- Chapter 1- Physical World
- Chapter 2- Units and Measurements
- Chapter 3- Motion in a Straight Line
- Chapter 4- Motion in a plane
- Chapter 5- Laws of motion
- Chapter 6- Work Energy and power
- Chapter 7- System of particles and Rotational Motion
- Chapter 8- Gravitation
- Chapter 9- Mechanical Properties Of Solids
- Chapter 10- Mechanical Properties Of Fluids
- Chapter 11- Thermal Properties of matter
- Chapter 12- Thermodynamics
- Chapter 13- Kinetic Theory
- Chapter 14- Oscillations

Chapter 1- Physical World |
Chapter 2- Units and Measurements |
Chapter 3- Motion in a Straight Line |
Chapter 4- Motion in a plane |
Chapter 5- Laws of motion |
Chapter 6- Work Energy and power |
Chapter 7- System of particles and Rotational Motion |
Chapter 8- Gravitation |
Chapter 9- Mechanical Properties Of Solids |
Chapter 10- Mechanical Properties Of Fluids |
Chapter 11- Thermal Properties of matter |
Chapter 12- Thermodynamics |
Chapter 13- Kinetic Theory |
Chapter 14- Oscillations |

**Answer
1** :

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,

The velocity (v) of the transverse wave in the string is given by the relation:

∴Time taken by the disturbance to reach the other end, t =

**Answer
2** :

Height of the tower, s = 300 m

Initial velocity of the stone, u = 0

Acceleration, a = g = 9.8 m/s^{2}

Speed of sound in air = 340 m/s

The time ( ) taken by the stone to strike the water in the pond can be calculated using the second equation of motion, as:

Time taken by the sound to reach the top of the tower,

Therefore, the time after which the splash is heard,

= 7.82 + 0.88 = 8.7 s

**Answer
3** :

Length of the steel wire, *l* =12 m

Mass of the steel wire, *m* =2.10 kg

Velocity of the transverse wave, *v* =343 m/s

Mass per unit length,

For tension *T*, velocity of thetransverse wave can be obtained using the relation:

∴*T* = *v*^{2} *µ*

= (343)^{2} × 0.175 =20588.575 ≈ 2.06 × 10^{4} N

Use the formula to explain why the speed of sound in air

(a) is independent of pressure,

(b) increases with temperature,

(c) increases with humidity.

**Answer
4** :

(a) Take the relation:

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:

For one mole of an ideal gas, the gas equation can be written as:

PV = RT

P = … (ii)

Substituting equation (ii) in equation (i), we get:

Where,

Mass, M = ρV is a constant

γ and R are also constants

We conclude from equation (iv) that .

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 be the speeds of sound in moist air and dry air respectively.

Let be the densities of moist air and dry air respectively.

Take the relation:

Hence, the speed of sound in moist air is:

And the speed of sound in dry air is:

On dividing equations (i) and (ii), we get:

However, the presence of water vapour reduces the density of air, i.e.,

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.

You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t)where x and t must appear in the combination x – v t or x + v t, i.e. y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:

(a) (x – vt)^{2}

(b)

(c)

**Answer
5** :

Answer: No;

(a) Does not represent a wave

(b) Represents a wave

(c) Does not represent a wave

The converse of the given statement is not true. The essential requirement for a function to represent a travelling wave is that it should remain finite for all values of x and t.

Explanation:

(a) For x = 0 and t = 0, the function (x – vt)^{2 }becomes 0.

Hence, for x = 0 and t = 0, the function represents a point and not a wave.

(b) For x = 0 and t = 0, the function

Since the function does not converge to a finite value for x = 0 and t = 0, it represents a travelling wave.

(c) For x = 0 and t = 0, the function

Since the function does not converge to a finite value for x = 0 and t = 0, it does not represent a travelling wave.

**Answer
6** :

(a) Frequency of the ultrasonic sound, ν = 1000 kHz = 10^{6} Hz

Speed of sound in air, *v*_{a} =340 m/s

The wavelength (λ_{r}) ofthe reflected sound is given by the relation:

(b) Frequency of the ultrasonic sound, ν = 1000 kHz = 10^{6} Hz

Speed of sound in water, *vw* =1486 m/s

The wavelength of the transmitted sound is given as:

=1.49 × 10^{–3} m

A hospital uses an ultrasonic scanner tolocate tumours in a tissue. What is the wavelength of sound in the tissue inwhich the speed of sound is 1.7 km s^{–1}? The operatingfrequency of the scanner is 4.2 MHz.

**Answer
7** :

Speed of sound in the tissue, *v* =1.7 km/s = 1.7 × 10^{3} m/s

Operating frequency of the scanner, ν = 4.2MHz = 4.2 × 10^{6} Hz

The wavelength of sound in the tissue is given as:

A transverse harmonic wave on a string is described by

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?

**Answer
8** :

(a) Yes; Speed = 20 m/s, Direction = Right to left

(b) 3 cm; 5.73 Hz

(c)

(d) 3.49 m

Explanation:

(a) The equation of a progressive wave travelling from rightto left is given by the displacement function:

*y* (*x*, *t*)= *a* sin (ω*t* + *kx* + Φ) … (*i*)

The given equation is:

On comparing both the equations, we findthat equation (*ii*) represents a travelling wave, propagating from rightto left.

Now, using equations (*i*) and (*ii*),we can write:

ω = 36 rad/s and *k* = 0.018m^{–1}

We know that:

and

Also,

*v* = νλ

Hence, the speed of the given travelling wave is 20 m/s.

(b) Amplitude of the given wave, *a* = 3 cm

Frequency of the given wave:

(c) On comparing equations (*i*) and (*ii*), we findthat the initial phase angle,

(d) The distance between two successive crests or troughs isequal to the wavelength of the wave.

Wavelength is given by the relation:

**Answer
9** :

All the waves have different phases.

The given transverse harmonic wave is:

For x = 0, the equation reduces to:

Now, plotting *y* vs. *t* graphsusing the different values of *t*, as listed in the given table.

| 0 |
| ||||||

| 3 | 0 | –3 | 0 |

For *x* = 0, *x* =2, and *x* = 4, the phases of the three waves will get changed.This is because amplitude and frequency are invariant for any change in *x*.The *y*–*t* plots of the three waves are shown in the givenfigure.

For the travelling harmonic wave

y (x, t) = 2.0 cos 2π (10t – 0.0080x + 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) ,

(d)

**Answer
10** :

Equation for a travelling harmonic wave is given as:

*y* (*x*, *t*)= 2.0 cos 2π (10*t* – 0.0080*x* + 0.35)

= 2.0 cos (20π*t* – 0.016π*x* +0.70 π)

Where,

Propagation constant, *k* =0.0160 π

Amplitude, *a *= 2 cm

Angular frequency, *ω*= 20 π rad/s

Phase difference is given by the relation:

(a) For *x* = 4 m = 400 cm

Φ = 0.016 π × 400 = 6.4 π rad

(b) For 0.5 m = 50 cm

Φ = 0.016 π × 50 = 0.8 π rad

(c) For

(d) For

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