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Sound : NCERT Exemplar

Short Answer Type Questions: -

Q.1     The given graph shows the displacement versus time relation for a disturbance travelling with a velocity of 1500 m/s. Calculate the wavelength of the disturbance. Sol.

Time to complete 1 wave = $2\mu s$ (Time between two consecutive crests)
= $2 \times {10^{ - 6}}s.$
Velocity of wave = 1500 m/s.
v= ${\lambda \over t}$
${\mkern 1mu} {\mkern 1mu} \lambda = V \times T$
= 1500 × 2 × ${10^{ - 6}}m$
$3 \times {10^3}{10^{ - 6}}m = 3{\mkern 1mu} \times {10^{ - 3}}m$ Sol.

Male voice have generally lower pitch and hence lower frequency of sound. From the graph, wave "a" is likely to be of a male voice.

Q.3     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.
Sol.

For the echo to be heard, the reflected sound should reach after 0.1 s. If we assume the speed of sound to be 340 m/s , the minimum distance for the reflecting surface should be 17 m. (t= ${{v \times t} \over 2}$).
Since the distance of the wall are much smaller no echo will be heard.

Q.4      Why do we hear the sound produced by the humming bees while the sound of vibrations of pendulum is not heard?
Sol.

The frequency of sound produced by humming bear is greater than 20 H3, while the frequency of vibrations produced by pendulum is less than 20 H3.
Audible range is 20 H3 to 20000 H3, so bees vibration can be heard, while the pendulum's vibration cannot be heard.

Q.5      If any explosion takes place at the bottom of a lake, what type of shock waves in water will be produced?
Sol.      Longitudinal waves will be produced.

Q.6    The 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 ms–1)
Sol.

Speed of sound V = 340 m/s
Time for to reach, t = 10 s.
Distance $= V \times t = 340 \times 10 = 3400m$ Sol.

Since angle of incidence = $90^\circ - {50^o} = {40^o}$

Long Answer Type Questions: -

Q.8      Represent graphically by two separate diagrams in each case.
(a) Two sound waves having the same amplitude but different frequencies.
(b) Two sound waves having the same frequency but different amplitudes.
(c) Two sound waves having different amplitudes and also different wavelengths.
Sol.

(a)  (b)  (c)  Q.9      Establish the relationship between speed of sound, its wavelength and frequency. If velocity of sound in air is 340 ms–1, calculate.
(a) wavelength when frequency is 256 Hz.
(b) frequency when wavelength is 0.85 m

Sol.       Velocity = ${{Distance} \over {Time}}$

A wave constitutes a compression and rarefaction.The distance between one wave and another is called it wavelength denoted by $\lambda$.
The time for one wave to move through $\lambda$is called time period
Velocity of wave $V = {\lambda \over T} = \lambda \times \nu$ since $\left( {\nu = {1 \over T}} \right)$
(a) $v = \nu {\mkern 1mu} \lambda$
$\lambda = {{340{\mkern 1mu} m/s} \over {256{\mkern 1mu} Hz}} = 1.33{\mkern 1mu} m$
(b) $\nu = {\upsilon \over \lambda } = {{340m/s} \over {0.85m}} = 400Hz$

Sol. Distance between two consecutive crests or trough is called wavelength.
Time to travel through this distance is time period.