First slide

Speed of wave

Question

A transverse wave is described by the equation $Y = Y_{0} \sin 2 π (ft - x / λ)$ The maximum particle velocity is equal to four times the wave velocity if

Moderate

A

$λ = {πY}_{0} / 4$
B

$λ = {πY}_{0} / 2$
C

${λ=πY}_{0}$
D

$λ = 2 {πY}_{0}$

Solution

$v = \frac{d Y}{d t} = Y_{0} c o s [2 π (f t - \frac{x}{λ})] \times 2 π f o r V = 2 π f Y_{0} c o s [2 π (f t - \frac{x}{λ}) T h e p a r t i c l e v e l o c i t y i s m a x i m u m, w h e n c o s [2 π (f t - \frac{x}{λ})] = 1 V_{max} = 2 π f Y_{0} W e k n o w t h a t Y = q s i n (r o t - k x T h e w a v e v e l o c i t y V i s g i v e n b Y V = \frac{ω}{k} = \frac{2 π f}{2 π / λ} = f λ Given that v_{max} = 4 V ∴ 2 π f Y_{0} = 4 f λ or λ = \frac{π Y_{0}}{2}$

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