- Messages
- 129
- Reaction score
- 203
- Points
- 53
Here you go Credits papajohn
Distance S2M =128 (from Pythagoras thereom)
Path difference S2M - S1M
128-100 = 28 (If you do not know this concept then take help from Pacific Physics Vol 2 )
Wavelenght at 1 Khz of sound wave is λ=v/f 330/1^103= 33cm
λ of 4kHz 330/4^103= 8.25 cm
Path difference for minima = odd number of the half the λ
Or simply (2n+1) * λ/2
Put n = 0 in the formula (2*0 +1) λ/2 = 56
Put N= 1 =2(1)+1) * λ/2 =18.7
Put N = 2 2(2) =1) * λ/2 = 11.2
Put N= 4 2(3) +1* λ/2 = 8
Sice λ 56 cm and 8 cm are not within the range of 8.25 to 33 cm, so minima is obtained for λ 11.2 and 18.7cm Therefore two minma are detected.
Why do we need to calculate path difference i didn't see any use.
And did you calculate those, i mean when n = 0 then it will be λ/2 which is 4.125, λ=8.25
I don't know how did you get 56
Last edited: