The diagram above illustrates a variation of the displacement y of a wave particle with time t. lf the velocity of wave is 250ms-1, calculate the distance between two successive particles which are in phase

Bukason.

10 years ago

V=lamda/time

x=lamda

:.v=x/t

x=vt

v=250m/s

t=0.02sec(it took the particle 0.02sec to complete a cycle in the diagram above

.

x=0.02x250

x(lamda)=5.0m

:. My answer is

{5.0m} nt 2.5m

BroTolu

4 years ago

250 × 0.02

= 5.0m

Akrano

3 years ago

in phase was used in the question that means from node to node or from anti-node to anti-node looking at the diagram drawn 0.01 is to be used not a whole wavelenght's measurement i.e(0.02) so using this your answer would be 2.5m

Tega204

2 years ago

V=F✓
✓=V/F
F= number of oscillations/ time taken
=2/0.04=50
✓=250/50=5m

(0.04 was ommited from the diagram)

friday.lucky

1 year ago

the solving is not correct

lycoperdon

11 years ago

v=2x/t

tyjesma

11 years ago

V=F*Λ...and F=1/T.. : ◦ V=Λ/T...v=wave speed,

Λ=Wave Length(landar) and T=time...

V=250, T=0.01, Λ=?...Λ=2.5