Stationary Waves

Characteristics of stationary waves

The waveform remains stationary.

Nodes and antinodes are formed alternately.

The points where displacement is zero are called nodes and the points where the displacement is maximum are called antinodes.

Pressure changes are maximum at nodes and minimum at antinodes.

All the particles except those at the nodes, execute simple harmonic motions of same period.

Amplitude of each particle is not the same, it is maximum at antinodes decreases gradually and is zero at the nodes.

The velocity of the particles at the nodes is zero. It increases gradually and is maximum at the antinodes.

Distance between any two consecutive nodes or antinodes is equal to ^λ₂ , whereas the distance between a node and its adjacent antinode is equal to λ/4 .

There is no transfer of energy. All the particles of the medium pass through their mean position simultaneously twice during each vibration.

Particles in the same segment vibrate in the same phase and between the neighbouring segments, the particles vibrate in opposite phase.

Standing waves in strings

In musical instruments like sitar, violin, etc. sound is produced due to the vibrations of the stretched strings. Here, we shall discuss the different modes of vibrations of a string which is rigidly fixed at both ends.

When a string under tension is set into vibration, a transverse progressive wave moves towards the end of the wire and gets reflected. Thus stationary waves are formed.

Sonometer

The sonometer consists of a hollow sounding box about a metre long. One end of a thin metallic wire of uniform cross-section is fixed to a hook and the other end is passed over a pulley and attached to a weight hanger as shown in figure. The wire is stretched over two knife edges P and Q by adding sufficient weights on the hanger. The distance between the two knife edges can be adjusted to change the vibrating length of the wire.

A transverse stationary wave is set up in the wire. Since the ends are fixed, nodes are formed at P and Q and antinode is formed in the middle.

The length of the vibrating l = λ/2

Thus λ = 2l. If n is the frequency of the vibrating segment, then,

n = v/λ = v/2l                     …... (1)

We know that, v = √T/m where T is the tension and m is the mass per unit length of the wire.

Thus,

          …... (2)

Modes of vibration of stretched string

When a wire AB of length l is made to vibrate in one segment then, l = λ₁/2

λ₁ = 2l. This gives the lowest frequency called fundamental frequency, n₁ = v/λ₁

So, n₁ = (1/2l) √T/m                …... (3)

(ii) Overtones in stretched string

If the wire AB is made to vibrate in two segments then l = λ₂/2 +  λ₂/2

So, λ₂ = l

But, n₂ = v/λ₂

So, n₂ = 1/l √T/m = 2n₁          …... (4)

n₂ is the frequency of the first overtone.

Since the frequency is equal to twice the fundamental, it is also known as second harmonic.

Similarly, higher overtones are produced, if the wire vibrates with more segments. If there are P segments, the length of each segment is

l/p = λ_p/2

Or, λ_p = 2l/P

So, frequency, n_p = (P/2l)√T/m = Pn₁          …... (5)

(i.e) P^th harmonic corresponds to (P–1)^th overtone.

Refer this video to know more about standing wave

Laws of transverse vibrations of stretched strings

The laws of transverse vibrations of stretched strings are (i) the law of length (ii) law of tension and (iii) the law of mass.

(i) For a given wire (m is constant), when T is constant, the fundamental frequency of vibration is inversely proportional to the vibrating length (i.e)

n ∝ 1/l

Or, nl = constant

(ii) For constant l and m, the fundamental frequency is directly proportional to the square root of the tension (i.e)

n ∝ √T.

(iii) For constant l and T, the fundamental frequency varies inversely as the square root of the mass per unit length of the wire

(i.e) n ∝ 1/√m.

Problem (JEE Main):

Equations of two progressive waves at a certain point in a medium are given by y₁ = a sin (ωt + φ₁) and y₂ = a sin (ωt + φ₂). If amplitude and time period of resultant wave formed by the superposition of these two waves is same as that of both the waves, then

φ₁ – φ₂ is,

(a) π/3                    (b) 2π/3

(c) π/6                    (d) π/4

Solution:

A² = a₁² + a₂² + 2a₁a₂ cos φ              

Here, φ =  φ₁ – φ₂, A = a₁ = a₂ = a

Substituting these values in equation (1), we get,

cos φ = – ½

Or, φ = 2π/3

Thus, from the above observation we conclude that, option (b) is correct.

A tube closed at one end and containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends, the fundamental frequency that can be excited is (in Hz) (I.I.T. 86)

(a) 1024

(b) 512

(c) 256

(d) 128

Question 2

Velocity of sound in air is 320 m/s. a pipe closed at one end has length of 1 m. neglecting end corrections, the air column in the pipe can resonates for sound of frequency (I.I.T 89)

(a) 80 Hz

(b) 320 Hz

(c) 400 Hz

(d) All (a), (b) & (c)

Question 3

A stationary sound wave is produced in a resonance column apparatus using an electrically excited tuning fork. If P and Q are consecutive nodes, which one of the following statements is correct?

(a) If P is a position of condensation, Q is a position of rarefaction

(b) If P is a position of condensation, Q also is a position of condensation

(c) If P is a position of condensation, Q is a position of normal density (of air)

(d) Both P and Q are positions of normal density (of air)

(e) Both P and Q are positions of rarefaction

Question 4

Which one of the following factors determines the pitch of a sound?

(a) The amplitude of the sound wave

(b) The distance of the sound wave from the source

(c) The frequency of the sound wave

(d) The phase of different parts of the sound wave