The fundamental frequency of vibration (n) of a string (or wire) is given by

n = (1/2L)√(T/m) where L is the length of the wire, T is the tension and ‘m’ is the linear density (mass per unit length) of the string.

You may get questions based on this relation. See the following MCQ:

(n + 4) α 1/60 and

(n – 4) α 1/62 where ‘n’ is the frequency of the tuning fork.

[Note that the frequency of the wire is greater than that of the fork when its length is smaller].

Dividing, (n + 4)/ (n – 4) = 62/60 from which n = 244Hz.

Now, consider the following questionwhich appeared in EAMCET (Med) 2003 question paper:

n = (1/2L)√(T/m) = (1/2L)√(T/πr

n

= (35/36) √[(8/1)( 1 /16) (2/1)]

= (35/36)×1

This means that n

Therefore, beat frequency = 360 – 350 = 10 Hz.

n = (1/2L)√(T/m) where L is the length of the wire, T is the tension and ‘m’ is the linear density (mass per unit length) of the string.

You may get questions based on this relation. See the following MCQ:

**A sonometer wire and a tuning fork are excited together. Four beats are heard when the length of the wire is 60 cm as well as 62 cm. The frequency of the tuning fork is ( in Hz)**

(a) 512 (b) 488 (c) 384 (d) 256 (e) 244

Since the frequency of vibration of the wire is inversely proportional to its length, we can write, in the two cases,(a) 512 (b) 488 (c) 384 (d) 256 (e) 244

(n + 4) α 1/60 and

(n – 4) α 1/62 where ‘n’ is the frequency of the tuning fork.

[Note that the frequency of the wire is greater than that of the fork when its length is smaller].

Dividing, (n + 4)/ (n – 4) = 62/60 from which n = 244Hz.

Now, consider the following questionwhich appeared in EAMCET (Med) 2003 question paper:

**Two uniform wires are vibrating simultaneously in their fundamental modes. The tensions, lengths, diameters and the densities of the two wires are in thr ratio 8:1, 36:35, 4:1 and 1:2 respectively. If the note of the higher pitch has a frequency 360 Hz, the number of beats produced per second is**

(a) 5 (b) 10 (c) 15 (d) 20

Since the frequency of vibration of a stretched wire is given by(a) 5 (b) 10 (c) 15 (d) 20

n = (1/2L)√(T/m) = (1/2L)√(T/πr

^{2}ρ) where L is the length of the wire, T is the tension and ‘m’ is the linear density (mass per unit length) which is πr^{2}ρ where ‘r’ is the radius and ρ is the density of the material of the wire, we have,n

_{1}/n_{2}= (L_{2}/L_{1}) √[(T_{1}/T_{2})( r_{2}/r_{1})^{2}(ρ_{2}/ρ_{1})]= (35/36) √[(8/1)( 1 /16) (2/1)]

= (35/36)×1

This means that n

_{2}is the higher frequency. Since the higher frquency is given as 360 Hz, we have n_{1}/360 = 35/36 from which n_{1}= 350 Hz.Therefore, beat frequency = 360 – 350 = 10 Hz.

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