Physics. Wave and Sound - 3 Session Session Objectives.

Physics

Wave and Sound - 3

Session

Session Objectives

Session Objective

Intensity of Sound waves - Sound Level

Standing Waves (Wave Function)

Position of Nodes & antinodes in standing waves

Periodic Sound Waves

Periodic Sound Waves

The general form of the wave function is

where n = Integer

y(x, t) f(x vt)

y (x n ), t y(x, t)

In a periodic wave, the displacement repeats

itself after a fixed periodic length , i.e.

Intensity of Sound waves - Sound Level

2 2 2I 2 dvA

where v = Velocity of the wave in that medium

= Frequency of particle oscillation

A = Amplitude

Intensity of Sound waves - Sound Level

Also the intensity level L of a sound of

intensity ‘I’ is represented as

100

L log

where I0 = Zero level of intensity or threshold intensity

12 20I 1 10 W m

Intensity level is generally expressed in ‘decibels’.

Standing Waves (Wave Function)

Two equal sine waves travel along a stretched string(length l) from opposite directions.

1y A sin( t kx)

2y A sin( t kx) 1 2y y y (2A cos kx) sin t

Position of Nodes in standing waves

For nodes, amplitude of the particles is zero.

2A coskx 0

1kx n

2 2

1or x n

2 2where n is an integer.

coskx 0

Position of Antinodes in standing waves

For antinodes, amplitude of the particles

is maximum.

2A cosk x is max imum

nor x

2

coskx 1

kx n

Class Test

Class Exercise - 1

A sine wave is traveling in a medium. A

particular particle has zero

displacement at a certain instant. The

particle closest to it having zero

displacement is at a distance of

(a) (b)4 3

(c) (d)2

Solution

Hence answer is (c).

Particles which are at distance

from each other, have same

displacement.

2

Class Exercise - 2

A sonometer wire of length vibrates in

fundamental mode when excited by a

tuning fork of frequency 416 Hz. If the

length is doubled, keeping other things

same, the string will vibrate with a

frequency of

(a) 416 Hz (b) 208 Hz

(c) 832 Hz (d) Stop vibrating

' '

1 Tn

2 m

Solution :-

Hence answer is (a).

Class Exercise - 3

A standing wave is produced on a

string clamped at one end and free

at the other. The length of the string

must be an integral multiple of

(a) (b)2

(c) (d)3 3

This is condition for standing waves.

Solution :-

Hence answer is (d).

Class Exercise - 4

Two strings A and B, made of same

material, are stretched by same tension.

The radius of string A is double the radius

of B. The transverse wave travels on A

with speed VA and on B with speed

VB. The ratio is

(a) (b) 2

(c) (d) 4

A

B

V

V1

2

1

4

Solution

Hence answer is (a).

1

Speed vr

Class Exercise - 5

The amplitude of sound is doubled and

the frequency is reduced to one-fourth.

The intensity of sound at the same point

will be

(a) increased by a factor of 2

(b) increased by a factor of 4

(c) decreased by a factor of 2

(d) decreased by a factor of 4

Solution

Hence answer is (d).

2 2I A

Class Exercise - 6

A standing wave having three nodes

and two antinodes is formed between

two atoms having a distance of 1.21 Å

between them. The wavelength of the

standing wave is

(a) 1.21 Å (b) 2.42 Å

(c) 6.05 Å (d) 3.63 Å

Solution

Hence answer is (a).

2 2

In this case

Class Exercise - 7

In a sinusoidal wave, the time required

for a particular point to move from the

maximum displacement to zero

displacement is 0.170 s. The frequency

of the wave is

(a) 1.47 Hz (b) 0.36 Hz

(c) 0.73 Hz (d) 2.94 Hz

Solution

Hence answer is (a).

T = 4 × 0.170 = 0.68 s

1 1

1.47 HzT 0.68

Class Exercise - 8

Particle displacements (in centimeters)

in a standing wave are given by

.

The distance between a node and the

next antinode's is

(a) 2.5 cm (b) 5 cm

(c) 7.5 cm (d) 10 cm

y(x, t) 2sin(0.1 x)cos(100 t)

Solution

Hence answer is (b).

Distance between a node and

an antinode is .4

Class Exercise - 9

Transverse waves of same frequency are

generated in two steel wires A and B.

The diameter of A is twice that of B and

tension in A is half that in B. The ratio of

velocities of waves in A and B is

(a) 1 : 2 (b)

(c) (d) 1: 2

1: 2 2

3 : 2 2

Solution

Hence answer is (c).

2

Tv

r

BA B A

Tr 2r ; T

2

2A A B

B B A

v T r

v T r

Putting the values, we get

A

B

v 1 1

v 8 2 2

Class Exercise - 10

y acos(k x t)A wave represented by is

superimposed with another wave to form

a standing wave, such that point x = 0 is

a node. Equation of the other wave is

(a) (b)

(c) (d)

a sin(kx t) acos(k x t)

acos(k x t) a sin(k x t)

Solution

Hence answer is (b).

For formation of stationary waves two

waves should travel in opposite

directions. Hence, the waves are (a)

and (c) but (a) is not fulfilling the given

condition completely.

Thank you