A stretched sonometer wire is in unison with a tuning fork. When the length of the wire is increased by 5%, the number of beats heard per second is 10. Find the frequency of the tuning fork.

Given: Length initial length of wire be, L 1 = 100 L and final length after increment, L 2 = 100 L + 5 L = 105 L Beats produced, f = 10 , let n 1 be the initial frequency and n 2 be the frequency of sonometer wire after increment. Therefore, beat produced, 10 = n 1 - n 2 .....(i) Now, using the law of length, frequency is inversely proportional to length, n 1 = k 1 L 1 (where k is constant). and n 2 = k 1 L 2 , dividing (i) and (ii) we get, n 1 n 2 = k L 1 k L 2 = L 2 L 1 = 105 100 .....(ii) Now solve (i) and (ii), put n 2 = 100 105 n 1 in (i) we get, 10 = n 1 - 100 105 n 1 n 1 = 210 Hz Therefore, frequency of tuning fork will be, n 1 = 210 H z

Explain the formation of stationary waves by the analytical method. Show that nodes and antinodes are equally spaced in stationary waves.

Consider two simple harmonic progressive waves having same amplitude ( a ), frequency ( n ) and period ( T ) travelling along same medium in opposite direction. The wave along positive direction of X -axis is given by : Y 1 = a s i n 2 π ( n t - x λ ) .....(i) Y 2 = a s i n 2 π ( n t + x λ ) .....(ii) Using the principle of super position the resultant displacement will be: Y = Y 1 + Y 2 Y = a s i n 2 π ( n t - x λ ) + a s i n 2 π ( n t + x λ ) ∴ Y = 2 a s i n 2 π 2 ( n t - x λ + n t + x λ ) . c o s 2 π 2 ( n t - x λ - n t - x λ ) Y = 2 a s i n 2 πnt . c os 2 π ( - x λ ) Y = [ 2 a cos 2 πx λ ] . sin 2 πnt Y = A . sin ( 2 πnt ) The points of medium which vibrates with minimum or zero amplitude are called nodes. The points of medium which vibrates with maximum amplitude are called antinodes. For nodes put A = 0 2 a cos 2 πx λ = 0 cos 2 πx λ = 0 2 πx λ = π 2 , 3 π 2 , 5 π 2 ∴ x = λ 4 , 3 λ 4 , 5 λ 4 For antinodes: A = ± 2 a 2 a cos 2 πx λ = ± 2 a cos 2 πx λ = ± 1 2 πx λ = 0 , π , 2 π ∴ x = 0 , λ 2 , λ Thus the distance between two successive nodes and antinodes and nodes equal to λ 2 . Therefore, antinodes and nodes are equally spaced.

Show that even as well as odd harmonics are present as overtones in the case of an air column vibrating in a pipe open at both the ends.

In fundamental mode, the vibration of air column open at both ends, there are two anti-nodes and one node as shown in the figure. Let λ be the wavelength of the wave, v be the velocity and L be the length. Now, as we can see L = λ 2 or λ = 2 L and n = v λ = v 2 L , this is the frequency of fundamental mode in a pipe open at both the ends. Now for 1 s t harmonic let λ 1 be the wavelength, from the figure, L = λ 1 and n 1 = v λ 1 = v L ∴ n 1 = 2 n Now for 2 n d harmonic let λ 2 be the wavelength, from the figure, L = 3 λ 2 2 and n 2 = v λ 2 = 3 v 2 L ∴ n 2 = 3 n Therefore, n , 2 n , 3 n ..... are the harmonics which are both even as well as odd.