__Modes of Oscillations:-__

**ν= (vn)/2L**where v=speed of the travelling wave, L=length of the string, n=any natural number.

**First Harmonic:-**- For n=1, mode of oscillation is known as
**Fundamental mode**. - Therefore ν
_{1}=v/(2L).This is the lowest possible value of frequency. - Therefore ν
_{1}is the lowest possible mode of the frequency. - 2 nodes at the ends and 1 antinode.

- For n=1, mode of oscillation is known as
**Second Harmonic:-**- For n=2, ν
_{2}=(2v)/ (2L) =v/L - This is second harmonic mode of oscillation.
- 3 nodes at the ends and 2 antinodes.

- For n=2, ν
**Third Harmonic:-**- For n=3,ν
_{3 }= (3v)/ (2L). - This is third harmonic mode of oscillation.
- 4 nodes and 3 antinodes.

- For n=3,ν

** Problem:- **Find the frequency of note emitted (fundamental note) by a string 1m long and stretched by a load of 20 kg, if this string weighs 4.9 g. Given, g = 980 cm s

^{–2}?

__Answer:-__

L = 100 cm T = 20 kg = 20 × 1000 × 980 dyne

m= 4.9/100 = 0.049 g cm^{-1}

Now the frequency of fundamental note produced,

ν= (1/2L) √ (T/m)

ν = 1/ (2×100) √(20x1000x980)/(0.049)

=100Hz

** Problem:- **A pipe 20 cm long is closed at one end, which harmonic mode of the pipeis resonantly excited by a 430 Hzsource? Will this same source can bein resonance with the pipe, if both ends are open? Speed of sound = 340ms

^{–1}?

__Answer:-__

The frequency of nth mode of vibration of a pipe closed at one endis given by

ν_{n}=(2n 1)ν/4L

ν=340ms^{-1}; L=20cm=0.2m;ν_{n}=430Hz.

Therefore 430= ((2n-1) x 340)/ (4×0.2)

=>n=1

Therefore, first mode of vibration of the pipe is excited, for open pipe sincen must be an integer, the same source cannot be in resonance with thepipe with both ends open.

**Nodes and Antinodes: system closed at one end**

- For a system which is closed at one end,only one node is formed at the closed end.
- Consider there is one fixed end(x=0) and one open end(x=L), and a string is attached between these two ends.
- At x=L, antinodes will be formed.This means amplitude will be maximum at this end.
- Condition for formation of antinodes is x= (n+ (1/2)) (λ/2).=>L=(n+ (1/2)) (λ/2)
- =>
**λ=(2L)/(n+(1/2))**.This expression shows that the values of wavelength is restricted. n=0, 1, 2, 3… - Corresponding frequencies will be ν= v/(2L)(n+1/2) (By using ν=v/λ);n=0,1,2,3, …

__Modes of oscillations:-__

- Fundamental frequency:- Also known as First Harmonic:-It corresponds to lowest possible value for n. That is n=0.
- The expression for fundamental frequency is ν
_{0}= v/(2L)x1/2 = v/(4L) **Odd Harmonics**- n=1 ; ν =v/(2L)(1+(1/2)) = (3v)/(4L) =3ν
_{0} - n=2; ν =v/(2L)(2+(1/2))= (5v)/(4L)=5ν
_{0} - n=3;ν =v/(2L)(3+(1/2))=(7v)/(4L)=7ν
_{0}

- n=1 ; ν =v/(2L)(1+(1/2)) = (3v)/(4L) =3ν
- For a system which is closed at one end and open at another end will get one fundamental frequency and all other odd harmonics.

Therefore nx550=1100 => n=2. It is the second harmonic.

- Fundamental Frequency: – ν
_{1}= (V)/ (4L)

Odd harmonics :- (3V)/ (4L), (5V)/ (4L), (7V)/ (4L) ….

=>ν_{1}=ν_{Y} =275Hz. Where ν_{Y} = Fundamental Frequency

=>nν_{1} = 1100 => n=1100/275=4

This should correspond to 4^{th} harmonic but 4^{th} harmonic cannot be present as only odd harmonics are present.

Therefore the same source cannot be resonated if the pipe is closed at one end.