# A wave of frequency 500Hz has a velocity 360m/s. The distance between two nearest points which are 60 degree out of phase is ....?please explain

Shivam Dimri
43 Points
11 years ago

firts finding the wavelength

k = speed/frequency

=> k =36/50

now phase difference = (2∏/k )* path difference

60 = (360/36) * 50 * path difference

distance bewtween two nearest points is = path difference = 6/50 metre

got it!!

Ryhhg
23 Points
6 years ago
firts finding the wavelengthk = speed/frequency=> k =36/50now phase difference = (2∏/k )* path difference60 = (360/36) * 50 * path differencedistance bewtween two nearest points is = path difference = 6/50 metre
Praneeth
15 Points
5 years ago

K=v/f
K=360/500=36/50
Phase difference=2pie/k ×dx
60=360×50×dx/36     dx=12cm
60=360/36 ×50×dxnger
dx=12cm
askIITians Faculty 628 Points
4 years ago
Dear student,
Please find the solution to your problem.

k = speed/frequency
=> k = 36/50
now phase difference = (2π/k ) x path difference
π/3 = (2π/36 x 50) x path difference
Hence, the distance bewtween two nearest points = path difference = 0.12 m = 12 cm

Thanks and regards,
Kushagra