lies in the audible range. 4. A doctor counts 72 heartbeats in 1 min. Calculate the frequency and period of the heartbeats. **Given Data** - No of heart beats = n = 72 - Time = t = 1 min = 60 sec **To Find** - (i) Frequency - (ii) Time period **Solution:** - (i) Frequency = f = ? - f = n/t - f = 72/60 - f = 1.2 Hz - (ii) Time period = T = ? - Time period = T = 1/f - T = 1 / 1.2 - T = 0.83s 5. A marine survey ship sends a sound wave straight to the seabed. It receives an echo 1.5 s later. The speed of sound in seawater is 1500ms−¹. Find the depth of the sea at this position. **Given Data** - Time for echo = T = 1.5s - Time for the wave to reach the seabed - T = (T/2) = 1.5 /2 - t = 0.75s - Speed of sound = v = 1500ms−1 **To Find** - Depth of sea = S = ? **Solution** - S = vt - S = 1500 × 0.75 - S = 1125m 6. A student clapped his hands hear a cliff and heard the echo after 5s. What is the distance of the cliff from the person if the speed of the sound, v, is taken as 346 ms−¹? **Given Data:** - Time for echo = T = 5s - Time of sound to reach the cliff = t = 5/2 = 2.5 - Speed of sound = v = 346ms−1 - Distance of cliff from person = S = ? **Solution:** - S = vt - S = 346 × 2.5 - S = 865m 7. A ship sends out ultrasound that returns from the seabed and is detected after 3.42s. If the speed of ultrasound through seawater is 1531ms−¹, what is the distance of the seabed from the ship? **Given Data:** - Time in which ultrasound return from seabed to ship = T = 3.42s - Speed of sound in water = v = 1531m/sec - Distance from the ship to seabed = d =? **Solution:** - d = vt - d = (1531) (3.42) - d = 5236m - Required distance would be half of the covered distance which is the depth of seabed from the ship. - d = 5236/2 - Depth of seabed = 2618m 8. The highest frequency sound humans can hear is about 20,000 Hz. What is the wavelength of sound in air at this frequency at a temperature of 20°C? What is the wavelength of the lowest sounds we can hear of about 20Hz? Assume the speed of sound in air at 20°C is 343 ms−1. **Given Data** - Frequency of loud sound = f1 = 20,000 Hz - Speed of sound = v = 343ms−1 - Frequency of lowest sound = f2 = 20Hz **To Find** - (i) Wavelength of loud sound = λ1 = ? - (ii) Wavelength of lowest sound = λ2 = ? **Solution** - (i) Wavelength of loud sound = λ1= ? - v = f1λ1 - 343 = (20000)λ1 - 343 / 20000 = λ1 - λ1 = 0.01715m - λ1 = 1.7 × 10-2m - (ii) Wavelength of lowest sound = λ2 = ? - v = f2λ2 - 343 = (20)λ2 - 343 / 20 = λ2 - λ2 = 17.2m 9. A sound wave has a frequency of 2 kHz and a wavelength of 35 cm. How long will it take to travel 1.5 km? **Given Data** - Frequency of sound wave = f = 2kHz = 2 × 103Hz - Wavelength of sound wave = λ = 35cm = (35/100) m = 0.35m - Distance covered by sound = S = 1.5 km = 1.5 × 103 m **To Find** - Time = T = ? **Solution:** - (i) v = fλ - v = 2 × 103 × 0.35 - v = 0.7 × 103ms−1 - v = 700ms−1 - (ii) S = vt - 1.5 × 103 = (0.7 × 103)t - 1.5 × 103 / (0.7 × 103) = t - t = 2.1sec