Basic Maths and Units and DimensionsEasy

Question

If the speed of the wave shown in the figure is 330 m/s in the given medium, then the equation of the wave propagating in the positive x-direction and velocity of particle P will be (all quantities are in M.K.S. units) :-
       

Options

A.y = 0.05 sin 2π (4000 t - 12.5 x),
330 √3 m/s upward
B.y = 0.05 sin 2π (4000 t - 122.5 x),
330 √3 m/s upward
C.y = 0.05 sin 2π (3300 t - 10 x),
330 √3 m/s upward
D.y = 0.05 sin 2π (3300 t - 10 x),
330 √3 m/s downward

Solution

2.5λ = 0.25 ⇒ λ = 0.1 m , a = 0.05
⇒ K = = 20π rad/m
v = 330 m/s then
W = Kv = 20π × 330 = 6600π rad/sec
Propagating in (+)x-axis then
y = 0.05sin(6600π - 20πx)
= 0.05sin2π(3300 - 10x)
Now
vp = -V(slope)
= -330 × (-√3)
vp = + 330 √3 m/s
= 330 √3 (upward)

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