Question

233.The length of a sonometer wire is 90cm and the stationary waves setup in the wire is represented by an equ y=6sin(pie x/30)cos(250 pie t)where x,y are in cm and t is in second.The distances of successive antinodes from one end of the wire are

Answer 1

L = 90cm

Stationary wave:  y = 6 Sin(π x/30)  Cos (250 π t)  cm
at x = 0,  | y | = 0.  This is one node and one end of the sonometer wire.
at x = 15 cm,  | y | = 6 as sin (πx/30)  = 1

antinodes are where  πx/30 = (2n+1) π/2         n=0,1,2,3,..
          ie., 15 cm, 45 cm, 75 cm

Distance between antinodes = 30 cm

Answer 2

Answer: 30

Explanation: From the given equation we can see the value of 'k'

We already know k=2π/wavelength

Therefore π/30=2π/wavelength

From this wavelength =60

(From here we can directly say distance is 30)

But let's do it in a methodical manner

We know wavelength=2L/n where L is the length of the string and n is the harmonic number. Value of l is given as 90 so by substituting the values we get n as 3. This means there are. 3 loops formed in the standing wave. So by logic, if the length is 90 and the no of loops is 3distance Frome one antinode to the other is 90/3= 30