‘NCERT Solutions for Class 11 Physics Chapter 15‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 11 Physics Chapter 15 Exercise Solution’ using the download button.
NCERT Class 11 Physics Textbook Chapter 15 With Answer PDF Free Download
Chapter 15: Waves
In the previous chapter, we studied the motion of objects oscillating in isolation. What happens in a system, which is a collection of such objects?
A material medium provides such an example. Here, elastic forces bind the constituents to each other and, therefore, the motion of one affects that of the other.
If you drop a little pebble in a pond of still water, the water surface gets disturbed. The disturbance does not remain confined to one place but propagates outward along a circle.
If you continue dropping pebbles in the pond, you see circles rapidly moving outward from the point where the water surface is disturbed.
It gives a feeling as if the water is moving outward from the point of disturbance.
If you put some cork pieces on the disturbed surface, it is seen that the cork pieces move up and down but do not move away from the center of the disturbance.
This shows that the water mass does not flow outward with the circles, but rather a moving disturbance is created.
Similarly, when we speak, the sound moves outward from us, without any flow of air from one part of the medium to another.
The disturbances produced in the air are much less obvious and only our ears or a microphone can detect them.
These patterns, which move without the actual physical transfer or flow of matter as a whole, are called waves. In this chapter, we will study such waves.
Waves transport energy and the pattern of disturbance has information that propagates from one point to another.
All our communications essentially depend on the transmission of signals through waves.
Speech means the production of sound waves in the air and hearing amounts to their detection. Often, communication involves different kinds of waves.
The aesthetic influence of waves on art and literature is seen from very early times, yet the
first scientific analysis of wave motion dates back to the seventeenth century.
Some of the famous scientists associated with the physics of wave motion are Christiaan Huygens (1629-1695), Robert Hooke, and Isaac Newton.
The understanding of the physics of waves followed the physics of oscillations of masses tied to springs and the physics of the simple pendulum.
Waves in elastic media are intimately connected with harmonic oscillations. (Stretched strings, coiled springs, air, etc)
Now let us consider the propagation of sound waves in the air. As the wave passes through the air, it compresses or expands a small region of the air.
This causes a change in the density of that region, say δρ, this change induces a change in pressure, δp, in that region.
Pressure is force per unit area, so there is a restoring force proportional to the disturbance, just like in spring.
In this case, the quantity similar to the extension or compression of the spring is the change in density.
If a region is compressed, the molecules in that region are packed together, and they tend to move out to the adjoining region.
| Author | NCERT |
| Language | English |
| No. of Pages | 28 |
| PDF Size | 950 KB |
| Category | Physics |
| Source/Credits | ncert.nic.in |
NCERT Solutions Class 11 Physics Chapter 15 Waves
Q1. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Ans:
Given,
Mass of the string, M = 2.50 kg
Tension in the string, T = 200 N
Length of the string, l = 20.0 m
Mass per unit length, μ = M /l = 5/40 = 0.125 kg / m
We know,
Velocity of the transverse wave , v = \sqrt{\frac{T}{\mu }}μT
= \sqrt{\frac{200}{0.125 }}0.125200 = 40 m/s
Therefore, the time taken by the transverse wave to reach the other side is 40 m/s.
Q2. A stone dropped from the top of a tower of height 300 m 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)
Ans:
Given,
Height of the bridge, s = 300 m
Initial velocity of the stone, u = 0
Acceleration, a = g = 9.8 m/s2
Speed of sound in air = 340 m/s
Let t be the time taken by the stone to hit the water’s surface
We know,
s = ut + ½ gt2
300 = 0 + ½ x 9.8 x t2
therefore, t = 7.82 s
The time taken by the sound to reach the bridge, t’= 300/340 = 0.88 s
Therefore, from the moment the stone is released from the bridge, the sound of it splashing the water is heard after = t +t’ = 7.82 s + 0.88 s = 8.7s
Q3. 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 the speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1
Ans:
Given,
Length of the steel wire, l = 12 m
Mass of the steel wire, m = 2.0 kg
Velocity of the transverse wave, v = 343 m/s
Mass per unit length, μ = M /l = 2.10/12 = 0.175 kg / m
We know,
Velocity of the transverse wave , v = \sqrt{\frac{T}{\mu }}μT
Therefore, T = v2 μ
= 3432 x 0.175 = 2.06 x 104 N
Waves Textbook With Solutions PDF Free Download