Note is a sound
A. Of mixture of several frequencies
B. Of mixture of two frequencies only
C. Of a single frequency
D. Always unpleasant to listen
A note of definite pitch is one where a listener can possibly (or relatively easily) discern the pitch.
A key of a mechanical piano struck gently and then struck again but much harder this time. In the second case
A. Sound will be louder but pitch will not be different
B. Sound will be louder and pitch will also be higher
C. Sound will be louder but pitch will be lower
D. Both loudness and pitch will remain unaffected
Yes, the pitch will not change because loudness depends on amplitude and not the pitch. The pitch depends on the frequency of the sound. By hitting it hard it will make its amplitude higher. But the frequency will be same. And the point is to be noted is that the loudness depends on the amplitude not the frequency.
In SONAR, we use
A. Ultrasonic waves
B. Infrasonic waves
C. Radio waves
D. Audible sound waves
Sonar (originally an acronym for Sound Navigation and Ranging) is a technique that uses sound propagation (usually underwater, as in submarine navigation) to navigate, communicate with or detect objects on or under the surface of the water.
Ultrasonic waves: The waves with frequencies higher than 20,000Hz are termed as ultrasonics.
Sound travels in air if
A. Particles of medium travel from one place to another
B. There is no moisture in the atmosphere
C. Disturbance moves
D. Both particles as well as disturbance travel from one place to another
Sound travels through the air in the form of vibrations. These vibrations cause particles of air to compress together, creating disturbance and this causes the air around them to move in such a way that they are driven in waves away from the source.
When we change feeble sound loud sound we increase its
A. Frequency
B. Amplitude
C. Velocity
D. Wavelength
With sound waves, amplitude is the extent to which air particles are displaced, and this amplitude of sounder sound amplitude is experienced as the loudness of sound.
Amplitude is directly proportional to the loudness of the sound.
In the curve (Fig.12.1) half the wavelength is
A. A B
B. B D
C. D E
D. A E
As one wavelength = one crest + one trough,
half the wavelength = one crest or
one trough = AC (or CE).
Further, AC = AB + BC = BC + CD = BD (as AB = CD)
Earthquake produces which kind of sound before the main shock wave begins
A. Ultrasound
B. Infrasound
C. Audible sound
D. None of the above
This is the reason why most probably animals are able to sense the earthquake much before its occurrence. And Also, the infrasound is used by the animals to communicate like rhinoceros and whale.
Infrasound can be heard by
A. Dog
B. Bat
C. Rhinoceros
D. Human beings
Infrasound, sometimes referred to as low-frequency sound, is sound that is lower in frequency than 20 Hz or cycles per second, the "normal" limit of human hearing.
Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting
A. Intensity of sound only
B. Amplitude of sound only
C. Frequency of the sitar string with the frequency of other musical instruments
D. Loudness of sound
The sitarist adjusts the frequencies of the sitar strings by adjusting the tension in them.
The given graph (Fig.12.2) shows the displacement versus time relation for a disturbance travelling with velocity of 1500 m s-1. Calculate the wavelength of the disturbance.
From the graph we get,
Time-period, T = 2 * 10-6s.
Frequency is the number of wave cycles or revolutions per second. The formula for period (T) in terms of frequency is given by:
Frequency, √ = = 5 * 105 Hz.
Wavelength Formula for any wave is given by:
Wavelength, λ =
Velocity of the wave = 1500 ms-1
Therefore, λ = 1500 * 2 * 10–6
= 3 * 103 * 10-6
= 3 * 10-3 m
Which of the above two graphs (a) and (b) (Fig.12.3) representing the human voice is likely to be the male voice? Give reason for your answer.
Graph (a) represents the male voice. Usually the male voice has less pitch as compared to female.
The faster the vibration of the source, the higher is the frequency and the higher is the pitch, as shown in fig. (b). Thus, a high pitch sound corresponds to more number of compression and rarefaction passing a fixed point per unit time.
In fig. (a), the vibration of the source is slower, the frequency is lower and therefore pitch is also lower.
A girl is sitting in the middle of a park of dimension 12 m × 12 m. On the left side of it there is a building adjoining the park and on right side of the park, there is a road adjoining the park. A sound is produced on the road by a cracker. Is it possible for the girl to hear the echo of this sound? Explain your answer.
To hear a distinct echo the time interval between the original sound and the reflected one must be at least 0.1 s.
If we take the speed of sound to be 344 m/s at a given temperature, say at 2o C in air, the sound must go to the obstacle and reach back the ear of the listener on reflection after 0.1 s. Hence, the total distance covered by the sound from the point of generation to the reflecting surface and back should be at least (344m/s) * 0.1s = 34.4 m.
Thus, for hearing distinct echoes, the minimum distance of the obstacle from the source of sound must be half of this distance, that is, 17.2 m.
But in this case, the distance traveled by the sound reflected from the building and then reaching to the girl will be (6 + 6) = 12 m. That is, the minimum distance of the obstacle from the source of sound is 6 m, which is much smaller than the required distance (17.2 m). Therefore, no echo can be heard.
Why do we hear the sound produced by the humming bees while the sound of vibrations of pendulum is not heard?
The audible range of sound for human beings extends for human beings extends from about 20 Hz to 20000 Hz, therefore, the sounds of frequencies below 20 Hz (called infrasonic sound or infra sound) are not audible and the sounds having frequencies higher than 20 kHz (called ultrasonic sound or ultrasound) are not audible to human beings.
The frequency of the vibration of the wings of a bee is much higher than the oscillations of a pendulum. Humming bees have the sound of higher frequency and it is in the audible range of sound for human beings.
If any explosion takes place at the bottom of a lake, what type of shock waves in water will take place?
If any explosion takes place at the bottom of a lake than the longitudinal waves will be produced.
Sound produced by a thunderstorm is heard 10 s after the lightning is seen. Calculate the approximate distance of the thunder cloud. (Given speed of sound = 340 m s-1)
Speed of sound = 340 ms-1
Time = 10 s
Speed =
Distance = Speed * Time
= 340 * 10 = 3400 m
For hearing the loudest ticking sound heard by the ear, find the angle x in the Fig.12.4.
From the given figure,
We get, Angle of incidence = Angle of reflection
Thus, 50o + x + x + 50o = 180o
2x + 100o = 180o
2x = 180o – 100o
2x = 80o
x = 40o
Therefore, Angle of reflection = 40o
Why is the ceiling and wall behind the stage of good conference halls or concert halls made curved?
The ceilings of concert halls, conference halls and cinema halls are curved so that sound after reflection reaches all corners of the hall. Sometimes a curved soundboard may be placed behind the stage so that the sound, after reflecting from the sound board, spreads evenly across the width of the hall.
Represent graphically by two separate diagrams in each case
(i) Two sound waves having the same amplitude but different frequencies?
(ii) Two sound waves having the same frequency but different amplitudes.
(iii) Two sound waves having different amplitudes and also different wavelengths.
(i) Two sound waves having the same amplitude but different frequencies:
(ii) Two sound waves having the same frequency but different amplitude:
(iii) Two sound waves having different amplitudes and also different wavelengths:
Establish the relationship between speed of sound, its wavelength and frequency. If velocity of sound in air is 340 m s-1, calculate
(i) Wavelength when frequency is 256 Hz.
(ii) Frequency when wavelength is 0.85 m.
Frequency and time period are related as follows: √ = (I)
We know, Speed, v =
v =
Here λ is the wavelength of the sound wave. It is the distance traveled by the sound wave in one time-period (T) of the wave.
Thus,
v = λ√ (From I)
That is, Speed = wavelength * Frequency
The speed of sound remains almost the same for all frequencies in a given medium under the same physical conditions.
(i) Given, Speed (v) = 340 m/s
Frequency (√) = 256 Hz
Wavelength (λ) =?
v = λ√
340 = 256λ
λ = 1.33 m
(ii) Given, Speed (v) = 340 m/s
Wavelength (λ) = 0.85 m
Frequency (√) =?
v = λ√
340 = √ (0.85)
√ = 400 Hz
Draw a curve showing density or pressure variations with respect to distance for a disturbance produced by sound. Mark the position of compression and rarefaction on this curve. Also define wavelengths and time-period using this curve.
Wavelength is the distance between two consecutive compressions or two consecutive rarefaction of a wave sound. Wave length is represented by Greek letter λ (lambda). The SI unit of wavelength is metre (m).
Time period is the time taken to travel the distance between any two consecutive compression or rarefaction from a fixed point. The time period of sound wave is represented by letter ‘T’. The SI unit of time period is second (s).