Class 11 Physics - Chapter Waves NCERT Solutions | Use the formula v = √ γP/&rh
Use the formula v = √ γP/ρ to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.
(a) Take the relation:
v = √ γP/ρ .... (i)
where density, ρ = Mass / Volume = M / V
Hense eqation one reduces to:
v = √ γPV/M ....(ii)
Now from the ideal gas equation for n = 1:
PV = RT
For constant T, PV = Constant
Since both M and γ are constants, v = Constant
Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.
(b) Take the relation:
v = √ γP/ρ .... (i)
For one mole of an ideal gas, the gas equation can be written as:
PV = RT
P = RT / V … (ii)
Substituting equation (ii) in equation (i), we get:
v = √ γRT / Vρ = √γRT / M ....(iii)
Where, Mass, M = ρV is a constant
γ and R are also constants
We conclude from equation (iii) that v ∝ √T
Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.
(c) Let Vm and Vd be the speeds of sound in moist air and dry air respectively.
Let ρm and ρd be the densities of moist air and dry air respectively.
Take the relation:
v = √ γP/ρ
Hence, the speed of sound in moist air is:
vm = √ γP/ρm .....(i)
And the speed of sound in dry air is:
vd = √ γP/ρd .....(ii)
On dividing equations (i) and (ii), we get:
However, the presence of water vapour reduces the density of air, i.e.,
ρd < ρm
therefore, Vm > Vd
Hence, the speed of sound in moist air is greater than it is in dry air. Thus, in a gaseous medium, the speed of sound increases with humidity.
More Questions From Class 11 Physics - Chapter Waves
- Q:-
A transverse harmonic wave on a string is described by
y(x,t) = 3.0 sin [36t + 0.018x + π /4]
Where x and y are in cm and t in s. The positive direction of x is from left to right.
(a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
- Q:- Explain why (or how):
(a) In a sound wave, a displacement node is a pressure antinode and vice versa,
(b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any eyes,
(c) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
(e) The shape of a pulse gets distorted during propagation in a dispersive medium. - Q:-
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium.
(a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation?
(b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to or 0.05 Hz?
- Q:-
A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s-1? (g= 9.8 m s-2)
- Q:-
A travelling harmonic wave on a string is described by
y(x,t) = 7.5sin [0.0050x + 12t + π/4]
(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.
- Q:-
A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m s-1 and in water 1486 m s-1.
- Q:-
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s-1.
- Q:-
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is 340 m s-1).
- Q:-
For the travelling harmonic wave
y (x, t) = 2.0 cos 2π (10t - 0.0080 x + 0.35)
Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) 4 m,
(b) 0.5 m,
(c) λ / 2 ,
(d) 3λ / 4
- Q:-
A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s-1.
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Some of the most profound statements on the nature of science have come from Albert Einstein, one of the greatest scientists of all time. What do you think did Einstein mean when he said : “The most incomprehensible thing about the world is that it is comprehensible”?
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Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3Å.
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The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.
(Fig. 3.28)
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On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h-1. For an observer on a stationary platform outside, what is the
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(b) speed of the child running opposite to the direction of motion of the belt ?
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(Figure 3.26)
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Fill in the blanks
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